# Double integral of $(y-2)^2$ over $\{(x,y)\in\mathbb{R}^2\colon x^2 +y^2 \le 6x\}$

Trying to solve this for hours... will appreciate any help

$$\int\int_{D}^{} (y-2)^2dxdy$$ $$D=\{(x,y)| x^2+y^2\leq6x\}$$

Tried doing it in a few ways... doesn't seem to work out.

• $x^2+y^2 \le 6x \Longrightarrow (x-3)^2+y^2\le 9$ – InterstellarProbe May 31 '18 at 18:05

Observe that the given region $D$ is a closed disk with centre $(3,0)$ and radius $3$, since \begin{align*} x^2 + y^2 & \le 6x \\ x^2 - 6x + y^2 & \le 0 \\ x^2 - 6x + 9 + y^2 & \le 9 \\ (x-3)^2 + y^2 & \le 3. \end{align*} To this end, consider the polar coordinates $x-3 = r\cos\theta$ and $y = r\sin\theta$, with $r\in [0,3]$ and $\theta\in[0,2\pi]$. Then $dxdy = rdrd\theta$ and \begin{align*} \iint_D (y-2)^2 dxdy & = \int_0^{2\pi}\int_0^3 (r\sin\theta - 2)^2 r\, drd\theta \\ & = \int_0^{2\pi}\int_0^3 r\Big(r^2\sin^2\theta - 4r\sin\theta + 4\Big)\, drd\theta \\ & = \int_0^{2\pi}\int_0^3 \Big(r^3\sin^2\theta - 4r^2\sin\theta + 4r\Big)\, drd\theta \\ & = \int_0^{2\pi}\int_0^3 r^3\sin^2\theta\, drd\theta - \int_0^{2\pi}\int_0^3 4r^2\sin\theta\, drd\theta + \int_0^{2\pi}\int_0^3 4r\, drd\theta \\ & = I_1 + I_2 + I_3. \end{align*} We use the identity $\sin^2\theta = \dfrac{1-\cos(2\theta)}{2}$ to compute the integral $I_1$: \begin{align*} I_1 & = \left(\int_0^3 r^3\, dr\right)\left(\int_0^{2\pi} \left(\frac{1 - \cos(2\theta)}{2}\right)\, d\theta\right) \\ & = \left(\frac{3^4}{4}\right) \left(\int_0^{2\pi}\frac{1}{2}\, d\theta - \int_0^{2\pi} \frac{\cos(2\theta)}{2}\, d\theta\right). \end{align*} The second integral in the parentheses vanishes since $\cos(2\theta)$ is an odd function about $\theta = \pi/2$ on $[0,\pi]$ and $\theta = 3\pi/2$ on $[\pi,2\pi]$ and so $$I_1 = \left(\frac{81}{4}\right)\left(\frac{2\pi}{2}\right) = \frac{81\pi}{4}.$$ The integral $I_2$ vanishes since $\sin\theta$ is an odd function about $\theta = \pi$ on $[0,2\pi]$. The integral $I_3$ is \begin{align*} I_3 & = \left(\int_0^{2\pi}\, d\theta\right)\left(\int_0^3 4r\, dr\right) \\ & = 2\pi\left(2r^2\Big|_0^3\right) \\ & = \left(2\pi\right)(2)(3^2) \\ & = 36\pi. \end{align*} Hence, $$\iint_D (y-2)^2\, dxdy = \frac{81\pi}{4} + 0 + 36\pi.$$