## stamp [CALCULUS III—DOUBLE INTEGRALS] Evaluate by converting to polar coordinates. (figure inside) one year ago one year ago

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$\int_0^3\int_0^{\sqrt{9-x^2}}(x^2+y^2)^{3/2}\ dydx$

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3. zepdrix

Let's start by drawing the region.$\large 0 \le y \le \sqrt{9-x^2}$Here are the boundaries on y, it appears to be the upper half of a circle, with radius 3.|dw:1363217680487:dw|

4. zepdrix

How bout the boundaries on x?$\large 0 \le x \le 3$This is telling us that our region is only the first quadrant.

5. zepdrix

|dw:1363217804038:dw|This is how our boundaries would change in polar.

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Hey I appreciate your help. Keep posting guidelines, I am currently finishing up at work but when I get home I will be looking at this again and begin solving.

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If you do decide to post a lot, leave the answer and final evaluations to me ;)

8. zepdrix

Changing to polar, we let $$\large x=r\cos\theta$$ and $$\large y=r\sin\theta$$ and simplify our integral down. An important thing to remember, is that when we change the differentials, a factor of $$\large r$$ will pop out as well.$\large dx\;dy \qquad \rightarrow \qquad r\;dr\;d\theta$

9. zepdrix

Haha XD I prolly won't post the final answer, just some steps :D They frown on that stuff here.. If I don't let you do some of the work, heh

10. zepdrix

So I think this is how our integral would change. $\large \int\limits_{\theta=0}^{\pi/2}\quad \int\limits_{r=0}^{3} \left((r \cos \theta)^2+(r \sin \theta)^2 \right)^{3/2}\left(r \;dr\;d \theta\right)$

11. zepdrix

Lemme know if you have trouble solving that. Don't forget your important trig identity, $$\large \cos^2x+\sin^2x=1$$