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I am trying to find the volume above $\ z = 0 $ under $\ z = x^2 + y^2 $ and inside the cylinder $\ x^2+y^2 = 2x $ using double integral.

the intersection between the cylnder and the paraboloid is the plane $\ z = 2x $ and the project on $\ xy $ plane is just circle at $\ 1,0 $ . so suppose $\ y $ values run from $\ -1 $ to $\ 1 $ and $\ x $ from $\ y=\sqrt{2x-x^2} $ to $\ y= - \sqrt{2x-x^2} $ I get

$$\ \int_{-1}^1 \int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} 2x \ dy \ dx $$

so doing such integral is obviously not the purpsoe of the excercise. So I tried to convert to polar form which should be

$$\ \int\int_R 2r \cos \theta \ r \ dr \ d\theta $$

but here I get stuck, I mean i can imagine the circle at $\ 1,0 $ but how do I know exactly what are the angles of $\theta $ and I know the smallest $\ r $ is the $\ 0 $ and the largest $\ 2 $ but not sure if this is what I should use?

bm1125
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    You could move the origin to the center of the cylinder. Or, equivalently, move the paraboloid and the cylinder so that the center of the cylinder is at the origin. Then do polar. Don't know if it's easier; the bounds become trivial, but the new equation for the paraboloid isn't entirely optimal. But in my experience, simple bound expressions are worth a lot. – Arthur May 09 '19 at 22:00
  • The new equation of the paraboloid , the cylinder and the intersection plane (respectively) will be $\ (x+1)^2+y^2=z, x^2+y^2 = 1 , \ z = 2x+2 $ then I calculated the integral $\ \int_0^{2\pi} \int_0^1 (2r \cos \theta +2) \ r \ dr \ d\theta $ which still wrong (get the wrong solution according to cheat sheet). – bm1125 May 09 '19 at 22:16

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Note that the base of your cylinder is the inside of the circle $$\ x^2+y^2 = 2x$$ which in polar coordinates is described by $$r=2\cos(\theta)$$

Also your upper surface $$z=x^2+y^2$$ is simply $$z=r^2$$.

Thus in polar coordinates you find the volume by $$v=2\int _{0}^{\pi/2}\int_{0}^{2\cos(\theta)}(r^2)rdrd\theta$$

Mohammad Riazi-Kermani
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  • Thank you sir! Just two more questions. 1. why did you run $\ \theta $ from $\ 0 $ to $\ \pi/2 $ (while the actual intersection seems like a whole circle) ? 2. I guess its related but why did you have to multiply by $\ 2 $ the whole integral? – bm1125 May 10 '19 at 07:49
  • I did not want to get negative values of $r$ so I restricted $\theta$ and used symmetry to get the whole volume by multiplying by $2$. – Mohammad Riazi-Kermani May 10 '19 at 10:34