Converting an integral into polar form

Question

$$\int_{0}^{1} \int_{0}^{\sqrt{2 - x^2}} \frac{x}{\sqrt{x^2 +y^2}} \ dy\ dx$$ How to convert this into polar form as there would be 2 parts? What is the use of limits x=0 to x=1 as i am finding no use of them in the solution

Answer 1

The domain of integration is given by: $$ D=\{(x,y)\in\mathbb{R}^2: 0\leq x\leq 1, 0\leq y\leq \sqrt{2-x^2}\},$$ but since: $$ D = T\cup C, $$ $$ T = \{(x,y)\in\mathbb{R}^2: 0\leq x\leq 1,0\leq y\leq x\},$$ $$ C = \{(x,y)\in\mathbb{R}^2: 0\leq x\leq 1,x\leq y\leq \sqrt{2-x^2}\}$$ where $T$ is a triangle and $C$ is a circular sector with radius $\sqrt{2}$ and amplitude $\frac{\pi}{4}$, we have:

$$ I = \int_T \frac{x}{\sqrt{x^2+y^2}}\,d\mu + \int_{C}\frac{x}{\sqrt{x^2+y^2}}\,d\mu $$ or: $$I = \int_0^{\pi/4} \int_0^{\sec \theta} \cos \theta\, r\, dr\, d\theta + \int_{\pi/4}^{\pi/2} \int_0^{\sqrt{2}} \cos \theta\, r \, dr\, d\theta,$$ or: $$ I = \int_{0}^{\pi/4}\frac{d\theta}{2\cos\theta}+\int_{\pi/4}^{\pi/2}\cos\theta\,d\theta =\frac{1}{2}\log(1+\sqrt{2})+1-\frac{1}{\sqrt{2}}.$$

Answer 2

Sketching the region of integration, you find that it consists of a sector with bounds $0 \le r \le \sqrt{2}$, $\pi/4 \le \theta \le \pi/2$ and a triangle with bounds $0 \le \theta \le \pi/4$, $0 \le r \le \sec\theta$. So your integral becomes

$$\int_0^{\pi/4} \int_0^{\sec \theta} \cos \theta\, r\, dr\, d\theta + \int_{\pi/4}^{\pi/2} \int_0^{\sqrt{2}} \cos \theta\, r \, dr\, d\theta.$$