Double integral to polar coordinates, bounds

Question

I want to find $\int_0^2 \int_ {\sqrt{2x-x^2}}^{\sqrt{4-x^2}} \sqrt{4-x^2-y^2}dydx$.

My idea is to do a transformation to polar coordinates;

$\iint \sqrt{4-r^2} rdrd\theta$, but I'm unsure about the bounds. Plotting the bounds in $f(x,y)$ I realize I'm dealing with the area between a smaller circle centered at (1, 0) and r=1 and a bigger circle centered at (0,0) with r = 2. (Under the ball $\sqrt{4-x^2-y^2}$ Any ideas? Thank you

Answer 1

HINT

Note that the integral is on the region $R$ between $C_1$ and $C_2$ and $y>0$ with

• $C_1: x^2+y^2=4\,$ is a circle centered at the origin with radius $2$
• $C_2: (x-1)^2+y^2=1$ is a circle centered at $(1,0)$ with radius $1$

$C_1$ and $C_2$ Plot

thus

$$\iint_R \sqrt{4-r^2} rdrd\theta=\int_{0}^{\frac{\pi}2} d\theta\int_0^2 \sqrt{4-r^2} rdr-\int_{0}^{\frac{\pi}2} d\theta\int_0^{2\cos \theta} \sqrt{4-r^2} rdr$$

indeed

• for $C_1$ we are integrating on a quarter of circle

• for $C_2$ we are integrating on half circle and we have that the circle is defined by

$$x^2+y^2=2x\implies r^2=2r\cos \theta \implies r=2\cos \theta$$