MATH 251 Lecture 24

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Triple Integrals

Given a region $R\in {\mathbb{ℝ}}^3=[a,b]\times [c,d]\times [e,f]$,

$\underset{R}{\iiint }f(x,y,z)\mathrm{d}V=\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}\underset{e}{\overset{f}{\int }}f(x,y,z)\mathrm{d}z\mathrm{d}y\mathrm{d}x$

Example with Simplification

$R=[-1,1]\times [0,2]\times [0,1]$

$f(x,y,z)=x\sin y+{\mathrm{e}}^z$

$\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2}{\int }}\underset{-1}{\overset{1}{\int }}(x\sin y+{\mathrm{e}}^z)\mathrm{d}x\mathrm{d}y\mathrm{d}z$

The integral can be split into a sum of two integrals:

$\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2}{\int }}\underset{-1}{\overset{1}{\int }}x\sin y\mathrm{d}x\mathrm{d}y\mathrm{d}z+\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2}{\int }}\underset{-1}{\overset{1}{\int }}{\mathrm{e}}^z\mathrm{d}x\mathrm{d}y\mathrm{d}z$

The first integral is over an odd region with respect to x, so for every point on one side, there is an equal and opposite point on the other side, so the entire integral will evaluate to 0:

$\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2}{\int }}\underset{-1}{\overset{1}{\int }}{\mathrm{e}}^z\mathrm{d}x\mathrm{d}y\mathrm{d}z$

The function inside this integral does not depend on x or y, so we can multiply the remaining integral with respect to z by the lengths of the intervals along x and y:

$(2-0)(1-(-1))\underset{0}{\overset{1}{\int }}{\mathrm{e}}^z\mathrm{d}z=4\underset{0}{\overset{1}{\int }}{\mathrm{e}}^z\mathrm{d}z$

This is easy to evaluate as normal:

$4(\mathrm{e}-1)$

Triple Integrals in Spherical Coordinates

Find the center of mass of a unit hemisphere: $x^2+y^2+z^2=1$, $z\ge 0$. Density is a constant $\rho (x,y,z)={\rho }_0$

Find the mass: $\frac{2}{3}\pi {\rho }_0$... that was easy.

Due to symmentry, $\overline{x}=\overline{y}=0$.

Therefore, we only need to calculate $\overline{z}=\frac{3}{2\pi {\rho }_0}\underset{R}{\iiint }z{\rho }_0\mathrm{d}V=\frac{3}{2\pi }\underset{R}{\iiint }z\mathrm{d}V$

This integral would be easier to evaluate by converting to cylindrical (polar) coordinates since $z=\sqrt{1-x^2-y^2}=\sqrt{1-r^2}$

$\frac{3}{2\pi }\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{1}{\int }}\frac{1}{2}(1-r^3)r^3\mathrm{d}r\mathrm{d}\theta =\dots =\frac{3}{8}$

Another Example

$f(x,y,z)=x^2-3xy-2z$, $R$ is bounded by $x+2y+3z\le 5$ in the first octant.

Let's evaluate in the order $\mathrm{d}y\mathrm{d}z\mathrm{d}x$

Find the bounds:

• $0\le y\le \frac{1}{2}(6-x-3z)$
• $0\le z\le \frac{1}{3}(6-x)$ (plug in 0 for $y$)
• $0\le x\le 6$ (plug in 0 for $z$ and $y$)

$\underset{0}{\overset{6}{\int }}\underset{0}{\overset{\frac{1}{3}(6-x)}{\int }}\underset{0}{\overset{\frac{1}{2}(6-x-3z)}{\int }}(x^2+3xy-2z)\mathrm{d}y\mathrm{d}z\mathrm{d}x$

And due to time constraints, we will not evaluate that integral.