Spherical Setup and Evaluation Workshop

From a Solid to an Integral

A spherical triple integral is only as reliable as its description of the solid. A useful setup starts with the geometry: identify the radial extent, the range of polar angles, and the range of azimuthal angles. Then transform the integrand, include the volume factor, and evaluate in a convenient order.

In this workshop, the same process is applied to a ball, a spherical cap, a wedge-shaped shell, and a region bounded by an off-center sphere. These examples illustrate a central habit: determine each bound from the solid, rather than guessing bounds from the appearance of the integrand.

Coordinate convention: Spherical coordinates are \((\rho,\phi,\theta)\), where \(\rho\) is distance from the origin, \(\phi\) is measured down from the positive \(z\)-axis, and \(\theta\) is the azimuthal angle in the \(xy\)-plane. Thus \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), \(z=\rho\cos\phi\), and \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\).

For a full turn, \(0\leq\theta\leq2\pi\); for a full sweep from the north pole to the south pole, \(0\leq\phi\leq\pi\). The radial coordinate satisfies \(\rho\geq0\). A particular solid usually restricts one or more of these ranges.

Why the Volume Factor Appears

The spherical volume factor can be understood from the lengths of three small coordinate changes. A change \(d\rho\) moves a distance \(d\rho\) radially. At fixed \(\rho\), changing \(\phi\) by \(d\phi\) moves approximately \(\rho\,d\phi\). At fixed \(\rho\) and \(\phi\), changing \(\theta\) by \(d\theta\) moves approximately \(\rho\sin\phi\,d\theta\), because the corresponding circle has radius \(\rho\sin\phi\).

These directions are perpendicular. The small volume is therefore approximately the product of the three side lengths:

dV = (d\rho)(\rho\,d\phi)(\rho\sin\phi\,d\theta) = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

This explains why transforming an integrand alone is not enough: the factor \(\rho^2\sin\phi\) must also be included. The formula applies when the spherical coordinates describe the region with the chosen ranges; boundaries on the axis or at the origin do not affect the integral over a three-dimensional solid.

A Setup Routine

For an integral \(\iiint_E f(x,y,z)\,dV\), use this sequence:

Describe the solid.
Identify its bounding spheres, planes, cones, and angular restrictions. A sketch or a cross-section helps distinguish which surface is the inner or outer boundary.

Choose the angular ranges.
Read polar and azimuthal restrictions from the cone, hemisphere, or wedge. Use the full angular range only when the solid actually includes it.

Set the radial limits.
At fixed angles, identify where a ray enters and leaves the solid. Solve the surface equations for \(\rho\), and check which intersection is nearer the origin.

Transform and evaluate.
Rewrite \(f\) using spherical coordinates, multiply by \(\rho^2\sin\phi\), and integrate over the chosen bounds. Keep the order of the differentials matched to those bounds.

Worked Examples

Worked Example: A Radial Integrand over a Ball

Evaluate \(\iiint_E (x^2+y^2)\,dV\), where \(E\) is the ball \(\rho\leq2\) centered at the origin.

The ball has the full angular ranges and \(0\leq\rho\leq2\). Since \(x^2+y^2=\rho^2\sin^2\phi\), the transformed integrand, including the volume factor, is \(\rho^4\sin^3\phi\). Therefore

\begin{aligned} \iiint_E (x^2+y^2)\,dV &=\int_0^{2\pi}\int_0^\pi\int_0^2 \rho^4\sin^3\phi\,d\rho\,d\phi\,d\theta\\ &=\left[\frac{\rho^5}{5}\right]_0^2 \left(\int_0^\pi\sin^3\phi\,d\phi\right) \left[\theta\right]_0^{2\pi}\\ &=\frac{32}{5}\cdot\frac{4}{3}\cdot2\pi =\frac{256\pi}{15}. \end{aligned}

For the middle factor, write \(\sin^3\phi=\sin\phi(1-\cos^2\phi)\); substituting \(u=\cos\phi\) gives \(\int_0^\pi\sin^3\phi\,d\phi=\int_{-1}^1(1-u^2)\,du=4/3\). As a check on the radial factor, differentiating \(\rho^5/5\) gives \(\rho^4\), and evaluating at \(2\) and \(0\) gives \(32/5\). The result is positive, as the integrand is nonnegative throughout the ball.

Worked Example: The Volume of a Spherical Cap

Find the volume inside the sphere \(\rho\leq2\) and above the plane \(z=1\).

The plane becomes \(\rho\cos\phi=1\), so a point on a ray meets the plane at \(\rho=\sec\phi\). The region exists where this intersection is at or below the sphere: \(\sec\phi\leq2\), or \(0\leq\phi\leq\pi/3\). The azimuthal angle makes a full turn. Thus \( \sec\phi\leq\rho\leq2\), and

\begin{aligned} V &=\int_0^{2\pi}\int_0^{\pi/3}\int_{\sec\phi}^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\frac{2\pi}{3}\int_0^{\pi/3} (8-\sec^3\phi)\sin\phi\,d\phi\\ &=\frac{2\pi}{3}\left( 8\left[-\cos\phi\right]_0^{\pi/3} -\left[\frac{\sec^2\phi-1}{2}\right]_0^{\pi/3} \right)\\ &=\frac{2\pi}{3}\left(8\cdot\frac12-\frac{4-1}{2}\right) =\frac{5\pi}{3}. \end{aligned}

To check the angular antiderivative, differentiating \((\sec^2\phi-1)/2\) gives \(\sec^2\phi\tan\phi=\sin\phi/\cos^3\phi\), which is \(\sec^3\phi\sin\phi\). As a geometric check, this cap has height \(1\) in a sphere of radius \(2\); the standard cap-volume expression \(\pi h^2(R-h/3)\) gives \(\pi(1)^2(2-1/3)=5\pi/3\), agreeing with the integral.

