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I have a sphere with the equation $x^2 + y^2 +z^2 = b^2$ with a another sphere $x^2 + y^2 +z^2 = a^2$ located inside of it so that $0<a<b$. I am trying to find the volume between these two spheres above the $xy$ plane, so as I understand they would both be hemispheres. I am having trouble when finding how to set up this integral. To for example get the limits for $z$ for both spheres I was going to set $z^2 = a^2 - x^2 - y^2$ equal to $z^2 = b^2 - x^2 - y^2$, but then I end up with $a^2 = b^2$. I imagine this is wrong or I should be switching to spherical coordinates.

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    $\begingroup$ While you CAN set up limits of integration, it's almost certainly easier to take the volume of the larger and subtract the volume of the smaller. Alternatively, switch to polar coordinates and do the integral there, where the limits of integration are much more natural. $\endgroup$ – John Hughes Nov 25 '13 at 16:01
  • $\begingroup$ @John I prefer to set up the limits as later on I will have to find the center of mass. $\endgroup$ – 0x41414141 Nov 25 '13 at 16:10
  • $\begingroup$ The title does not correspond to the question. The question has to do with the volume integral definition and not with the cm. $\endgroup$ – John Alexiou Nov 25 '13 at 17:19
  • $\begingroup$ @ja72 I changed it. $\endgroup$ – 0x41414141 Nov 25 '13 at 17:25
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    $\begingroup$ This question appears to be off-topic because it is trivial. $\endgroup$ – Felix Marin Nov 25 '13 at 17:27
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Setting the equations equal to each other is finding where the spheres intersect, which doesn't work (since they don't!).

Switch to spherical coordinates -- both rectangular and cylindrical coordinates are going to require you to break up the integral in ways you don't want.

In spherical coordinates $\rho$ is going from $a$ to $b$ and $\phi$ from $0$ to $\pi/2$ (since you're only above the $xy$ plane) while $\theta$ goes from $0$ to $2\pi$.

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    $\begingroup$ The volume could be found like this? $$\int_0^{2{\pi}} \int_0^{\frac{\pi}{2}} \int_0^b b dp d{\phi} d{\theta} - \int_0^{2{\pi}} \int_0^{\frac{\pi}{2}} \int_0^a a dp d{\phi} d{\theta}$$ $\endgroup$ – 0x41414141 Nov 25 '13 at 16:57
  • $\begingroup$ Unfortunately, no. The cost of the nice limits of integration is that you have to multiply by the volume element $\rho^2 \sin \phi$. Also, there's no need for an $a$ or $b$ in the integrand. $\endgroup$ – hunter Nov 25 '13 at 17:33
  • $\begingroup$ Yes I thought of that and came up with $$\int_0^{2{\pi}} \int_0^{\frac{\pi}{2}} \int_a^b p^2\sin{\phi} dp d{\phi} d{\theta}$$ $\endgroup$ – 0x41414141 Nov 25 '13 at 17:51
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ As $\tt\mbox{@bh3244}$ already explained in a comment: \begin{align} V & = \int_{0}^{2\pi}\dd\phi\int_{0}^{\pi/2}\dd\theta\,\sin\pars{\theta}\int_{a}^{b} \dd r\,r^{2}\,,\quad \left\lbrace% \begin{array}{rcl} \int_{0}^{2\pi}\dd\phi & = & 2\pi \\[1mm] \int_{0}^{\pi/2}\dd\theta\,\sin\pars{\theta} & = & 1 \\[1mm] \int_{a}^{b}\dd r\,r^{2} & = & {1 \over 3}\pars{b^{3} - a^{3}} \end{array}\right. \end{align} $$\color{#0000ff}{\large% V = {2\pi \over 3}\pars{b^{3} - a^{3}}} $$ The "above the $xy$ plane" condition is equivalent to $z > 0$ which in spherical coordinates means $\pars{~0 < \theta < \pi/2~}$. In addition, the result is just $\it\mbox{half the volume between two spheres}$: $$ \ds{{4\pi b^{3}/3 - 4\pi a^{3}/3 \over 2} = {2\pi \over 3}\pars{b^{3} - a^{3}}} $$

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