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Let $T$ be the region within the sphere $x^2+y^2+z^2=4$ and within the cylinder within the sphere $x^2+(y-1)^2=1$. Use polar coordinates to calculate the volume of $T$.

What I am thinking is we have

$$z=\pm \sqrt{4-r^2}$$ after converting $(x,y)\rightarrow (r\cos \theta, r \sin \theta)$, setting up the integral we have

$$\int_0^{2\pi } \int_0^r \int_{-z}^zrdzdrd\theta =\int_0^{2\pi}\int_0^{2\cos \theta}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}rdzdrd\theta $$

My question is did I set up the integral or did I completely blow it? Also I'm unsure whether $r=2 \cos\theta$ is the right limit for $r$, so if someone could confirm or correct this with a small explanation, that would be most helpful. Thanks!

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    $\begingroup$ $x+(y-1)^2=1$ is not a cylinder. $\endgroup$ – Emilio Novati Feb 12 '20 at 13:49
  • $\begingroup$ I'm sorry, it's $x^2+(y-1)^2 = 1$. I made a typo. $\endgroup$ – DerpyMcDerp Feb 12 '20 at 13:57
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Your limit $r=2\cos \theta$ is wrong but without consequence on the final result given the symmetry of the figure. The cylinder intersects the $y$ axis at $$y=r\sin \theta=2$$ so the correct limit is $y=2\sin \theta$ that we reach for $\theta=\frac{\pi}{2}$. Your limit is for a cylinder with center on the $X$ axis ( that gives the same volume).

Anyway, using the symmetry of the region you can use the limits: $$ 0<\theta<\frac{\pi}{2}\quad 0<r<2\sin \theta \quad 0<z<\sqrt{4-r^2} $$ and express the volume as $$ V=4\int_0^{\frac{\pi}{2}}\int_0^{2\sin \theta}\int_0^{\sqrt{4-r^2}} rdzdrd\theta $$

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  • $\begingroup$ I'm sorry for being awfully slow, but why is it exactly we integrate from $-\pi/2$ to $\pi/2$ instead of $0$ to $2 \pi$? $\endgroup$ – DerpyMcDerp Feb 12 '20 at 15:23
  • $\begingroup$ The fact that $r$ is not negative, implies that we have some limits for the values of $\theta$. if we want use $0 \le \theta<2\pi$ than we must use the convention that when $r<0$ we invert the direction on the line given by the angle $\theta$ and in doing so we ''redraw'' the same curve. $\endgroup$ – Emilio Novati Feb 12 '20 at 16:31

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