Multivariable calculus and vector analysis

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 $f\left(x,y,z\right)=kz$ for some constant $k$.

If $W$ is the cube, the mass is the triple integral

$$\begin{array}{llll}\hfill {\iiint }_{W}kzdV& ={\int \; <!--nolimits\; -->}_0^4{\int \; <!--nolimits\; -->}_0^4{\int \; <!--nolimits\; -->}_0^{4}kzdxdydz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^4{\int \; <!--nolimits\; -->}_0^4\left({\left.kxz\right|}_{x=0}^{x=4}\right)dydz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^4{\int \; <!--nolimits\; -->}_0^{4}4kzdydz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^4\left({\left.4kzy\right|}_{y=0}^{y=4}\right)dz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{4}16kzdz={\left.8k{z}^2\right|}_{z=0}^{z=4}=128k& \hfill & \end{array}$$

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

Example 2

Evaluate the integral

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^x{\int \; <!--nolimits\; -->}_0^{1+x+y}f\left(x,y,z\right)dzdydx& & \hfill \end{array}$$

where $f\left(x,y,z\right)=1$.

Solution:

$$\begin{array}{llll}\hfill & {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^x{\int \; <!--nolimits\; -->}_0^{1+x+y}dzdydx& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^x\left({\left.z\right|}_{z=0}^{z=1+x+y}\right)dydx& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^x\left(1+x+y\right)dydx& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\left.\left(y+yx+\frac{y^2}{2}\right)\right|}_{y=0}^{y=x}dx& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left(x+x^2+\frac{x^2}{2}\right)dx& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1\left(x+\frac{3x^2}{2}\right)dx& \hfill & \\ \hfill & =\frac{x^2}{2}+{\left.\frac{x^3}{2}\right|}_0^1=\frac{1}{2}+\frac{1}{2}=1& \hfill & \end{array}$$

Note: when we integrate $f\left(x,y,z\right)=1$, the integral ${\iiint }_{W}dV$ is the volume of the solid $W$.

Example 3a

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

Solution: The “ice cream cone” is between these two surfaces:

$$\begin{array}{lll}\hfill \sqrt{x^2+y^2}\le z\le \sqrt{1-x^2-y^2}.& & \hfill \end{array}$$

This gives the range of $z$ as a function of $x$ and $y$.

Now we need to find the maximal range of $x$ and $y$. 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 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

$$\begin{array}{lll}\hfill x^2+y^2=\frac{1}{2}.& & \hfill \end{array}$$

We have shown that in the “ice cream cone”

$$\begin{array}{lll}\hfill x^2+y^2\le \frac{1}{2}.& & \hfill \end{array}$$

We write this as

$$\begin{array}{lll}\hfill -\sqrt{1∕2-x^2}\le y\le \sqrt{1∕2-x^2}& & \hfill \end{array}$$

This gives the range of $y$ as a function of $x$.

Finally, 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 $x^2\le 1∕2$. The maximal range of $x$ is

$$\begin{array}{lll}\hfill -1∕\sqrt{2}\le x\le 1∕\sqrt{2}.& & \hfill \end{array}$$

We have determined all the limits on the interated integral. The integral is

$$\begin{array}{lll}\hfill {\iiint }_{W}fdV={\int \; <!--nolimits\; -->}_{-1∕\sqrt{2}}^{1∕\sqrt{2}}{\int \; <!--nolimits\; -->}_{-\sqrt{1∕2-x^2}}^{\sqrt{1∕2-x^2}}{\int \; <!--nolimits\; -->}_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}}f\left(x,y,z\right)dzdydx& & \hfill \end{array}$$

Example 3b

Find the volume of the ice cream cone.

Solution: Simply set $f=1$ in Example 3a.

Volume is

$$\begin{array}{lll}\hfill {\iiint }_{W}dV={\int \; <!--nolimits\; -->}_{-1∕\sqrt{2}}^{1∕\sqrt{2}}{\int \; <!--nolimits\; -->}_{-\sqrt{1∕2-x^2}}^{\sqrt{1∕2-x^2}}{\int \; <!--nolimits\; -->}_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}}dzdydx& & \hfill \end{array}$$

We won’t attempt to evaluate this integral in rectangular coordinates.

