Finding the volume between two concentrical hemispheres

Question

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.

Answer 1

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 $2\pi$.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 $\it\mbox{half the volume between two spheres}$: $$ \ds{{4\pi b^{3}/3 - 4\pi a^{3}/3 \over 2} = {2\pi \over 3}\pars{b^{3} - a^{3}}} $$