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## 15.4: Triple Integrals

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• Page ID 2612

• Gilbert Strang & Edwin “Jed” Herman

## Learning Objectives

• Recognize when a function of three variables is integrable over a rectangular box.
• Evaluate a triple integral by expressing it as an iterated integral.
• Recognize when a function of three variables is integrable over a closed and bounded region.
• Simplify a calculation by changing the order of integration of a triple integral.
• Calculate the average value of a function of three variables.

Previously, we discussed the double integral of a function $$f(x,y)$$ of two variables over a rectangular region in the plane. In this section we define the triple integral of a function $$f(x,y,z)$$ of three variables over a rectangular solid box in space, $$\mathbb{R}^3$$. Later in this section we extend the definition to more general regions in $$\mathbb{R}^3$$.

## Integrable Functions of Three Variables

We can define a rectangular box $$B$$ in $$\mathbb{R}^3$$ as

$B = \big\{(x,y,z)\,|\,a \leq x \leq b, \, c \leq y \leq d, \, e \leq z \leq f \big\}. \nonumber$

We follow a similar procedure to what we did in previously. We divide the interval $$[a,b]$$ into $$l$$ subintervals $$[x_{i-1},x_i]$$ of equal length $$\Delta x$$ with

$\Delta x = \dfrac{x_i - x_{i-1}}{l}, \nonumber$

divide the interval $$[c,d]$$ into $$m$$ subintervals $$[y_{i-1}, y_i]$$ of equal length $$\Delta y$$ with

$\Delta y = \dfrac{y_j - y_{j-1}}{m}, \nonumber$

and divide the interval $$[e,f]$$ into $$n$$ subintervals $$[z_{i-1},z_i]$$ of equal length $$\Delta z$$ with

$\Delta z = \dfrac{z_k - z_{k-1}}{n} \nonumber$

Then the rectangular box $$B$$ is subdivided into $$lmn$$ subboxes:

$B_{ijk} = [x_{i-1}, x_i] \times [y_{i-1}, y_i] \times [z_{i-1},z_i], \nonumber$

as shown in Figure $$\PageIndex{1}$$.

For each $$i, \, j,$$ and $$k$$, consider a sample point $$(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$$ in each sub-box $$B_{ijk}$$. We see that its volume is $$\Delta V = \Delta x \Delta y \Delta z$$. Form the triple Riemann sum

$\sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f ( x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)\,\Delta x \Delta y \Delta z. \nonumber$

We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.

## Definition: The triple integral

The triple integral of a function $$f(x,y,z)$$ over a rectangular box $$B$$ is defined as

$\lim_{l,m,n\rightarrow\infty} \sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f ( x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)\,\Delta x \Delta y \Delta z = \iiint_B f(x,y,z) \,dV \nonumber$ if this limit exists.

When the triple integral exists on $$B$$ the function $$f(x,y,z)$$ is said to be integrable on $$B$$. Also, the triple integral exists if $$f(x,y,z)$$ is continuous on $$B$$. Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, $$f$$ is bounded on $$B$$ and continuous except possibly on the boundary of $$B$$. The sample point $$(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$$ can be any point in the rectangular sub-box $$B_{ijk}$$ and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.

Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s theorem for triple integrals exists.

## Fubini’s Theorem for Triple Integrals

If $$f(x,y,z)$$ is continuous on a rectangular box $$B = [a,b] \times [c,d] \times [e,f]$$, then

$\iint_B f(x,y,z) \,dV = \int_e^f \int_c^d \int_a^b f(x,y,z) \,dx \, dy \, dz. \nonumber$

This integral is also equal to any of the other five possible orderings for the iterated triple integral.

For $$a, b, c, d, e$$ and $$f$$ real numbers, the iterated triple integral can be expressed in six different orderings:

\begin{align} \int_e^f \int_c^d \int_a^b f(x,y,z)\, dx \, dy \, dz = \int_e^f \left( \int_c^d \left( \int_a^b f(x,y,z) \,dx \right) dy \right) dz \\ = \int_c^d \left( \int_e^f \left( \int_a^b f(x,y,z) \,dx \right)dz \right) dy \\ = \int_a^b \left( \int_e^f \left( \int_c^d f(x,y,z) \,dy \right)dz \right) dx \\ = \int_e^f \left( \int_a^b \left( \int_c^d f(x,y,z) \,dy \right) dx \right) dz \\ = \int_c^d \left( \int_a^b \left( \int_c^d f(x,y,z) \,dz\right)dx \right) dy \\ = \int_a^b \left( \int_c^d \left( \int_e^f f(x,y,z) \,dz \right) dy \right) dx \end{align} \nonumber

For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).

## Example $$\PageIndex{1}$$: Evaluating a Triple Integral

Evaluate the triple integral $\int_{z=0}^{z=1} \int_{y=2}^{y=4} \int_{x=-1}^{x=5} (x + yz^2)\, dx \, dy \, dz. \nonumber$

The order of integration is specified in the problem, so integrate with respect to $$x$$ first, then y , and then $$z$$.

\begin{align*}&\int_{z=0}^{z=1} \int_{y=2}^{y=4} \int_{x=-1}^{x=5} (x + yz^2) \,dx \,dy \,dz \\ &= \int_{z=0}^{z=1} \int_{y=2}^{y=4} \left. \left[ \dfrac{x^2}{2} + xyz^2\right|_{x=-1}^{x=5}\right]\,dy \,dz &&\text{Integrate with respect to x.}\\ &= \int_{z=0}^{z=1} \int_{y=2}^{y=4} \left[12+6yz^2\right] \,dy \,dz &&\text{Evaluate.}\\ &= \int_{z=0}^{z=1} \left[ \left.12y+6\dfrac{y^2}{2}z^2 \right|_{y=2}^{y=4} \right] dz &&\text{Integrate with respect to y.} \\ &= \int_{z=0}^{z=1} [24+36z^2] \, dz &&\text{Evaluate.} \\ &= \left[ 24z+36\dfrac{z^3}{3} \right]_{z=0}^{z=1} &&\text{Integrate with respect to z.}\\ &=36. &&\text{Evaluate.}\end{align*}

## Example $$\PageIndex{2}$$: Evaluating a Triple Integral

Evaluate the triple integral

$\iiint_B x^2 yz \,dV \nonumber$

where $$B = \big\{(x,y,z)\,|\, - 2 \leq x \leq 1, \, 0 \leq y \leq 3, \, 1 \leq z \leq 5 \big\}$$ as shown in Figure $$\PageIndex{2}$$.

