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

  • $\begingroup$ 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. $\endgroup$ – ty. Feb 14 '18 at 20:17


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


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


  • 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

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

enter image description here

  • $\begingroup$ Thank you! Could you elaborate on you how you arrive at $-\pi/2$ and $\pi/2$ aswell as $0$ to $2cos(\theta)$? $\endgroup$ – novo Feb 14 '18 at 20:24
  • $\begingroup$ @novo I've just added some comment to explain, make a sketch of the region to better visualize how it works. $\endgroup$ – user Feb 14 '18 at 20:26
  • $\begingroup$ @novo sorry I just realize that we are only integrating on the upper part, I fix $\endgroup$ – user Feb 14 '18 at 20:30
  • 1
    $\begingroup$ @Aladdin I'm subtracting the half red form one quarter blue. $\endgroup$ – user Oct 11 '19 at 14:35
  • 1
    $\begingroup$ @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 $2$. $\endgroup$ – user Oct 11 '19 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.