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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

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  • $\begingroup$ 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
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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

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

enter image description here

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  • $\begingroup$ 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
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    $\begingroup$ @Aladdin I'm subtracting the half red form one quarter blue. $\endgroup$ – user Oct 11 at 14:35
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    $\begingroup$ @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 at 15:45

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