The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate $$y$$ first, then $$x$$, and then $$z$$.

\begin{align*}\iiint\limits_{B} x^2 yz \,dV &= \int_1^5 \int_{-2}^1 \int_0^3 [x^2 yz] \,dy \, dx \, dz \\&= \int_1^5 \int_{-2}^1 \left[ \left. x^2 \dfrac{y^3}{3} z\right|_0^3 \right] dx \, dz \\&= \int_1^5 \int_{-2}^1 \dfrac{y}{2} x^2 z \,dx \, dz \\&= \int_1^5 \left[ \left. \dfrac{9}{2} \dfrac{x^3}{3} z \right|_{-2}^1 \right] dz \\&= \int_1^5 \dfrac{27}{2} z \, dz \\&= \left. \dfrac{27}{2} \dfrac{z^2}{2} \right|_1^5 = 162.\end{align*}

Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to $$x$$ first, then $$z$$, then $$y$$

\begin{align*}\iiint\limits_{B} x^2yz \,dV &= \int_0^3 \int_1^5 \int_{-2}^1 [x^2yz] \,dx\, dz\, dy \\&= \int_0^3 \int_1^5 \left[ \left. \dfrac{x^3}{3} yz \right|_{-2}^1 \right] dz \,dy \\&= \int_0^3 \int_1^5 3yz \; dz \,dy \\&= \int_0^3 \left.\left[ 3y\dfrac{z^2}{2} \right|_1^5 \right] \,dy \\&= \int_0^3 36y \; dy \\&= \left. 36\dfrac{y^2}{2} \right|_0^3 =18(9-0) =162.\end{align*}

## Exercise $$\PageIndex{1}$$

$\iiint_B z \, \sin \, x \, \cos \, y \, dV\nonumber$

where $$B = \big\{(x,y,z)\,|\,0 \leq x \leq \pi, \, \dfrac{3\pi}{2} \leq y \leq 2\pi, \, 1 \leq z \leq 3 \big\}$$.

Follow the steps in the previous example.

$\iiint_B z \, \sin \, x \, \cos \, y \, dV = 8 \nonumber$

## Triple Integrals over a General Bounded Region

We now expand the definition of the triple integral to compute a triple integral over a more general bounded region $$E$$ in $$\mathbb{R}^3$$. The general bounded regions we will consider are of three types. First, let $$D$$ be the bounded region that is a projection of $$E$$ onto the $$xy$$-plane. Suppose the region $$E$$ in $$\mathbb{R}^3$$ has the form

$E = \big\{(x,y,z)\,|\,(x,y) \in D, u_1(x,y) \leq z \leq u_2(x,y) \big\}. \nonumber$

For two functions $$z = u_1(x,y)$$ and $$u_2(x,y)$$, such that $$u_1(x,y) \leq u_2(x,y)$$ for all $$(x,y)$$ in $$D$$ as shown in the following figure.

## Triple Integral over a General Region

The triple integral of a continuous function $$f(x,y,z)$$ over a general three-dimensional region

$E = \big\{(x,y,z)\,|\,(x,y) \in D, \, u_1(x,y) \leq z \leq u_2(x,y) \big\} \nonumber$

in $$\mathbb{R}^3$$, where $$D$$ is the projection of $$E$$ onto the $$xy$$-plane, is

$\iiint_E f(x,y,z) \,dV = \iint_D \left[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) \,dz \right] \, dA. \nonumber$

Similarly, we can consider a general bounded region $$D$$ in the $$xy$$-plane and two functions $$y = u_1(x,z)$$ and $$y = u_2(x,z)$$ such that $$u_1(x,z) \leq u_2(x,z)$$ for all $$(x,z)$$ in $$D$$. Then we can describe the solid region $$E$$ in $$\mathbb{R}^3$$ as

$E = \big\{(x,y,z)\,|\,(x,z) \in D, \, u_1(x,z) \leq z \leq u_2(x,z) \big\} \nonumber$ where $$D$$ is the projection of $$E$$ onto the $$xy$$-plane and the triple integral is

$\iiint_E f(x,y,z)\,dV = \iint_D \left[\int_{u_1(x,z)}^{u_2(x,z)} f(x,y,z) \,dy \right] \, dA. \nonumber$

Finally, if $$D$$ is a general bounded region in the $$xy$$-plane and we have two functions $$x = u_1(y,z)$$ and $$x = u_2(y,z)$$ such that $$u_1(y,z) \leq u_2(y,z)$$ for all $$(y,z)$$ in $$D$$, then the solid region $$E$$ in $$\mathbb{R}^3$$ can be described as

$E = \big\{(x,y,z)\,|\,(y,z) \in D, \, u_1(y,z) \leq z \leq u_2(y,z) \big\} \nonumber$ where $$D$$ is the projection of $$E$$ onto the $$xy$$-plane and the triple integral is

$\iiint_E f(x,y,z)\,dV = \iint_D \left[\int_{u_1(y,z)}^{u_2(y,z)} f(x,y,z) \, dx \right] \, dA. \nonumber$

Note that the region $$D$$ in any of the planes may be of Type I or Type II as described in previously. If $$D$$ in the $$xy$$-plane is of Type I (Figure $$\PageIndex{4}$$), then

$E = \big\{(x,y,z)\,|\,a \leq x \leq b, \, g_1(x) \leq y \leq g_2(x), \, u_1(x,y) \leq z \leq u_2(x,y) \big\}. \nonumber$

Then the triple integral becomes

$\iiint_E f(x,y,z) \,dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) \,dz \, dy \, dx. \nonumber$

If $$D$$ in the $$xy$$-plane is of Type II (Figure $$\PageIndex{5}$$), then

$E = \big\{(x,y,z)\,|\,c \leq x \leq d, h_1(x) \leq y \leq h_2(x), \, u_1(x,y) \leq z \leq u_2(x,y) \big\}. \nonumber$

$\iiint_E f(x,y,z) \,dV = \int_{y=c}^{y=d} \int_{x=h_1(y)}^{x=h_2(y)} \int_{z=u_1(x,y)}^{z=u_2(x,y)} f(x,y,z)\,dz \, dx \, dy. \nonumber$

## Example $$\PageIndex{3A}$$: Evaluating a Triple Integral over a General Bounded Region

Evaluate the triple integral of the function $$f(x,y,z) = 5x - 3y$$ over the solid tetrahedron bounded by the planes $$x = 0, \, y = 0, \, z = 0$$, and $$x + y + z = 1$$.

Figure $$\PageIndex{6}$$ shows the solid tetrahedron $$E$$ and its projection $$D$$ on the $$xy$$-plane.

