Solving a double integration $ \int_{0}^1\int_0^{\sqrt{2y-y^2}}\ dxdy$ using polar coordinates

Question

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 $\pi/4$.

Answer 1

Yes that's correct indeed the equation

$$x=\sqrt{2y-y^2}\implies x^2 +(y-1)^2 = 1$$

represents a circle centered at $(0,1)$ with radius equal to $1$ and the given integral corresponds to the area of a quarter that is ineed equal to $\frac{\pi}4$

Answer 2

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.

$x = r\cos \theta\\ y = r\sin\theta + 1$

$\int_{-\frac {\pi}{2}}^{0}\int_0^{1} r\ dr\ d\theta\\ \frac {\pi}{4}$

Answer 3

Your answer is correct but you actually did not solve the required integral.

What you have done with your limits of integration is smart but wrong, because you changed your limits to find twice the area which makes your integration simple.

Note that they require the integral which represent the quarter circle not the semi circle.

You have to do two integrals one for $0\le \theta \le \pi /4$ and one for $\pi /4\le \theta \le \pi /2$ and your $r$ has different limits due the straight line $y=1$ for the second part.