# Double integral to polar coordinates, bounds

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

• You get the bounds simpl by putting the value or taking the limit for each coordinate. For example say $r=\sqrt{x^2+y^2}$ and say that $x,y \in[0,2]$. Then you look at the maximum and minimum of $r$ as a function of $x$ and $y$. The solution will be an interval because r(x,y) is continuous. So in this case $r\in [0,2]$. You do this for all coordinates. However you must make sure that the integrand is correct. For example for $r=1, \theta=0$ the integrand is 0. -- Perhaps it is easier to use linearity of the integral and make two integrals. I.e. the difference of two balls. – ty. Feb 14 '18 at 20:17

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

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

• Thank you! Could you elaborate on you how you arrive at $-\pi/2$ and $\pi/2$ aswell as $0$ to $2cos(\theta)$? – novo Feb 14 '18 at 20:24
• @novo I've just added some comment to explain, make a sketch of the region to better visualize how it works. – user Feb 14 '18 at 20:26
• @novo sorry I just realize that we are only integrating on the upper part, I fix – user Feb 14 '18 at 20:30
• @Aladdin I'm subtracting the half red form one quarter blue. – user Oct 11 '19 at 14:35
• @Aladdin Look at sketch in my answer. As you can see for a point on the shaded area we need r between $2\cos \theta$ and $2$. – user Oct 11 '19 at 15:45