We can describe the solid region tetrahedron as

$E = \big\{(x,y,z)\,|\,0 \leq x \leq 1, \, 0 \leq y \leq 1 - x, \, 0 \leq z \leq 1 - x - y \big\}. \nonumber$

Hence, the triple integral is

$\iiint_E f(x,y,z) \,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y} (5x - 3y) \,dz \, dy \, dx. \nonumber$

To simplify the calculation, first evaluate the integral $$\displaystyle \int_{z=0}^{z=1-x-y} (5x - 3y) \,dz$$. We have

$\int_{z=0}^{z=1-x-y} (5x - 3y) \,dz = (5x - 3y)z \bigg|_{z=0}^{z=1-x-y} = (5x - 3y)(1 - x - y).\nonumber$

Now evaluate the integral

$\int_{y=0}^{y=1-x} (5x - 3y)(1 - x - y) \,dy, \nonumber$

$\int_{y=0}^{y=1-x} (5x - 3y)(1 - x - y)\,dy = \dfrac{1}{2}(x - 1)^2 (6x - 1).\nonumber$

Finally evaluate

$\int_{x=0}^{x=1} \dfrac{1}{2}(x - 1)^2 (6x - 1)\,dx = \dfrac{1}{12}.\nonumber$

Putting it all together, we have

$\iiint_E f(x,y,z)\,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y}(5x - 3y)\,dz \, dy \, dx = \dfrac{1}{12}.\nonumber$

Just as we used the double integral $\iint_D 1 \,dA \nonumber$ to find the area of a general bounded region $$D$$ we can use $\iiint_E 1\,dV \nonumber$ to find the volume of a general solid bounded region $$E$$. The next example illustrates the method.

## Example $$\PageIndex{3B}$$: Finding a Volume by Evaluating a Triple Integral

Find the volume of a right pyramid that has the square base in the $$xy$$-plane $$[-1,1] \times [-1,1]$$ and vertex at the point $$(0, 0, 1)$$ as shown in the following figure.

In this pyramid the value of $$z$$ changes from 0 to 1 and at each height $$z$$ the cross section of the pyramid for any value of $$z$$ is the square

$[-1 + z, \, 1 - z] \times [-1 + z, \, 1 - z].\nonumber$

Hence, the volume of the pyramid is $\iiint_E 1\,dV\nonumber$ where

$E = \big\{(x,y,z)\,|\,0 \leq z \leq 1, \, -1 + z \leq y \leq 1 - z, \, -1 + z \leq x \leq 1 - z \big\}.\nonumber$

Thus, we have

\begin{align*} \iiint_E 1\,dV &= \int_{z=0}^{z=1} \int_{y=-1+z}^{y=1-z} \int_{x=-1+z}^{x=1-z} 1\,dx \, dy \, dz \\[5pt] &= \int_{z=0}^{z=1} \int_{y=-1+z}^{y=1-z} (2 - 2z)\, dy \, dz \\[5pt] &= \int_{z=0}^{z=1}(2 - 2z)^2 \,dz = \dfrac{4}{3}. \end{align*}

Hence, the volume of the pyramid is $$\dfrac{4}{3}$$ cubic units.

## Exercise $$\PageIndex{3}$$

Consider the solid sphere $$E = \big\{(x,y,z)\,|\,x^2 + y^2 + z^2 = 9 \big\}$$. Write the triple integral $\iiint_E f(x,y,z) \,dV\nonumber$ for an arbitrary function $$f$$ as an iterated integral. Then evaluate this triple integral with $$f(x,y,z) = 1$$. Notice that this gives the volume of a sphere using a triple integral.

Follow the steps in the previous example. Use symmetry.

\begin{align*} \iiint_E 1\,dV = 8 \int_{x=-3}^{x=3} \int_{y=-\sqrt{9-z^2}}^{y=\sqrt{9-z^2}}\int_{z=-\sqrt{9-x^2-y^2}}^{z=\sqrt{9-x^2-y^2}} 1\,dz \, dy \, dx \\ = 36 \pi \,\text{cubic units}. \end{align*}

## Changing the Order of Integration

As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.

## Example $$\PageIndex{4}$$: Changing the Order of Integration

Consider the iterated integral

$\int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \int_{z=0}^{z=y} f(x,y,z)\,dz \, dy \, dx. \nonumber$

The order of integration here is first with respect to z , then y , and then x . Express this integral by changing the order of integration to be first with respect to $$x$$, then $$z$$, and then $$y$$. Verify that the value of the integral is the same if we let $$f (x,y,z) =xyz$$.

The best way to do this is to sketch the region $$E$$ and its projections onto each of the three coordinate planes. Thus, let

$E = \big\{(x,y,z)\,|\,0 \leq x \leq 1, \, 0 \leq y \leq x^2, \, 0 \leq z \leq y \big\}.\nonumber$

$\int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \int_{z=0}^{z=x^2} f(x,y,z) \,dz \, dy \, dx = \iiint_E f(x,y,z)\,dV.\nonumber$

We need to express this triple integral as

$\int_{y=c}^{y=d} \int_{z=v_1(y)}^{z=v_2(y)} \int_{x=u_1(y,z)}^{x=u_2(y,z)} f(x,y,z)\,dx \, dz \, dy.\nonumber$

Knowing the region $$E$$ we can draw the following projections (Figure $$\PageIndex{8}$$):

on the $$xy$$-plane is $$D_1 = \big\{(x,y)\,|\, 0 \leq x \leq 1, \, 0 \leq y \leq x^2 \big\} = \{ (x,y) \,|\, 0 \leq y \leq 1, \, \sqrt{y} \leq x \leq 1 \big\},$$

on the $$yz$$-plane is $$D_2 = \big\{(y,z) \,|\, 0 \leq y \leq 1, \, 0 \leq z \leq y^2 \big\}$$, and

on the $$xz$$-plane is $$D_3 = \big\{(x,z) \,|\, 0 \leq x \leq 1, \, 0 \leq z \leq x^2 \big\}$$.

