Using polar coordinates find the value of the double integral:
$$
\int_{0}^1\int_0^{\sqrt{2y-y^2}}\ dxdy.
$$
My attempt was as follows :
For the limits, $\theta$ will vary from $0$ to $\pi/2$ and $r$ will vary from $0$ to $2\sin(\theta)$. The integrand will be $r\ drd\theta$. The answer of this double integration will give me the area of right half of the circle, then multiplying this result by $1/2$ will give me the area of the right quarter of the circle as needed.
So this attempt is correct? My answer is $\pi/4$.
Best Answer
We are constrained by:
$x = \sqrt {2y - y^2},x = 0, y=1$
In polar coordinates we need to break this in two. one region bound by:
$r = 2\sin\theta\\ \theta \in [0, \frac \pi{4})$
$\int_0^{\frac {\pi}{4}}\int_0^{2\sin\theta} r\ dr\ d\theta\\ t - \sin\theta\cos\theta|_0^{\frac {\pi}{4}}\\ \frac \pi 4 - \frac 12$
one bound by: $r = \csc \theta\\ r \in [\frac {\pi}4, \frac {\pi}{2}]$
$+\int_{\frac {\pi}{4}}^{\frac {\pi}{2}}\int_0^{\csc\theta} r \ dr\ d\theta\\ -\frac 12 \cot\theta |_{\frac {\pi}{4}}^{\frac {\pi}{2}}\\ \frac 12$
or we could translate our polar coordinates.
$x = r\cos \theta\\ y = r\sin\theta + 1$
$\int_{-\frac {\pi}{2}}^{0}\int_0^{1} r\ dr\ d\theta\\ \frac {\pi}{4}$