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Let $C$ be the subset of the plane given by $$ C \colon= \{ \ (x,y) \in \mathbb{R}^2 \ | \ 0 \leq x^2 + y^2 \leq 1 \}.$$

Then what is the value of the double integral $$ \int_{C} \int (x^2 + y^2) \ dx \ dy?$$

My work:

In $C$, we have $-1 \leq x \leq 1$ and $-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$. So we have $$ \int_C \int (x^2 + y^2) \ dx \ dy = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \ (x^2 + y^2) \ dy \ dx = 2 \int_{-1}^1 (x^2 \sqrt{1-x^2} + \frac{ (\sqrt{1-x^2})^3}{3} ) \ dx = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx. $$ Now let $x= \sin t$. Then we obtain $$ \int_C \int (x^2 + y^2) \ dx \ dy = \frac{4}{3} \int_0^1 (2x^2 + 1) \sqrt{1-x^2} \ dx = \frac{4}{3} \int_{t = 0}^{\pi/2} (2\sin^2 t + 1) \cos^2 t \ dt \ \ = \frac{1}{3} \int_{t=0}^{\pi/2} (2\sin^2 2t + 4 \cos^2 t) \ dt = \frac{4}{3} \int_{t=0}^{\pi/2} ( 1 - \cos 4t + 2(1+ \cos 2t) ) \ dt = \frac{1}{3} \int_{t=0}^{\pi/2} (3 + 2 \cos 2t - \cos 4t) \ dt = \frac{\pi}{2}. $$ Am I right?

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It is easier to use polar coordinates? –  Mhenni Benghorbal Jul 27 at 10:48
2  
I didn't check the computations you made, but the value of the integral is correct. –  Git Gud Jul 27 at 10:52
    
You could check the result with latest version (10) of Mathematica: Integrate[x^2+y^2, {x, y} \[Element] Disk[{0, 0}]]. –  murray Jul 27 at 20:55

2 Answers 2

Use polar coordinates,

We know that

$$r^2 = x^2 + y^2$$

So our double integral becomes

$$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$

Now solve.

EDIT

I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral.

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I think it is worth making note that $dx dy = r dr d\theta$ is a well known result when converting Cartesian to polar coordinates. (When I first encountered polar coordinates, I was somewhat confused by the appearance of the extra $r$.) –  jpmc26 Jul 27 at 18:49
    
That is because of the jacobian –  Varun Iyer Jul 27 at 19:20

If you use polar coordinates you will get the following

$$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta $$

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