Now we can describe the same region $$E$$ as $$\big\{(x,y,z) \,|\, 0 \leq y \leq 1, \, 0 \leq z \leq y^2, \, \sqrt{y} \leq x \leq 1 \big\}$$, and consequently, the triple integral becomes

$\int_{y=c}^{y=d} \int_{z=v_1(y)}^{z=v_2(y)} \int_{x=u_1(y,z)}^{x=u_2(y,z)} f(x,y,z)\,dx \, dz \, dy = \int_{y=0}^{y=1} \int_{z=0}^{z=x^2} \int_{x=\sqrt{y}}^{x=1} f(x,y,z)\,dx \, dz \, dy \nonumber$

Now assume that $$f (x,y,z) = xyz$$ in each of the integrals. Then we have

\begin{align*} \int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \int_{z=0}^{z=y^2} xyz \, dz \, dy \, dx &= \int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \left. \left[xy \dfrac{z^2}{2} \right|_{z=0}^{z=y^2} \right] \, dy \, dx \\[5pt] &= \int_{x=0}^{x=1} \int_{y=0}^{y=x^2} \left( x \dfrac{y^5}{2}\right) dy \, dx \\[5pt] &= \int_{x=0}^{x=1} \left. \left[ x\dfrac{y^6}{12} \right|_{y=0}^{y=x^2}\right] dx \\[5pt] &= \int_{x=0}^{x=1} \dfrac{x^{13}}{12} dx = \left. \dfrac{x^{14}}{168}\right|_{x=0}^{x=1} \\[5pt] &= \dfrac{1}{168}, \end{align*}

\begin{align*} \int_{y=0}^{y=1} \int_{z=0}^{z=y^2} \int_{x=\sqrt{y}}^{x=1} xyz \, dx \, dz \, dy &= \int_{y=0}^{y=1} \int_{z=0}^{z=y^2} \left.\left[yz \dfrac{x^2}{2} \right|_{\sqrt{y}}^{1} \right] dz \, dy \\[5pt] &= \int_{y=0}^{y=1} \int_{z=0}^{z=y^2} \left( \dfrac{yz}{2} - \dfrac{y^2z}{2} \right) dz \, dy \\[5pt] &= \int_{y=0}^{y=1} \left. \left[ \dfrac{yz^2}{4} - \dfrac{y^2z^2}{4} \right|_{z=0}^{z=y^2} \right] dy \\[5pt] &= \int_{y=0}^{y=1} \left(\dfrac{y^5}{4} - \dfrac{y^6}{4} \right) dy \\[5pt] &= \left. \left(\dfrac{y^6}{24} - \dfrac{y^7}{28} \right) \right|_{y=0}^{y=1} \\[5pt] &= \dfrac{1}{168}. \end{align*} \nonumber

## Exercise $$\PageIndex{4}$$

Write five different iterated integrals equal to the given integral

$\int_{z=0}^{z=4} \int_{y=0}^{y=4-z} \int_{x=0}^{x=\sqrt{y}} f(x,y,z) \, dx \, dy \, dz.\nonumber$

Follow the steps in the previous example, using the region $$E$$ as $$\big\{(x,y,z) \,|\, 0 \leq z \leq 4, \, 0 \leq y \leq 4 - z, \, 0 \leq x \leq \sqrt{y} \big\}$$, and describe and sketch the projections onto each of the three planes, five different times.

$(i) \, \int_{z=0}^{z=4} \int_{x=0}^{x=\sqrt{4-z}} \int_{y=x^2}^{y=4-z} f(x,y,z) \, dy \, dx \, dz, \, (ii) \, \int_{y=0}^{y=4} \int_{z=0}^{z=4-y} \int_{x=0}^{x=\sqrt{y}} f(x,y,z) \,dx \, dz \, dy, \,(iii) \, \int_{y=0}^{y=4} \int_{x=0}^{x=\sqrt{y}} \int_{z=0}^{Z=4-y} f(x,y,z) \,dz \, dx \, dy, \, \nonumber$

$(iv) \, \int_{x=0}^{x=2} \int_{y=x^2}^{y=4} \int_{z=0}^{z=4-y} f(x,y,z) \,dz \, dy \, dx, \, (v) \int_{x=0}^{x=2} \int_{z=0}^{z=4-x^2} \int_{y=x^2}^{y=4-z} f(x,y,z) \,dy \, dz \, dx \nonumber$

## Example $$\PageIndex{5}$$: Changing Integration Order and Coordinate Systems

$\iiint_{E} \sqrt{x^2 + z^2} \,dV, \nonumber$

where $$E$$ is the region bounded by the paraboloid $$y = x^2 + z^2$$ (Figure $$\PageIndex{9}$$) and the plane $$y = 4$$.

The projection of the solid region $$E$$ onto the $$xy$$-plane is the region bounded above by $$y = 4$$ and below by the parabola $$y = x^2$$ as shown.

$E = \big\{(x,y,z) \,|\, -2 \leq x \leq 2, \, x^2 \leq y \leq 4, \, -\sqrt{y - x^2} \leq z \leq \sqrt{y - x^2} \big\}.\nonumber$

The triple integral becomes

$\iiint_E \sqrt{x^2 + z^2} \,dV = \int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} \int_{z=-\sqrt{y-x^2}}^{z=\sqrt{y-x^2}} \sqrt{x^2 + z^2} \,dz \, dy \, dx.\nonumber$

This expression is difficult to compute, so consider the projection of $$E$$ onto the $$xz$$-plane. This is a circular disc $$x^2 + z^2 \leq 4$$. So we obtain

$\iiint_E \sqrt{x^2 + z^2} \,dV = \int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} \int_{z=-\sqrt{y-x^2}}^{z=\sqrt{y-x^2}} \sqrt{x^2 + z^2} \,dz \, dy \, dx = \int_{x=-2}^{x=2} \int_{z=-\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} \int_{y=x^2+z^2}^{y=4} \sqrt{x^2 + z^2} \,dy \, dz \, dx.\nonumber$

Here the order of integration changes from being first with respect to $$z$$ then $$y$$ and then $$x$$ to being first with respect to $$y$$ then to $$z$$ and then to $$x$$. It will soon be clear how this change can be beneficial for computation. We have

$\int_{x=-2}^{x=2} \int_{z=\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} \int_{y=x^2+z^2}^{y=4} \sqrt{x^2 + z^2} \,dy \, dz \, dx = \int_{x=-2}^{x=2} \int_{z=-\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} (4 - x^2 - z^2) \sqrt{x^2 + z^2} \,dz \, dx.\nonumber$

Now use the polar substitution $$x = r \, \cos \, \theta, \, z = r \, \sin \, \theta$$, and $$dz \, dx = r \, dr \, d\theta$$ in the $$xz$$-plane. This is essentially the same thing as when we used polar coordinates in the $$xy$$-plane, except we are replacing $$y$$ by $$z$$. Consequently the limits of integration change and we have, by using $$r^2 = x^2 + z^2$$,

$\int_{x=-2}^{x=2} \int_{z=-\sqrt{4-x^2}}^{z=\sqrt{4-x^2}} (4 - x^2 - z^2) \sqrt{x^2 + z^2}\,dz \, dx = \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=2} (4 - r^2) rr \, dr \, d\theta = \int_0^{2\pi} \left. \left[ \dfrac{4r^3}{3} - \dfrac{r^5}{5} \right|_0^2 \right] \, d\theta = \int_0^{2\pi} \dfrac{64}{15} \,d\theta = \dfrac{128\pi}{15}\nonumber$

## Average Value of a Function of Three Variables

Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.

