Answer to Question #111448 in Calculus for sohaib

Question

Evaluate the following integral by first converting to an integral in polar coordinates.
∫
0
−
2
∫
√
4
−
y
2
−
√
4
−
y
2
x
2
d
x
d
y
∫−20∫−4−y24−y2x2dxdy

Answer 1

$$\displaystyle\int_{-2}^0\displaystyle\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}x^2dxdy$$

$x=r\cos\theta, y=r\sin\theta$

$0\leq r\leq 2, -\pi\leq \theta\leq 0$

$$\displaystyle\int_{-2}^0\displaystyle\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}x^2dxdy=$$

$$=\displaystyle\int_{-\pi}^0\displaystyle\int_{0}^{2}rr^2\cos^2\theta drd\theta=$$

$$=\displaystyle\int_{-\pi}^0\big[{r^4 \over 4}\big]\begin{array}{cc} 2 \\ 0 \end{array}\cos^2\theta d\theta=$$

$$=2\displaystyle\int_{-\pi}^0 (1+\cos(2\theta)) d\theta=$$

$$=\big[2\theta+\sin(2\theta)\big]\begin{array}{cc} 0\\ -\pi \end{array}=2\pi$$