integrationmultivariable-calculus
Evaluate triple integral
$$\iiint_B xyz \, \mathrm{d}V$$
where $B$ is the portion of the unit ball in the first octant (i.e. all coordinates positive).
Our professor told us the answer is $1/48$, and $B$ are the bounds of integration, but I'm not sure how to graph this or evaluate this triple integral.
You are very close. The integral in cylindrical coordinates has the right value, but the lower limit for $z$ should be $r$, not $0$. Notice that in spherical coordinates the angle $\phi$ should be integrated from $0$ to $\pi/4$. This restrict us to the volume inside the cone.
Addendum: The volume is the region between the cone and the sphere in octant I ($x,y,z\ge 0$). See the figure below.
In cylindrical coordinates we have $$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r}^{\sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta.$$
In spherical coordinates we have $$\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi.$$
you can check if this is correct: $$\int_{\theta=0}^{\pi/2}\int_{r=0}^3\int_{x=0}^{\frac{rcos\theta}{3}}r^2sin\theta dxdrd\theta$$
Best Answer
Use spherical coordinates for the sphere in the first orthant:
$$\;0\le r\le1,\,0\le\theta\le\frac\pi2,\,0\le\phi\le\frac\pi2\;$$
Observe that here $\;\theta\;$ is the azimuth angle and $\;\phi\;$ is the altitude, or vertical, angle. Then the integral is
$$\int_0^1\int_0^{\pi/2}\int_0^{\pi/2}\overbrace{r\cos\theta\sin\phi}^{=x}\cdot \overbrace{r\sin\theta\sin\phi}^{=y}\cdot \overbrace{r\cos\phi}^{=z}\cdot \overbrace{r^2\sin\phi}^{=\text{Jacobian}}\, d\phi\,\mathrm d\theta\,\mathrm dr=$$
$$=\int_0^1\int_0^{\pi/2}\int_0^{\pi/2} r^5\cos\theta\sin\theta\cos\phi\sin^3\phi\, d\phi\,\mathrm d\theta\,\mathrm dr=$$
$$=\left.\frac16r^6\right|_0^1\cdot\left.\frac12\sin^2\theta\right|_0^{\pi/2}\cdot\left.\frac14\sin^4\phi\right|_0^{\pi/2}=\frac16\cdot\frac12\cdot\frac14=\frac1{48}$$