If $$f(x,y,z)$$ is integrable over a solid bounded region $$E$$ with positive volume $$V \, (E),$$ then the average value of the function is

$f_{ave} = \dfrac{1}{V \, (E)} \iiint_E f(x,y,z) \, dV. \nonumber$

Note that the volume is

$V \, (E) = \iiint_E 1 \,dV. \nonumber$

## Example $$\PageIndex{6}$$: Finding an Average Temperature

The temperature at a point $$(x,y,z)$$ of a solid $$E$$ bounded by the coordinate planes and the plane $$x + y + z = 1$$ is $$T(x,y,z) = (xy + 8z + 20) \, \text{°}\text{C}$$. Find the average temperature over the solid.

Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane $$x + y + z = 1$$ has intercepts $$(1,0,0), \, (0,1,0),$$ and $$(0,0,1)$$. The region $$E$$ looks like

$E = \big\{(x,y,z) \,|\, 0 \leq x \leq 1, \, 0 \leq y \leq 1 - x, \, 0 \leq z \leq 1 - x - y \big\}.\nonumber$

Hence the triple integral of the temperature is

$\iiint_E f(x,y,z) \,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y} (xy + 8z + 20) \, dz \, dy \, dx = \dfrac{147}{40}. \nonumber$

The volume evaluation is

$V \, (E) = \iiint_E 1\,dV = \int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=0}^{z=1-x-y} 1 \,dz \, dy \, dx = \dfrac{1}{6}. \nonumber$

Hence the average value is

$T_{ave} = \dfrac{147/40}{1/6} = \dfrac{6(147)}{40} = \dfrac{441}{20} \, \text{°}\text{C} \nonumber$.

## Exercise $$\PageIndex{6}$$

Find the average value of the function $$f(x,y,z) = xyz$$ over the cube with sides of length 4 units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.

$$f_{ave} = 8$$

## Key Concepts

• To compute a triple integral we use Fubini’s theorem, which states that if $$f(x,y,z)$$ is continuous on a rectangular box $$B = [a,b] \times [c,d] \times [e,f]$$, then $\iiint_B f(x,y,z) \,dV = \int_e^f \int_c^d \int_a^b f(x,y,z) \, dx \, dy \, dz \nonumber$ and is also equal to any of the other five possible orderings for the iterated triple integral.
• To compute the volume of a general solid bounded region $$E$$ we use the triple integral $V \, (E) = \iiint_E 1 \,dV. \nonumber$
• Interchanging the order of the iterated integrals does not change the answer. As a matter of fact, interchanging the order of integration can help simplify the computation.
• To compute the average value of a function over a general three-dimensional region, we use $f_{ave} = \dfrac{1}{V \, (E)} \iiint_E f(x,y,z) \,dV. \nonumber$

## Key Equations

• Triple integral

$\lim_{l,m,n \rightarrow \infty} \sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \,\Delta x \Delta y \Delta z = \iiint_B f(x,y,z) \,dV \nonumber$

## Math Insight

Triple integral examples.

A cube has sides of length 4. Let one corner be at the origin and the adjacent corners be on the positive $x$, $y$, and $z$ axes.

If the cube's density is proportional to the distance from the xy-plane, find its mass.

Solution : The density of the cube is $f(x,y,z) = kz$ for some constant $k$.

If $\dlv$ is the cube, the mass is the triple integral \begin{align*} \iiint_\dlv kz\,dV &= \int_0^4 \int_0^4 \int_0^4 kz\,dx\,dy\,dz\\ &= \int_0^4 \int_0^4 \left(\left.kxz \right|_{x=0}^{x=4}\right) dy\,dz\\ &= \int_0^4 \int_0^4 4 k z \,dy\,dz\\ &= \int_0^4 \left(\left. 4kzy \right|_{y=0}^{y=4}\right) dz\\ &= \int_0^4 16 kz dz = \left.\left.8kz^2\right|_{z=0}^{z=4}\right. = 128k \end{align*}

If distance is in cm and $k=1$ gram per cubic cm per cm, then the mass of the cube is 128 grams.

Evaluate the integral \begin{align*} \int_0^1 \int_0^x \int_0^{1+x+y} f(x,y,z) dz \, dy\, dx \end{align*} where $f(x,y,z)=1$.

Solution : \begin{align*} &\int_0^1 \int_0^x \int_0^{1+x+y} dz \, dy\, dx\\ &\qquad= \int_0^1 \int_0^x \left(z\Big|_{z=0}^{z=1+x+y}\right)dy\, dx\\ &\qquad= \int_0^1 \int_0^x (1+x+y) dy\, dx\\ &\qquad= \int_0^1 \biggl[y + yx + \frac{y^2}{2}\biggr]_{y=0}^{y=x} dx\\ &\qquad= \int_0^1 \biggl(x + x^2 + \frac{x^2}{2}\biggr) dx\\ &\qquad= \int_0^1 \biggl(x + \frac{3x^2}{2} \biggr) dx\\ &\qquad= \biggl[\frac{x^2}{2} + \frac{x^3}{2} \biggr]_0^1 = \frac{1}{2} + \frac{1}{2} = 1 \end{align*}

Note: when we integrate $f(x,y,z)=1$, the integral $\iiint_\dlv dV$ is the volume of the solid $\dlv$.

Set up the integral of $f(x,y,z)$ over $\dlv$, the solid “ice cream cone” bounded by the cone $z=\sqrt{x^2+y^2}$ and the half-sphere $z = \sqrt{1-x^2-y^2}$, pictured below.

Ice cream cone region. The ice cream cone region is bounded above by the half-sphere $z=\sqrt{1-x^2-y^2}$ and bounded below by the cone $z=\sqrt{x^2+y^2}$.

