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Calculate the mass: $$D = \{1 \leq x^2 + y^2 \leq 4 , y \leq 0\},\quad p(x,y) = y^2.$$

So I said:

$M = \iint_{D} {y^2 dxdy} = [\text{polar coordinates}] = \int_{\pi}^{2\pi}d\theta {\int_{1}^{2} {r^3sin^2\theta dr}}$.

But when I calculated that I got the answer $0$ which is wrong, it should be $\frac{15\pi}{8}$. Can someone please tell me what I did wrong?

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up vote 2 down vote accepted

You have the integral $$M=\int_\pi^{2\pi} d\theta\int_1^2 r^3 \sin^2\theta dr $$ this seems fine. One thing is sure this integral is not zero: Indeed your can write it as a product of two integrals $$M=(\int_\pi^{2\pi} \sin^2\theta d\theta) (\int_1^2 r^3 dr)$$ and both those integrals give strictly positive numbers.

I would advise you to compute these two integrals separately and check you get stricly positive numbers.

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I calculated it with wolfram and it gives me a correct answer. Well at least my direction is alright, I'll try and calculate it again. Much appreciated. – TheNotMe Jul 6 '13 at 9:21

You have the set-up correct, but you have incorrectly computed the integral

Let's work it out together.

$\int_{\pi}^{2\pi}d\theta {\int_{1}^{2} {r^3\sin^2\theta dr}}$

$\int_{\pi}^{2\pi} {\int_{1}^{2} {r^3\sin^2\theta drd\theta}}$

$\int_{\pi}^{2\pi} \sin^2\theta d\theta {\int_{1}^{2} {r^3dr}}$

$\int_{\pi}^{2\pi} \sin^2\theta d\theta (\frac{2^4}{4} - \frac{1^4}{4})$

$\int_{\pi}^{2\pi} \sin^2\theta d\theta (3\frac{3}{4})$

$\frac{1}{2}((2\pi - \sin(2\pi)\cos(2\pi) - \pi +\sin(\pi)\cos(\pi)) (3\frac{3}{4})$

note that the integral of $\sin^2(x)$ = $\frac{1}{2}(x - \sin(x)\cos(x))$

$\frac{1}{2}(\pi)(3\frac{3}{4}) = \frac{15\pi}{8}$

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