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The solid bounded below by the hemisphere $\rho=1,$ $ 0\leq z$ and above by the cardoid of revolution $\rho=1+\cos\phi$ I am new to the triple integral in spherical coordinates. I know that limits of $\rho$ will be $1\leq\rho\leq 1+\cos\phi$, the limits on $\theta$ will be $0\leq\theta\leq2\pi$. So the integral will be $$\int_{0}^{2\pi}\int_{?}^{?}\int_{1}^{1+\cos\phi}(\rho)^2\sin\phi d\rho d\phi d\theta$$ But how to find the limits of $\phi$ in this case, I don't know. Can anyone help! Thanks in advance!

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Notice that for the cardioid of revolution, $\rho\ge1$ for $0\le\phi\le\pi/2$ and $\rho<1$ for $\pi/2<\phi\le\pi$. It intersects the hemisphere only in the $xy$ plane. Therefore, $0\le\phi\le\pi/2$.

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  • $\begingroup$ Sir the limits on the $\rho$ and $\theta$ I have written are correct? $\endgroup$ – Noor Aslam Jan 8 at 12:47
  • $\begingroup$ @NoorAslam They are correct $\endgroup$ – Shubham Johri Jan 8 at 18:01

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