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Evaluate the given integral by changing to polar coordinates.

$\int\int_D x^2y \, dA$, where $D$ is the top half of the disk with center the origin and radius $5$.

when i plugged everything in, I got the double integral $$\int_0^\pi \int_0^5 r^4\cos^2\theta \sin \theta \,dr \,d\theta.$$ then I used $u$ substitution and got $5\int u^2 \,du =\left. \frac{5u^3}{3}\right\vert_{\theta=0}^{\theta=\pi} =-5/3 - 5/3 = -10/3$ but that's definitely wrong. I looked at multiple examples, what am I doing wrong here? I'm pretty sure i have the right method but something's going wrong.

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  • $\begingroup$ i just saw that derivative of cosine is -sine >.< $\endgroup$ – 2316354654 Oct 31 '17 at 7:26
  • $\begingroup$ but that just changes my answer to positive 10/3, and the answer is 1250/3... what am I doing wrong??? $\endgroup$ – 2316354654 Oct 31 '17 at 7:29
  • $\begingroup$ math.meta.stackexchange.com/q/5020/306553 mathjax reference $\endgroup$ – Siong Thye Goh Oct 31 '17 at 7:41
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You have forgotten to integrate with respect to $r$ and I have no idea where does your leading $5$ comes from.

$$\int_0^5r^4 \,dr =\frac{5^5}{5}=5^4$$

\begin{align} \int_0^\pi \cos^2 \theta \sin \theta \, d\theta &= \left. -\frac{\cos^3\theta}{3} \right\vert_0^\pi \end{align}

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  • $\begingroup$ thanks!! ...... $\endgroup$ – 2316354654 Oct 31 '17 at 7:51
  • $\begingroup$ remark: btw, it is interesting that your limit of the integral is opposite from mine. I go from smaller to bigger number while you do the opposite. but yup, doesn't matter in this case as two negatives make things positive, but if you involve odd number of integrals, your sign might be opposite from others. $\endgroup$ – Siong Thye Goh Oct 31 '17 at 8:01
  • $\begingroup$ when i said 5 to 0 and pi to 0 in my original post i meant to have the 5 and pi on top. So I guess I should be saying "0 to 5 and 0 to pi" if i want the bigger numbers on top? $\endgroup$ – 2316354654 Oct 31 '17 at 8:04
  • $\begingroup$ oops, sorry, i misinterpreted you. $\endgroup$ – Siong Thye Goh Oct 31 '17 at 8:05

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