Double integral (polar coordinates) limits

Question

$\int \int_{R} (x^2+y^2)^{-2}dA$; R={$(x,y);x^2+y^2 \leq 2, x \geq 1$}

How to solve by polar coordinates? I couldn't push the boundaries of integration via polar coordinates

Answer 1

We need the equation for the line in polar coordinates.

$$x = r\cos\theta \to 1 = r\cos\theta \implies r = \sec\theta$$

Drawing a picture, we can see that the integral becomes

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{\sec\theta}^{\sqrt{2}} \frac{1}{r^3}drd\theta = \frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2\theta - \frac{1}{2}d\theta = \frac{1}{4}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos2\theta d\theta = \frac{1}{4}$$

Answer 2

The area of interest is a section of a disc with a chord and an arc as bounds. These curves intersect at $(1,1)$ and $(1,-1)$, so that the angle $(\theta)$ range is $(-\frac{\pi}{4}, \frac{\pi}{4})$ The upper limit for $r$ is $\sqrt{2}$. To get the lower limit for $r$, look at a radius at angle $\theta$. The part of the radius which intersects $x=1$ starts at $sec(\theta)$,the lower limit.

The integral then becomes $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\int_{sec(\theta)}^\sqrt{2}\frac{1}{r^3}drd\theta$.