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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.

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  • $\begingroup$ are we supposed to convert to spherical or polar? if so, how? $\endgroup$ – Math19384 Apr 7 '18 at 18:52
  • $\begingroup$ youtube.com/watch?v=zFy-OpajEtA $\endgroup$ – NicNic8 Apr 7 '18 at 18:53
  • $\begingroup$ @NicNic8 i know how to do a triple integral. i'm not sure how to write the bounds of integration for this. $\endgroup$ – Math19384 Apr 7 '18 at 18:55
  • $\begingroup$ I see. It looks like someone gave you a good answer below. But let's see if we can see how to think about this problem in a good way? A simpler problem is to find the area of a 1/4 of a circle. Can you determine the bounds for that problem? Alternatively, we might know that the volume of a sphere is 4/3 pi r^3. And one quadrant would divide the volume by 8, so the answer is 4/24 pi r^3 = 1/6 pi r^3. $\endgroup$ – NicNic8 Apr 8 '18 at 19:31
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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

$$\int_0^1\int_0^{\pi/2}\int_0^{\pi/2}\overbrace{r\cos\theta\sin\phi}^{=x}\cdot \overbrace{r\sin\theta\sin\phi}^{=y}\cdot \overbrace{r\cos\phi}^{=z}\cdot \overbrace{r^2\sin\phi}^{=\text{Jacobian}}\, d\phi\,\mathrm d\theta\,\mathrm dr=$$

$$=\int_0^1\int_0^{\pi/2}\int_0^{\pi/2} r^5\cos\theta\sin\theta\cos\phi\sin^3\phi\, d\phi\,\mathrm d\theta\,\mathrm dr=$$

$$=\left.\frac16r^6\right|_0^1\cdot\left.\frac12\sin^2\theta\right|_0^{\pi/2}\cdot\left.\frac14\sin^4\phi\right|_0^{\pi/2}=\frac16\cdot\frac12\cdot\frac14=\frac1{48}$$

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