# Double integral with change of coordinates

Evaluate the following double integral:

$$\iint_D \sqrt{x^2+y^2}\,dx\,dy$$

where $$D=\{(x,y)\in\mathbb{R^2},\ x^2+y^2-4x<0\}$$. I'm trying to use the polar coordinates by making the substitution $$x=\rho\cos\theta$$ and $$y=\rho\sin\theta$$.

After replacing in the domain, I find out that $$0<\rho<4\cos\theta$$. But what about $$\theta$$ ?

• The radius of the circle is 2 so that 0<𝜌<2 and the 𝜃 goes from 0 to 2*pi as mentioned from JVV. – Wrench Dec 23 '19 at 11:49

First, let's look at the region $$D$$: $$x^2+y^2-4x < 0 \iff (x-2)^2 + y^2 < 4$$ Hence, $$D$$ is the interior of the circle with centre $$(2,0)$$ and radius $$2$$. In polar coordinates, this is the region $$0<\rho<4\cos\theta$$ and $$-\pi/2<\theta<\pi/2$$. (See here for explanation).
Therefore, the integral is: \begin{align} \iint_D\sqrt{x^2+y^2}\,dx\,dy &= \int_{-\pi/2}^{\pi/2}\int_0^{4\cos\theta}\rho^2\,d\rho\,d\theta \\ &= \int_{-\pi/2}^{\pi/2} \left[\frac{\rho^3}{3}\right]_0^{4\cos\theta}\,d\theta \\ &= \frac{64}{3} \int_{-\pi/2}^{\pi/2}\cos^3\theta\,d\theta \\ &= \frac{64}{3} \int_{-\pi/2}^{\pi/2}(\cos\theta - \cos\theta\sin^2\theta)\,d\theta \\ &= \frac{128}{3} \left[\sin\theta-\frac{\sin^3\theta}{3}\right]_{0}^{\pi/2} \\ &= \frac{256}{9} \end{align}