How do you do change of variables for triple integrals?

Question

I am evaluating the function over the following bounds.$$\int_0^2\int_0^{\sqrt{4-x^2}}\int_0^{\sqrt{4-x^2-y^2}}z\sqrt{4-x^2-y^2}\,\mathrm dz\,\mathrm dy\,\mathrm dx$$

I'm not sure how to combine triple integrals and change of variables. Can someone run me through the steps for this problem?

thanks

Answer 1

the bound is a globe (the one eighth in the first octant). you can change to spherical coordinates.

$x=r\cos\theta\cos\phi$

$y=r\sin\theta\cos\phi$

$z=r\sin\phi$

$\mathrm{d}V=\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=r^2\sin\phi\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$

Answer 2

As you said in a comment to Lance's answer, the integral is taken over the first octant, since we have $x,y,z\ge0$. The sphere has radius $2$, so we have $0\le r\le 2$. Now $\theta$ is the angle of rotation about the $z$-axis. It's just like $\theta$ in polar coordinates. Since we're only in the first octant, $0\le\theta\le\frac{pi}{2}$. Now $\phi$ is the angle between the vector and the $z-axis$. There are a couple of different systems for defining it. Some people restrict $0\le\theta\le \pi$. Others restrict $-\frac{\pi}{2}\le\theta\le \frac{\pi}{2}.$ In any event, because we are only taking the hemisphere above the $xy$-plane, $\phi$ will have a range of length $\frac{\pi}{2}.$

Using the system given in Lance's answer, we will have $0\le \theta \le \frac{pi}{2}.$ (Notice that when $\theta=0, z=r\sin \theta= 0,$ and when $\theta=\frac{\pi}{2}, z=r\sin \theta =r.$) Finally, the integral becomes $$\int_0^2\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}{r\sin \phi}\sqrt{4-r^2cos^2\phi}r^2\sin\phi\, \mathrm d\phi\, \mathrm d\theta\, \mathrm dr $$

Answer 3

\begin{align} & \int_0^2 \left( \int_0^{\sqrt{4-x^2}} \left( \int_0^{\sqrt{4-x^2-y^2}}z\sqrt{4-x^2-y^2}\,\mathrm dz\right) \,\mathrm dy \right)\,\mathrm dx \\[10pt] & \text{The innermost integral is easy:} \\ & \int_0^{\sqrt{4-x^2-y^2}} z \underbrace{\sqrt{4-x^2-y^2}}_\text{No $z$ appears here.} \,\, \mathrm dz \\[10pt] = {} & \sqrt{4-x^2-y^2} \int_0^{\sqrt{4-x^2-y^2}} z\,\mathrm d z \\ & \text{This can be done because the factor that} \\ & \text{was pulled out does not depend on $z$.} \\[10pt] = {} & \sqrt{4-x^2-y^2} \cdot \frac{4-x^2-y^2} 2 = \frac 1 2 (4-x^2-y^2)^{3/2}. \\[10pt] & \text{So now we have} \\ & \frac 1 2 \int_0^2 \int_0^{\sqrt{4-x^2}} (4-x^2-y^2)^{3/2} \, \mathrm dy \, \mathrm dx \\[10pt] = {} & \frac 1 2 \iint\limits_{\{\,(x,y)\,:\, x^2+y^2\,\le\,4 \, x,y\,\ge\,0 \,\}} (4-x^2-y^2) \, \mathrm d(x,y) \\[10pt] = {} & \frac 1 2 \int_0^{\pi/2} \underbrace{\left( \int_0^2 (4-r^2)^{3/2} r\, dr \right)}_\text{No $\theta$ appears here.} \, d\theta \\[10pt] = {} & \frac 1 2 \cdot \frac \pi 2 \int_0^2 (4-r^2)^{3/2} r\,dr \quad \text{This works because no $\theta$ was in $\int_0^{\pi/2}\cdots\,d\theta$.} \\[10pt] = {} & \frac \pi 4 \int_4^0 u^{3/2} \left( \frac{-du} 2\right) = \cdots \end{align}