Question

Let D= x^2+y^2 ≤ 4x Using polar coordinates, What is the integral: ∬y^2/ (x^2+y^2)dxdy?

Answer 1

In polar coordinates, the inequality changes to

`x^2+y^2\le4x\implies r^2\le4r\cos\theta\implies r\le4\cos\theta`

which is a circle of radius 2 and centered at (2, 0). The set D is then

`D=\left\{(r,\theta)\mid0\le r\le4\cos\theta\land0\le\theta\le\pi\right\}`

The integral is then

`\displaystyle\iint_D(y^2)/(x^2+y^2)\,\mathrm dx\,\mathrm dy=\int_0^\pi\int_0^(4\cos\theta)(r^2\sin^2\theta)/(r^2)r\,\mathrm dr\,\mathrm d\theta`

`=\displaystyle\int_0^\pi\int_0^(4\cos\theta)r\sin^2\theta\,\mathrm dr\,\mathrm d\theta`

`=\displaystyle\frac12\int_0^\pi((4\cos\theta)^2-0^2)\sin^2\theta\,\mathrm d\theta`

`=\displaystyle8\int_0^\pi\cos^2\theta\sin^2\theta\,\mathrm d\theta`

There are several ways to compute the remaining integral; one would be to invoke the double-angle formula,

`\sin(2\theta)=2\sin\theta\cos\theta`

so that the integral is

`=\displaystyle8\int_0^\pi\frac{\sin^2(2\theta)}4\,\mathrm d\theta`

`=\displaystyle2\int_0^\pi\sin^2(2\theta)\,\mathrm d\theta`

Then invoke another double-angle formula,

`\sin^2\theta=\frac{1-\cos(2\theta)}2`

to change the integral to

`=\displaystyle\int_0^\pi1-\cos(4\theta)\,\mathrm d\theta`

`=(\pi-\cos(4\pi))-(0-\cos0)=\boxed{\pi}`