# Thread: Triple Integral in Spherical Coords.

1. ## Triple Integral in Spherical Coords.

Hey,
I'm having difficulty with a specific problem. It was given on my professor's previous exam and is meant to be worked by hand but I'm getting a ridiculous integral that I don't think is solvable by hand in 10 mins (which it should be).

It seems my formatting was lost so I attached a picture :s
The cutoff part says (x^2+y^2)^1/2

Sorry for the terrible picture quality.

2. I'm having difficulty reading the scanned handwriting, so I'm not sure how much of this you have already done. I'd start by converting x and y to polar coordinates and writing $\displaystyle r = \sqrt{x^2+y^2}$. Then the limits for z are $\displaystyle r\leqslant z\leqslant\sqrt{4-r^2}$. That implies that $\displaystyle r^2\leqslant 4-r^2$, or $\displaystyle r\leqslant\sqrt2$.

So if you do the z-integral, and convert x and y to polars, then you get

\displaystyle \begin{aligned}\int\!\!\!\int\!\!\!\int_Ex^2y^2z\, dzdxdy &= \int_0^{2\pi}\int_0^{\sqrt2}r^2\cos^2\theta*r^2\si n^2\theta \Bigl[\tfrac12z^2\Bigr]_r^{\sqrt{4-r^2}}\,rdrd\theta\\ &= \int_0^{2\pi}\tfrac12\cos^2\theta\sin^2\theta\Bigl[\tfrac23r^6 - \tfrac14r^8\Bigr]_0^{\sqrt2} = \frac23\int_0^{2\pi}\cos^2\theta\sin^2\theta\,d\th eta. \end{aligned}

Now use the fact that $\displaystyle \cos^2\theta\sin^2\theta = \tfrac14\sin^2(2\theta) = \tfrac18(1-\cos(4\theta))$ to do the theta-integral. I get the final answer as $\displaystyle \frac{\pi}6$.