Integrals in Cylindrical and Spherical Coordinates

These integrals are computed just like ordinary triple integrals except we multiply the integrand by $r$ (in cylindrical coordinates) or $\rho^2\sin(\phi)$ (in spherical coordinates). See the next section for a more general way of doing this. It is sometimes useful to use plots to decide how to represent the region of integration.

Let's do an integral in cylindrical coordinates.

\begin{displaymath}
\int_{-2}^2 \int_0^{\sqrt{4-x^2}}\int_0^{3}yz \mathrm{d}z \mathrm{d}y \mathrm{d}x.
\end{displaymath}

Note that $r$ goes from 0 to 2 and $\theta$ goes from 0 to $\pi$.


\begin{maximasession}
f(x,y,z) := y*z;
[x,y,z] : [r*cos(theta), r*sin(theta), z]...
...te(f(x,y,z)*r, z,0,3), r,0,2), theta,0,%pi); \\
\o3. 24 \\
\end{maximasession}

Now let's do an integral in spherical coordinates.


\begin{displaymath}
\int_{-1}^1 \int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}} xz \mathrm{d}z \mathrm{d}y \mathrm{d}x.
\end{displaymath}

Note that $\rho$ goes from 0 to 1, $\theta$ goes from 0 to $\pi$, and $\phi$ goes from 0 to $\pi/2$.


\begin{maximasession}
kill(f,x,y,z);
f(x,y,z) := x*z;
[x,y,z] : [rho*sin(phi)*co...
...,theta,0,%pi), phi,0,%pi/2); \\
\o7. {{\pi^2}\over{40}} \\
\end{maximasession}





G. Jay Kerns 2009-12-01