Solution : We'll use the shadow method to set up the bounds on the integral. This means we'll write the triple integral as a double integral on the outside and a single integral on the inside of the form \begin{gather*} \iint_{\textit{shadow}} \int_{\textit{bottom}}^{\textit{top}} f(x,y,z). \end{gather*} We'll let the $z$-axis be the vertical axis so that the cone $z=\sqrt{x^2+y^2}$ is the bottom and the half-sphere $z = \sqrt{1-x^2-y^2}$ is the top of the ice cream cone $\dlv$. Hence, $\dlv$ is the region between these two surfaces: \begin{align} \sqrt{x^2+y^2} \le z \le \sqrt{1-x^2-y^2}. \label{zinequalities} \end{align} These inequalities give the range of $z$ as a function of $x$ and $y$ and thus form the bounds of the inner integral, which will be an integral with respect to $z$ of the form \begin{gather*} \int_{\textit{bottom}}^{\textit{top}} f(x,y,z)dz= \int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}} f(x,y,z)dz. \end{gather*}

The whole region $\dlv$ is the set of points satisfying the inequalities \eqref{zinequalities} while $x$ and $y$ range over the shadow of the ice cream cone that is parallel to the $xy$-plane, as illustrated by the cyan circle below.

Ice cream cone region with shadow. The ice cream cone region is bounded above by the half-sphere $z=\sqrt{1-x^2-y^2}$ and bounded below by the cone $z=\sqrt{x^2+y^2}$. The two surfaces intersect along a circle defined by $x^2+y^2=1/2$ and $z=1/\sqrt{2}$, which is the widest part of the ice cream cone. Therefore, the shadow of the ice cream cone region parallel to the $xy$-plane is the disk of radius $1/\sqrt{2}$ described by $x^2+y^2 \le 1/2$.

The shadow parallel to the $xy$-plane is the maximal range of $x$ and $y$ over all points inside $\dlv$. Inside the ice cream cone, the maximal range of $x$ and $y$ occurs where the two surfaces meet, i.e., where the “ice cream” (the half-sphere) meets the cone. From the figure, you can see that the surfaces meet in a circle, and the range of $x$ and $y$ is the disk that is the interior of that circle.

The surfaces meet when $\sqrt{x^2+y^2} = \sqrt{1-x^2-y^2}$, which means $x^2+y^2 = 1-x^2-y^2$ or \begin{align*} x^2+y^2 = \frac{1}{2}. \end{align*} In other words, for any point $(x,y,z)$ in the ice cream cone, the inequality \begin{align} x^2+y^2 \le \frac{1}{2} \label{shadow} \end{align} is satisfied. This inequality describes the shadow of the ice cream cone, which is the set of points $(x,y)$ that lie in a disk of radius $1/\sqrt{2}$, as illustrated below.

Now we've reduced the rest of the task of finding bounds for the triple integral to the much simpler task of finding bounds for a double integral over the shadow described by inequality \eqref{shadow}. We'll let $y$ be the inner integral of the double integral, meaning we need to describe the range of $y$ in the shadow as a function of $x$. To do this, we simply rewrite inequality \eqref{shadow} in terms of $y$ as \begin{align*} -\sqrt{1/2 - x^2} \le y \le \sqrt{1/2 - x^2}. \end{align*} This range of $y$ as a function of $x$ gives the bounds on the inner integral of the double integral.

Finally, for the bounds on the outer integral, we need the maximal range of $x$ alone. Given that $x^2+y^2 \le 1/2$, the maximal range occurs when $y=0$ so that $x^2 \le 1/2$. We can write the maximal range of $x$ is \begin{align*} -1/\sqrt{2} \le x \le 1/\sqrt{2}. \end{align*} The double integral with respect to $x$ and $y$ becomes \begin{gather*} \iint_{\textit{shadow}} \cdots dy\,dx = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \int_{-\sqrt{1/2-x^2}}^{\sqrt{1/2-x^2}} \cdots dy\,dx \end{gather*}

We have determined all the limits on the iterated integral. Putting the bottom/top limits together with the shadow limits, the ice cream cone can be described by the inequalities \begin{gather*} -1/\sqrt{2} \le x \le 1/\sqrt{2}\\ -\sqrt{1/2 - x^2} \le y \le \sqrt{1/2 - x^2}\\ \sqrt{x^2+y^2} \le z \le \sqrt{1-x^2-y^2} \end{gather*} and the integral of the function $f(x,y,z)$ over $\dlv$ is \begin{align} \iiint_\dlv f\, dV = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \int_{-\sqrt{1/2-x^2}}^{\sqrt{1/2-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}} f(x,y,z) dz\,dy\,dx. \label{icecreamintegral} \end{align}

Find the volume of the ice cream cone of Example 3a..

Solution : Simply set $f(x,y,z)=1$ in equation \eqref{icecreamintegral}.

The volume of the ice cream cone $\dlv$ given by the integral \begin{align*} \iiint_\dlv dV = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \int_{-\sqrt{1/2-x^2}}^{\sqrt{1/2-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}} dz\,dy\,dx. \end{align*} We won't attempt to evaluate this integral in rectangular coordinates . Once you've learned how to change variables in triple integrals , you can read how to compute the integral using spherical coordinates .

Find volume of the tetrahedron bounded by the coordinate planes and the plane through $(2,0,0)$, $(0,3,0)$, and $(0,0,1)$.

A tetrahedron. The tetrahedron is bounded by the coordinate planes ($x=0$, $y=0$, and $z=0$) and the plane through the three points (2,0,0), (0,3,0), and (0,0,1).

Solution : We know the equation for three of the surfaces of the tetrahedron, as they are the equations for the coordinates planes: $x=0$, $y=0$, and $z=0$. As an initial step, we can find the equation for the angled plane. You can follow the procedure in the second forming plane example to calculate that the plane is given by the equation \begin{align} 3x + 2y + 6z = 6. \label{plane_equation} \end{align}

To find the limits of the tetrahedron, we'll use the shadow method again, but this time, we'll think of the $y$-axis as being the vertical axis. You can imagine the sun that is casting the shadow as being at some point far on the positive $y$-axis.

With this orientation, the shadow of the tetrahedron is the maximal range of $x$ and $z$ over the tetrahedron. Since the tetrahedron gets wider in the $x$ and $z$ directions as $y$ decreases, the shadow of the tetrahedron is exactly the base of the tetrahedron in the $xz$-plane (the plane $y=0$), which is the triangle pictured below.

We approach the integral over this shadow as a double integral. In this shadow (and consequently in the tetrahedron itself), the total range of $z$ is \begin{align*} 0 \le z \le 1. \end{align*} To find the range of $x$ for each value of $z$, you can calculate from the figure of the shadow that the upper limit of $x$ is the line $z=1-x/2$ or $x=2(1-z)$. Given that the lower limit on $x$ is zero, the range of $x$ in the shadow for a given $z$ is \begin{align*} 0 \le x \le 2(1 - z). \end{align*} Alternatively, you could see that the upper limit on $x$ corresponds to the plane given by equation \eqref{plane_equation} when $y=0$. Plugging $y=0$ into equation \eqref{plane_equation} yields $3x + 6z=6$ or $x = 2(1-z)$.