Worked Example: A Wedge of a Spherical Shell

Find the volume between the spheres \(\rho=1\) and \(\rho=2\), with \(\pi/3\leq\phi\leq\pi/2\) and \(0\leq\theta\leq\pi/3\).

The radial bounds describe the shell, the polar bounds select a band toward the equator, and the azimuthal bounds select a wedge. Integrating the volume factor gives

\begin{aligned} V &=\int_0^{\pi/3}\int_{\pi/3}^{\pi/2}\int_1^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\left[\frac{\rho^3}{3}\right]_1^2 \left[-\cos\phi\right]_{\pi/3}^{\pi/2} \left[\theta\right]_0^{\pi/3}\\ &=\frac{8-1}{3}\left(0-\left(-\frac12\right)\right)\frac{\pi}{3} =\frac{7}{3}\cdot\frac12\cdot\frac{\pi}{3} =\frac{7\pi}{18}. \end{aligned}

The antiderivative \(\rho^3/3\) differentiates to \(\rho^2\), and its endpoint difference is \(8/3-1/3=7/3\). The angular factors are \(1/2\) and \(\pi/3\). Their product is \(7\pi/18\). Each factor is positive, as a volume must be.

Worked Example: A Region Inside an Intersecting Sphere

Find the volume inside the sphere \(x^2+y^2+(z-2)^2\leq4\).

Substitute \(x^2+y^2=\rho^2\sin^2\phi\) and \(z=\rho\cos\phi\) into the sphere equation. After expanding and cancelling \(\rho^2\cos^2\phi\), its boundary is \(\rho^2-4\rho\cos\phi=0\). The nonzero intersection is \(\rho=4\cos\phi\). Because \(\rho\geq0\), points in the sphere have \(0\leq\phi\leq\pi/2\); the azimuthal angle is unrestricted. Hence \(0\leq\rho\leq4\cos\phi\), and

\begin{aligned} V &=\int_0^{2\pi}\int_0^{\pi/2}\int_0^{4\cos\phi} \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\frac{2\pi}{3}\int_0^{\pi/2}(4\cos\phi)^3\sin\phi\,d\phi\\ &=\frac{128\pi}{3}\int_0^{\pi/2}\cos^3\phi\sin\phi\,d\phi\\ &=\frac{128\pi}{3}\left[-\frac{\cos^4\phi}{4}\right]_0^{\pi/2} =\frac{128\pi}{3}\cdot\frac14 =\frac{32\pi}{3}. \end{aligned}

Differentiating \(-\cos^4\phi/4\) gives \(\cos^3\phi\sin\phi\), and the endpoint difference is \(0-(-1/4)=1/4\). The sphere has radius \(2\), so its volume is \(4\pi(2^3)/3=32\pi/3\), which confirms the result. The radial upper bound would be incorrect for \(\phi>\pi/2\), because it would then be negative.

A Common Setup Error

A frequent mistake is to use the correct surface intersection but reverse the radial limits. In the cap example, the ray meets the plane at \(\rho=\sec\phi\) and the sphere at \(\rho=2\). For \(0\leq\phi\leq\pi/3\), \(\sec\phi\leq2\), so the plane is the inner boundary and the sphere is the outer boundary.

\text{Incorrect: }\quad \int_0^{\pi/3}\int_0^{2\pi}\int_2^{\sec\phi} \rho^2\sin\phi\,d\rho\,d\theta\,d\phi

This reversed integral is negative: its radial factor is \((\sec^3\phi-8)/3\leq0\), even though it is intended to represent a volume. The correction is to put the nearer boundary first and the farther one second:

\text{Correct: }\quad \int_0^{\pi/3}\int_0^{2\pi}\int_{\sec\phi}^{2} \rho^2\sin\phi\,d\rho\,d\theta\,d\phi.

Checking one ray, such as \(\phi=0\), makes the order clear: it runs from \(\rho=1\) at the plane to \(\rho=2\) at the sphere. Also check that the angular range is possible and that the integrand includes the full factor \(\rho^2\sin\phi\).

Why Careful Setup Matters

These examples share a method, but not identical bounds. A ball supplies a constant radial limit; a cap supplies a plane-dependent radial limit and a restricted polar angle; a wedge restricts azimuth; and an off-center sphere produces a radial bound that varies with \(\phi\). Recognizing which feature of the solid controls each bound is more important than memorizing any one integral.

Before evaluating, verify that each bound describes the intended solid, that lower limits do not exceed upper limits, and that the integrand has been transformed before multiplying by the volume factor. After evaluating, check signs and endpoint arithmetic; when practical, compare with a geometric volume or a symmetry check.

Check Your Understanding

For each question, identify the geometric feature that determines the bounds before attempting any integration.

For the region inside \(\rho=3\) and above \(z=3/2\), what are the ranges of \(\rho\), \(\phi\), and \(\theta\)?
Why must the volume element include \(\rho^2\sin\phi\), even when the original integrand is the constant \(1\)?
For the shell \(1\leq\rho\leq2\), \(\pi/4\leq\phi\leq\pi/2\), \(0\leq\theta\leq\pi/2\), write the volume integral in spherical coordinates.
When solving a spherical surface equation for a radial bound, why should you check the sign of the resulting expression over the proposed angular range?
In a region between two radial surfaces, how can checking a single ray help you determine the correct order of the radial limits?