How may I use this C loop to solve that integral?

Question

Let C be the curve of polar equation $r = 2cos^2(\theta)$ and D the area bounded by the loop C which is situated in the half-plane $x \ge 0$ region.

How may I calculate the D's area and use it to evaluate:

\begin{equation*} \int\int_{R} x^{11}y^{20} dA, \end{equation*}

where R is the region of the plane bounded by the curve C? Does C has a easy Cartesian equation in polynomial form??

As I'm stuck, so any tip will be helpful

Thanks in advance!

Answer 1

The region looks like this. Because $x\ge0$, we get that $-\pi/2\le\theta\le\pi/2$. For full credit calculate enough many points on it so that you can explain why the region has approximately this shape.

To calculate the integral in polar coordinates you thus need to remember that

1. $0\le r\le 2\cos^2\theta$,
2. $-\pi/2\le\theta\le\pi/2$,
3. $x=r\cos\theta$,
4. $y=r\sin\theta$,
5. $dA=r\,dr\,d\theta$.

I would use either a table of integrals or a CAS (or Wolfram Alpha) to calculate the resulting $\theta$-integral. The standard technique would also take me to the finish line, but I am feeling lazy and leaving that to you. The tables usually give trigonometric integrals over the interval $[0,\pi/2]$. As the exponent of $y$ is even, and $\cos^2\theta$ is also an even function we have a $(x,y)\leftrightarrow (x,-y)$ -symmetry, and may as well calculate the integral over $[0,\pi/2]$ and then multiply the result by two.

If you want to write this equation in cartesian coordinates, then first multiply it by $r^2$ to get $$ r^3=2r^2\cos^2\theta=2(r\cos\theta)^2. $$ Then observe that $r\cos\theta=x$ and $r=\sqrt{x^2+y^2}$. This gives you $$ (x^2+y^2)^{3/2}=2x^2. $$ I'm not sure how useful that is :-) You can make it polynomial by squaring it.