For each value of $x$ and $z$ in the shadow, we need to integrate $y$ from the bottom to the top (viewing $y$ as the vertical axis). The bottom from this perspective is in the plane $y=0$, and the top is the angled plane of equation \eqref{plane_equation}, which we can solve for $y$ to write as $y=3(1-x/2 -z)$. Hence, for a given $z$ and $x$, the range of $y$ is \begin{align*} 0 \le y \le 3\left(1 - \frac{x}{2} - z\right). \end{align*}

To find the volume, we integrate the function 1 over this region: \begin{align*} &\int_0^1 \int_0^{2(1-z)} \int_0^{3(1- x/2 - z)} dy \, dx \, dz\\ &\qquad = \int_0^1 \int_0^{2(1-z)} 3\left(1 - \frac{x}{2} - z \right) dx \, dz\\ &\qquad = \int_0^1 3\left.\left[x - \frac{x^2}{4} -zx\right]_{x=0}^{x=2(1-z)}\right. dz\\ &\qquad = \int_0^1 3\left(2(1-z) - (1-z)^2 - 2z(1-z)\right) dz\\ &\qquad = \int_0^1 3(1 - 2z +z^2) dz\\ &\qquad = 3 \left.\left[ z - z^2 + \frac{z^3}{3} \right]_0^1\right.\\ &\qquad = 3\left(1 - 1 +\frac{1}{3}\right) = 3\left(\frac{1}{3}\right) = 1. \end{align*}

Change the order of $x$ and $y$ in the integral we derived above, \begin{align*} \int_0^1 \int_0^{2(1-z)} \int_0^{3(1- x/2 - z)} dy \, dx \, dz, \end{align*} so that the order will be $dx \, dy \, dz$.

Solution : One way to change the order of integration is to build up the graph of the tetrahedron from the limits of the integral, and then repeat the procedure of Example 4 but let the shadow be cast from the positive $x$-axis. Instead, we'll illustrate an alternative procedure of calculating the new limits directly from the inequalities of the old limits.

If $y$ will be middle integral, we need limits of $y$ in terms of $z$ (independent of $x$).

For given $z$, how large can $y$ range? From the above limits, we know \begin{align*} 0 \le y \le 3\left(1 - \frac{x}{2} - z\right). \end{align*} The range is largest when $x=0$, so \begin{align*} 0 \le y \le 3\left(1 - z\right) \end{align*}

Then, given $z$ and $y$, we need to know the range of the $x$. The following relationship must still be true: \begin{align*} y \le 3\left(1 - \frac{x}{2} - z\right). \end{align*} We can rewrite this relationship in terms of $x$ as \begin{align*} \frac{3x}{2} \le 3 - 3z -y, \end{align*} or \begin{align*} x \le 2\left(1 - z - \frac{y}{3}\right). \end{align*}

Since we also know $x \ge 0$, the new limits of integration are \begin{align*} \int_0^1\int_0^{3(1-z)} \int_0^{2(1-z-y/3)} dx\, dy \, dz. \end{align*}

## More examples

More examples of calculating triple integrals can be found in the pages describing the shadow method and cross section method of determining integration bounds.

Multivariable calculus.

• Previous: The cross section method for determining triple integral bounds
• Next: Introduction to changing variables in double integrals
• Next: The length of a curve*

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## Section 15.5 : Triple Integrals

3. Evaluate $$\displaystyle \iiint\limits_{E}{{6{z^2}\,dV}}$$ where $$E$$ is the region below $$4x + y + 2z = 10$$ in the first octant.

Show All Steps   Hide All Steps

Okay, let’s start off with a quick sketch of the region $$E$$ so we can get a feel for what we’re dealing with.

We’ve given the sketches with a set of “traditional” axes as well as a set of “box” axes to help visualize the surface and region.

For this problem the region $$E$$ is just the region that is under the plane shown above and in the first octant. In other words, the sketch of the plane above is exactly the top of the region $$E$$. The bottom of the region is the $$xy$$-plane while the sides are simply the $$yz$$ and $$xz$$-planes.

So, from the sketch above we know that we’ll have the following limits for $$z$$.

where we got the upper $$z$$ limit simply by solving the equation of the plane for $$z$$.

With these limits we can also get the triple integral at least partially set up as follows.

Next, we’ll need limits for $$D$$ so we can finish setting up the integral. For this problem $$D$$ is simply the region in the $$xy$$-plane (since we are integrating with respect to $$z$$ first) that we used to graph the plane in Step 1. That also, in this case, makes $$D$$ the bottom of the region.

Here is a sketch of $$D$$.

The hypotenuse of $$D$$ is simply the intersection of the plane from Step 1 and the $$xy$$-plane and so we can quickly get its equation by plugging $$z = 0$$ into the equation of the plane.

Given the nature of this region as well as the function we’ll be integrating it looks like we can use either order of integration for $$D$$. So, to keep the limits at least a little nicer we’ll integrate $$y$$’s and then $$x$$’s.

Here are the limits for the double integral over $$D$$.

The upper $$x$$ limit was found simply by plugging $$y = 0$$ into the equation of the hypotenuse and solving for $$x$$ to determine where the hypotenuse intersected the $$x$$-axis.

With these limits plugged into the integral we now have,

Okay, now all we need to do is evaluate the integral. Here is the $$z$$ integration.

Do not multiply out the integrand of this integral.

As noted in the last step we do not want to multiply out the integrand of this integral. One of the bigger mistakes students make with multiple integrals is to just launch into a simplification mode after the integral and multiply everything out.

Sometimes of course that must be done but, in this case, note that we can easily do the $$y$$ integration with a simple Calculus I substitution. Here is that work.

We gave the substitution used in this step but are leaving it to you to verify the details of the substitution.

Again, notice that we can either do some “simplification” or we can just do another substitution to finish this integral out. Here is the final integration step for this problem. We’ll leave it to you to verify the substitution details.

So, once we got the limits all set up, the integration for this problem wasn’t too bad provided we took advantage of the substitutions of course. That will often be the case with these problems. Getting the limits for the integrals set up will often, but not always, be the hardest part of the problem. Once they get set up the integration is often pretty simple.

Also, as noted above do not forget about your basic Calculus I substitutions. Using them will often allow us to avoid some messy algebra that will be easy to make a mistake with.

1. Calculus III

Solution Here is a set of practice problems to accompany the Triple Integrals section of the Multiple Integrals chapter of the notes for Paul Dawkins Calculus III course at Lamar University.