Example 4

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

Solution:

The first step is to find the equation for the plane. You can follow the procedure in forming plane example 2 to calculate that the plane is

$$\begin{array}{lll}\hfill 3x+2y+6z=6.& & \hfill \text{ (1)}\end{array}$$

In the tetrahedron, the total range of $z$ is

$$\begin{array}{lll}\hfill 0\le z\le 1& & \hfill \end{array}$$

Now we need to find the maximal total range of $x$ for each value of $z$. For any value of $z$, $x$ has the largest range when $y=0$. So look at the cross section of the tetrahedron in the $xz$-plane (i.e., where $y=0$).

The lower limit of $x$ is zero, and the upper limit corresponds to the plane given by equation (1) when $y=0$. Plugging $y=0$ into equation (1), we obtain $3x+6z=6$, or $x=2\left(1-z\right)$. Hence, for a given $z$, the range of $x$ is

$$\begin{array}{lll}\hfill 0\le x\le 2\left(1-z\right).& & \hfill \end{array}$$

Last, we need to find the range of $y$ for a given value of $z$ and $x$. The lowest $y$ gets is zero. The upper value of $y$ is given by equation (1) for the upper plane, which we can solve for $y$ to write as $y=3\left(1-x∕2-z\right)$. Hence, for a given $z$ and $x$, the range of $y$ is

$$\begin{array}{lll}\hfill 0\le y\le 3\left(1-\frac{x}{2}-z\right).& & \hfill \end{array}$$

To find the volume, we integrate the function 1 over this region:

$$\begin{array}{llll}\hfill & {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\left(1-z\right)}{\int \; <!--nolimits\; -->}_0^{3\left(1-x∕2-z\right)}dydxdz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\left(1-z\right)}3\left(1-\frac{x}{2}-z\right)dxdz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{1}3{\left[x-\frac{x^2}{4}-zx\right]}_{x=0}^{x=2\left(1-z\right)}dz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{1}3\left(2\left(1-z\right)-{\left(1-z\right)}^2-2z\left(1-z\right)\right)dz& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{1}3\left(1-2z+z^2\right)dz& \hfill & \\ \hfill & =3{\left[z-z^2+\frac{z^3}{3}\right]}_0^1& \hfill & \\ \hfill & =3\left(1-1+\frac{1}{3}\right)=3\left(\frac{1}{3}\right)=1& \hfill & \end{array}$$

Example 5

Change the order of $x$ and $y$ in the integral we derived above,

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\left(1-z\right)}{\int \; <!--nolimits\; -->}_0^{3\left(1-x∕2-z\right)}dydxdz,& & \hfill \end{array}$$

so that the order will be $dxdydz$.

Solution: 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

$$\begin{array}{lll}\hfill 0\le y\le 3\left(1-\frac{x}{2}-z\right).& & \hfill \end{array}$$

The range is largest when $x=0$, so

$$\begin{array}{lll}\hfill 0\le y\le 3\left(1-z\right)& & \hfill \end{array}$$

Then, given $z$ and $y$, we need to know the range of the $x$. The following relationship must still be true.

$$\begin{array}{lll}\hfill y\le 3\left(1-\frac{x}{2}-z\right)& & \hfill \end{array}$$

Rewrite in terms of $x$:

$$\begin{array}{lll}\hfill \frac{3x}{2}\le 3-3z-y& & \hfill \end{array}$$

or

$$\begin{array}{lll}\hfill x\le 2\left(1-z-\frac{y}{3}\right)& & \hfill \end{array}$$

Since we also know $x\ge 0$, the new limits of integration are

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{3\left(1-z\right)}{\int \; <!--nolimits\; -->}_0^{2\left(1-z-y∕3\right)}dxdydz.& & \hfill \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

3a. Double integrals

3b. Triple Integrals

• Triple integrals
• Triple integral examples

3c. Changing variables

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material