Double integral over a circular region using rectangular coord's


by sbpf
Tags: circular, coord, double, integral, rectangular, region
sbpf
sbpf is offline
#1
Oct15-12, 05:04 PM
P: 3
I would like to compute
$$ \iint \limits_{x^2 + y^2 \le 3} \! x^2 + y^2 \, \mathrm{d} A $$
using rectangular coord's.

First, I'll compute the iterated integral using polar coordinates so that I can check my work.

Limits:
$$ 0 \le \theta \le 2\pi \\
0 \le r \le 3 $$
so
$$ \iint \limits_{x^2 + y^2 \le 3} \! x^2 + y^2 \, \mathrm{d} A \
= \int_0^{2\pi} \!\!\! \int_0^3 \! ((r\cos\theta)^2 + (r\sin\theta)^2)r \, \mathrm{d}r\mathrm{d}\theta \
= \int_0^{2\pi} \!\!\! \int_0^3 \! r^3 \, \mathrm{d}r\mathrm{d}\theta \
= \frac{81\pi}{2} $$

Hopefully that's right.

Now for rectangular coord's:

I'll choose ##\mathrm{d}y\mathrm{d}x## as the order of integration.

Limits:
$$ -\sqrt{3} \le x \le \sqrt{3} \\
-\sqrt{3-x^2} \le y \le \sqrt{3-x^2} $$

so
$$ \iint \limits_{x^2 + y^2 \le 3} \! x^2 + y^2 \, \mathrm{d} A \
= \int_{-\sqrt{3}}^{\sqrt{3}} \! \int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}} \! x^2+y^2 \, \mathrm{d}y\mathrm{d}x \
= \frac{9\pi}{2} $$

so I've made a mistake somewhere...

To be honest, I'm not sure if I'm even close here. I'm guessing either my limits are wrong or I have absolutely no clue what I'm doing, so a point in the right direction would be greatly appreciated. Also, is there any significance to this particular double integral? We've just started covering multiple integration in class so I'm still trying grasp the concepts surrounding it.

Thanks.

Edit: I think I messed up my limits for the polar one previously by going to ##3## instead of ##\sqrt{3}##, could someone please verify that this newer one is correct?
$$ \iint \limits_{x^2 + y^2 \le 3} \! x^2 + y^2 \, \mathrm{d} A \\
= \int_0^{2\pi} \!\!\! \int_0^\sqrt{3} \! ((r\cos\theta)^2 + (r\sin\theta)^2)r \, \mathrm{d}r\mathrm{d}\theta \\
= \int_0^{2\pi} \!\!\! \int_0^\sqrt{3} \! r^3 \, \mathrm{d}r\mathrm{d}\theta \\
= \int_0^{2\pi} \! \frac{r^4}{4} \, \biggr|_{r=0}^\sqrt{3} \, \mathrm{d}\theta \\
= \int_0^{2\pi} \! \frac{9}{4} \, \mathrm{d}\theta \\
= \frac{9}{4}\theta \, \biggr|_{0}^{2\pi} \\
= \frac{9\pi}{2} $$
which is what I got for the rectangular coord's one, so I guess I was partially right?
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