15.2 Evaluation of triple integrals in spherical coordinates

Spherical Coordinates Setup

Triple Integral Setup and Change of Variables

Spherical coordinates use three variables to locate a point in 3D space:

• $\rho$ (rho): the distance from the origin to the point (always $\rho \geq 0$)
• $\theta$ (theta): the azimuthal angle in the $xy$-plane, measured from the positive $x$-axis
• $\phi$ (phi): the polar angle measured down from the positive $z$-axis (so $\phi = 0$ points straight up)

The conversion from rectangular to spherical coordinates is:

• $x = \rho \sin\phi \cos\theta$
• $y = \rho \sin\phi \sin\theta$
• $z = \rho \cos\phi$

And going the other direction: $\rho = \sqrt{x^2 + y^2 + z^2}$.

When you change variables in a triple integral, you need the Jacobian determinant, which for spherical coordinates works out to $\rho^2 \sin\phi$. This factor accounts for how volume elements stretch and compress in curved coordinates. A small "box" in spherical coordinates doesn't have uniform side lengths the way a box in rectangular coordinates does; $\rho^2 \sin\phi$ corrects for that distortion.

The general form of a triple integral in spherical coordinates is:

$\iiint_E f(\rho,\theta,\phi)\, \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi$

Never forget the $\rho^2 \sin\phi$ factor. Leaving it out is one of the most common mistakes on exams.

Limits of Integration and Order of Integration

Setting up the correct limits is where most of the work happens. Here's how to approach it:

1. Sketch the region (or at least visualize it). Identify whether it's a full sphere, a hemisphere, a cone, a wedge, etc.
2. Determine $\phi$ limits. These control how far the region extends from the positive $z$-axis. A full sphere runs from $\phi = 0$ to $\phi = \pi$. An upper hemisphere uses $0$ to $\frac{\pi}{2}$. A cone with half-angle $\alpha$ uses $0$ to $\alpha$.
3. Determine $\theta$ limits. These control the sweep around the $z$-axis. A full revolution is $0$ to $2\pi$. A half-space or wedge uses a smaller interval.
4. Determine $\rho$ limits. These describe how far from the origin the region extends, and they can depend on $\theta$ and $\phi$. For a sphere of radius $R$, $\rho$ runs from $0$ to $R$. For a region between two spheres, $\rho$ runs from the inner radius to the outer radius.

The standard integration order is $d\rho\, d\theta\, d\phi$ ($\rho$ innermost, $\phi$ outermost), but you can rearrange it if a different order makes the limits simpler.

For example, the integral over a full ball of radius 3 looks like:

$\int_0^{\pi} \int_0^{2\pi} \int_0^{3} \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi$

Notice that all three sets of limits are constants here, which makes the integral straightforward to evaluate by iterating one variable at a time.

Symmetry and Shells

Symmetry Considerations

Symmetry can cut your work in half (or more). Before computing anything, check for it.

Azimuthal symmetry (symmetry about the $z$-axis): If the region and the integrand are both unchanged as you rotate around the $z$-axis, the $\theta$-integral just contributes a constant factor. You can integrate $\theta$ over a smaller interval and multiply. For instance, a cone whose axis is the $z$-axis has this symmetry. If the full $\theta$-range is $0$ to $2\pi$, you could integrate from $0$ to $\pi$ and multiply by 2, or from $0$ to $\frac{\pi}{2}$ and multiply by 4.

Reflectional symmetry about the $xy$-plane: If the region is symmetric above and below the $xy$-plane and the integrand is even in $z$ (equivalently, even under $\phi \to \pi - \phi$), you can integrate $\phi$ from $0$ to $\frac{\pi}{2}$ and double the result. A sphere centered at the origin has this symmetry.

Odd-function shortcut: If the integrand is odd with respect to a plane of symmetry of the region, the integral is zero. This can save you from computing anything at all.

Spherical Shells

A spherical shell is a thin hollow sphere at radius $\rho$ with thickness $d\rho$. Its surface area is $4\pi\rho^2$, so its volume element is $4\pi\rho^2\, d\rho$.

When a solid has full spherical symmetry (the integrand depends only on $\rho$, and the region is a complete shell or ball), you can collapse the $\theta$ and $\phi$ integrals into that $4\pi\rho^2$ factor and reduce the problem to a single integral in $\rho$:

$V = \int_a^b 4\pi\rho^2 \, d\rho$

where $a$ and $b$ are the inner and outer radii.

Worked example: Find the volume of the region between spheres of radius 2 and radius 5.

$V = \int_2^5 4\pi\rho^2 \, d\rho = 4\pi \left[\frac{\rho^3}{3}\right]_2^5 = 4\pi\left(\frac{125}{3} - \frac{8}{3}\right) = 4\pi \cdot \frac{117}{3} = 156\pi$

This shell technique also works for integrals like mass or charge when the density depends only on $\rho$. Just replace the integrand accordingly: $\int_a^b f(\rho)\, 4\pi\rho^2\, d\rho$.