Integrating $\int_1^2 \int_0^ \sqrt{2x-x^2} \frac{1}{((x^2+y^2)^2} dydx $ in polar coordinates

Solution 1:

I think you will find it easiest if you write the integral in the form $$\int_?^?\int_?^? \frac{1}{r^3}\,dr\,d\theta\ ,$$ that is, with $\theta$ on the "outside" integral. Looking at your diagram, the minimum $\theta$ value in the region of integration is $\theta=0$, which occurs along the $x$ axis. The maximum $\theta$ value occurs at the point $(1,1)$, giving $\theta=\pi/4$. So we have $$\int_0^{\pi/4}\int_?^? \frac{1}{r^3}\,dr\,d\theta\ .$$ For any fixed $\theta$ value, the values of $r$ in the region go from the vertical line $x=1$ to the semicircle. At $x=1$ we have $r=\sec\theta$ as you have noted. The easiest way to get the maximum $r$ value is to draw a line from the origin at angle $\theta$ until it hits the semicircle, and a line from there to $(2,0)$. Then you can see in the diagram a right-angled triangle (because the angle in a semicircle is a right angle) with hypotenuse $2$, and so we get $r_{\rm max}=2\cos\theta$. So the integral is $$\int_0^{\pi/4}\int_{\sec\theta}^{2\cos\theta} \frac{1}{r^3}\,dr\,d\theta\ .$$

For sketching the graph, $$y=\sqrt{2x-x^2}\quad\Rightarrow\quad x^2-2x+y^2=0 \quad\Rightarrow\quad (x-1)^2+y^2=1\ ,$$ and not forgetting that $y\ge0$.