# 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
ill re upload

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 $\displaystyle \frac{\pi}6$.