Evaluate an integral in polar form

Question

This problem is driving me crazy :

$$\int_{0}^{1}\int_0^{\sqrt{2y-y^2}}{dx.dy}$$

Someone please solve this problem in details by sketching the required boundary and how did he calculate it. The final answer is $\pi/4$ Thanks in advance

Edit : I am still learning polar coordinates , so i have failed to convert this double integral boundary to polar form !

Answer 1

Hint:

Use the fact that $$x=r\cos\theta\\y=r\sin\theta\\r^2=x^2+y^2$$

You have $\int_0^1\int_0^{\sqrt{2y-y^2}}dxdy$

From above we can say that $y=0,y=1$ and $x=0$, $x=\sqrt{2y-y^2}\implies x^2+(y-1)^2=1$, so it is a circle of radius $1$

Now can you continue from here?

Answer 2

note: $$\int_0^1\int_0^{\sqrt{2y-y^2}}dxdy=\int_0^1\int_0^{\sqrt{1-(1-y)^2}}dxdy$$ so if $$x=\sqrt{1-(1-y)^2}$$ $$x^2=1-(1-y)^2$$ $$x^2+(y-1)^2=1$$ so it is a circle with centre $(0,1)$ and a radius of $1$ $$I=\int_0^{2\pi}\int_0^1 rdrd\theta=\pi$$

Answer 3

Try the following: $\int_0^1 \int_{0}^{\sqrt{2y-y^2}} dx dy = \int_0^1 x|_{0}^{\sqrt{2y-y^2}} =\int_0^1 \sqrt{2y-y^2} dy = \int_0^1 \sqrt{1-(y-1)^2} dy$

Now Substitute $u:=y-1$, which leads you to $\int_{-1}^{0}\sqrt{1-u^2} du$

This is a standard integral, which is computed multiple times here in the forum (and on the whole internet), so we get: $\int_{-1}^{0}\sqrt{1-u^2} du = \frac{1}{2}(u \sqrt{1-u^2} + arcsin(u))|_{-1}^{0}= \frac{1}{2}arcsin(-1) = \frac{\pi}{4}$