2. PDF Lecture 17: Triple integrals

As in two dimensions, triple integrals can be evaluated by iterated 1D integral computations. Here is a simple example: If E is the box {x ∈ [1, 2], y ∈ [0, 1], z ∈ [0, 1]} and f(x, y, z) = 24x2y3z. 0 0 R 24x2y3z dz 0 R dy dx . We start from the core 12x3y3, R 24x2y3z dz 0 =

3. PDF Self-Help Work Sheets C11: Triple Integration Problems for Fun and Practice

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4. PDF 14.3 Triple Integrals

EXAMPLE 1 By triple integrals find the volume of a box and a prism (Figure 14.12). j j j d ~ = 5 j j dxdydz and jjjdV= j f dxdydz box z=O y=O x=O prism z=o ,!=o x = o The inner integral for both is 5 dx = 2. Lines in the x direction have length 2, cutting through the box and the prism.

5. PDF Triple Integrals

Learn how to calculate triple integrals in this PDF document from the Illinois Institute of Technology. You will find examples, formulas, and applications of this advanced calculus topic. Whether you are a student, a teacher, or a curious learner, this PDF will help you master triple integrals.

6. PDF Triple Integral Practice To Set Up A Triple Integral

Practice Problems (solutions follow) For each of the following, set up the triple integral: ZZZ E f(x;y;z) dV. 1. E lies under the plane z = 1+x+y and above the region in the xy-plane bounded by the curves y = p x, y = 0 and x = 1. 2. E is bounded by the cylinder y2 + x2 = 9 and the planes z = 0, y = 3z, and x = 0 in the rst octant. 3.

7. 14.4E: Triple Integrals (Exercises)

Evaluate the triple integral with order dxdydz. 11. D is bounded by the planes x = 0, x = 2, z = − y and by z = y2 / 2. Evaluate the triple integral with orders dydzdx and dzdydx to verify that you obtain the same volume either way. Answer: 12. D is bounded by the planes z = 0, y = 9, x = 0 and by z = √y2 − 9x2.

8. PDF Handout on Triple Integrals

The purpose of this handout is to provide a few more examples of triple integrals. In particular, I provide one example in the usual x-y-z coordinates, one in cylindrical coordinates and one in spherical coordinates. Example 1 : Here is the problem: Integrate the function f(x, y, z) = z over the tetrahedral pyramid in space where 0 ≤ x. 0 ≤ y.

9. PDF Unit 17: Triple integrals

The triple integral which is more natural when considering physical units as volume is measured in cubic meters for example. The triple integral also allows for exibility: we can replace 1 with a function f(x;y;z). If interpreted as a charge density, then the integral is the total charge. 17.5. The problem of computing volumes has been tackled ...

10. PDF Lecture 3: Triple Integrals (Ii)

6 LECTURE 3: TRIPLE INTEGRALS (II) (In my opinion better to use this than double integral of bigger minus smaller) Example Find Vol(E) where Eis the region enclosed by the surfaces y= x2;z= 0;z= 1 y (1) Picture: Note: y= x2 (no z) is a cylinder in the zdirection parallel to the parabola y= x2. And z= 1 y(no x) is a plane in the x direction.

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Triple Integrals What to know: Be able to set up a triple integral on a bounded domain of 3 in any of the 6 possible orders Know the formula for volume and the one for mass from the applications. Triple integrals on box-shaped solids

12. PDF Triple Integrals

EXAMPLE 1 Compute the triple integral of (x; y; z) = 8xyz over the solid between z = 0 and z = 1 and over the region : = 0 = 2 = 1 = 3 Solution: To do so, we use (1) to write 8xyz dV = ZZR Z 1 8xyz dz dA 0 = 4xyz2 1 dA 0 = 4xy dA We then evaluate the resulting double integral over R: Z 1 Z 3 8xyz dV = 4xy dydx = 5 0 2

13. PDF Triple Integrals

the y-axis, so the outer integral (of these two integrals) will be Zp 4 x2 p 4 x2 something dy. Along each vertical slice, z goes from x2 + y2 to 8 (x2 + y2), so we get the nal iterated integral Z 2) 1 Remember that we can think of a triple integral as either a single integral of a double integral or a

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Examples of changing the order in triple integrals Example 1: A tetrahedron T is de ned by the inequalities x; y; z 0 and 2x + 3y + z 6. The tetrahedron has three faces which are triangles in the coordinate planes. For example, the face of T in the xy-plane is given by x; y 0 and 2x + 3y 6.

15. PDF Summary of Chapter 12 Practice Problems

Chapter 12. Double and Triple Integrals 12.1 The Double Integral over a Rectangle Let f = f(x, y) be continuous on the Rectangle R: a < x < b, c < y < d. The double integral of f over R= ( ) is a sample point in . where Notation: double integral of f over R= I f ( x , y ) dxdy

16. PDF DOUBLE AND TRIPLE INTEGRALS

(5.1) where J = [a; b] is an interval on the real line, have been studied. Here we study double integrals Z Z f(x; y) dx dy (5.2) Ω where Ω is some region in the xy-plane, and a little later we will study triple integrals Z Z Z f(x; y; z) dx dy dz T (5.3) where T is a solid (volume) in the xyz-space. 5.2 Double Integrals 5.2.1 Properties

17. 15.4: Triple Integrals

The sample point $$(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$$ can be any point in the rectangular sub-box $$B_{ijk}$$ and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.

18. Triple integral examples

Triple integral examples Example 1 A cube has sides of length 4. Let one corner be at the origin and the adjacent corners be on the positive x, y, and z axes. If the cube's density is proportional to the distance from the xy-plane, find its mass. Solution : The density of the cube is f(x, y, z) = kz for some constant k.

19. Calculus III

Back to Problem List. 1. Evaluate ∫ 3 2 ∫ 4 −1 ∫ 0 1 4x2y−z3dzdydx ∫ 2 3 ∫ − 1 4 ∫ 1 0 4 x 2 y − z 3 d z d y d x. Show All Steps Hide All Steps.

20. Calculus III

Use a triple integral to determine the volume of the region below z = 6−x z = 6 − x, above z = −√4x2 +4y2 z = − 4 x 2 + 4 y 2 inside the cylinder x2+y2 = 3 x 2 + y 2 = 3 with x ≤ 0 x ≤ 0. Solution

21. Calculus III

Section 15.5 : Triple Integrals. Back to Problem List. 3. Evaluate ∭ E 6z2dV ∭ E 6 z 2 d V where E E is the region below 4x+y+2z = 10 4 x + y + 2 z = 10 in the first octant. Show All Steps Hide All Steps. Start Solution.