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[Find the volume of] The solid bounded below by the sphere $\rho=2\cos\phi$ and above by the cone $z=\sqrt{x^2+y^2}$.

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I'm a bit skeptical on how to decide bounds for the integral in spherical coordinates. I tried to calculate the bounds $\sqrt{x^2+y^2}$ into spherical which gave me $\pi/4$ so i did from $0$ to $\pi/4$ but when i looked for answers in the book it was $\pi/4$ to $\pi/2$. Please try to answer in a bit simple terms if possible as i'm only in 9th grade.

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  • $\begingroup$ @Debalanced, the text book is indeed correct. If you look closely, it clearly indicates the cone inside the hemisphere $\endgroup$ Commented May 9 at 13:52
  • $\begingroup$ @RandomMathEnthusiast Yea, I realized that a bit ago. I didn't look close enough. $\endgroup$ Commented May 9 at 13:53
  • $\begingroup$ Does the question require us to calculate the volume between the cone and the hemisphere? $\endgroup$ Commented May 9 at 13:55
  • $\begingroup$ Just looking at it on Desmos, it appears that $0\leq \phi \leq \frac{\pi}{4}$ gives you the top half of the sphere, while $\frac{\pi}{4}\leq \phi \leq \frac{\pi}{2}$ as shown in the book gives you the bottom half, which is correct. $\endgroup$ Commented May 9 at 13:56
  • $\begingroup$ @RandomMathEnthusiast Yes, it requires volume between the cone and hemisphere $\endgroup$ Commented May 9 at 13:58

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