Nonradial Integrals over Spherical Caps

Why Spherical Caps Require Careful Bounds

A spherical cap is a portion of a ball cut off by a plane. Spherical coordinates fit the curved boundary naturally, but a plane such as \(z=h\) does not usually give a constant radial lower bound. The plane condition must be translated into spherical coordinates, and its effect on the allowed angles must also be included.

For a radial integrand, the angular dependence may be limited to the volume element and the region's bounds. For a nonradial integrand, expressions such as \(x\), \(x^2\), or \(z\) also change with the angles. A reliable setup accounts for all of these factors before integrating.

The Cap and Its Spherical Description

Use spherical coordinates \((\rho,\phi,\theta)\), with \(\rho\) the distance from the origin, \(\phi\) the angle measured down from the positive \(z\)-axis, and \(\theta\) the azimuthal angle in the \(xy\)-plane. The coordinate relations and volume element are

x=\rho\sin\phi\cos\theta,\qquad y=\rho\sin\phi\sin\theta,\qquad z=\rho\cos\phi,\qquad dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

Consider the upper cap \(D\) inside the ball of radius \(R\), above the plane \(z=h\), where \(0<h<R\):

D=\{(x,y,z):x^2+y^2+z^2\le R^2,\ z\ge h\}.

The sphere gives \(0\le \rho\le R\). The plane condition becomes \(\rho\cos\phi\ge h\). Since \(h>0\), points in the cap have \(\cos\phi>0\), and the plane condition can be written \(\rho\ge h/\cos\phi\). This radial lower bound is possible only when \(h/\cos\phi\le R\), or \(\cos\phi\ge h/R\). Therefore \(0\le\phi\le\arccos(h/R)\). The cap goes all the way around the \(z\)-axis, so \(0\le\theta\le2\pi\).

Bounds for an upper cap: For \(0<h<R\), the region inside the ball \(\rho\le R\) and above \(z=h\) is described by \[ 0\le\theta\le2\pi,\qquad 0\le\phi\le\arccos(h/R),\qquad \frac{h}{\cos\phi}\le\rho\le R. \] The radial lower limit depends on \(\phi\); the polar-angle upper limit ensures that the lower limit does not exceed the sphere.

From a Cartesian Integrand to an Integral

If the integrand is \(f(x,y,z)\), first substitute the spherical-coordinate expressions for \(x\), \(y\), and \(z\). Then multiply by the spherical volume factor \(\rho^2\sin\phi\). For this cap the resulting setup is

\iiint_D f(x,y,z)\,dV = \int_0^{2\pi} \int_0^{\arccos(h/R)} \int_{h/\cos\phi}^{R} f(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi) \,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

The order shown is convenient because the radial bounds are explicit once \(\phi\) is fixed. It is important not to replace the lower radial bound by \(h\): the equation of the plane is \(\rho\cos\phi=h\), so its radial intersection is \(h/\cos\phi\).

Why the Bounds and Integrand Work

The coordinate relations describe each point of the cap, while the inequalities determine exactly which coordinate values correspond to points in the cap. In particular, \(\rho\le R\) is the sphere condition, and \(\rho\cos\phi\ge h\) is the plane condition. Solving the latter for \(\rho\) gives its lower bound. Requiring this lower bound to lie within the ball determines the range of \(\phi\). The full azimuthal range accounts for every direction around the \(z\)-axis.

At each point, the integrand is evaluated at that point's Cartesian coordinates, and the factor \(\rho^2\sin\phi\) supplies the volume scaling for spherical coordinates. Integrating over the stated bounds therefore adds the values of \(f\) throughout the cap. This setup relies on the standard spherical-coordinate volume element and on using bounds that describe the region without omission or overlap, apart from boundary surfaces that do not affect the integral.

Translate the integrand.
Replace \(x,y,z\) by their spherical-coordinate expressions.

Translate the region.
Use the sphere for the outer radial bound and the plane for the inner radial bound.

Restrict the polar angle.
Require the plane's radial intersection to lie inside the sphere.

Include the volume element.
Multiply the transformed function by \(\rho^2\sin\phi\), then integrate.

Worked Integrals over the Same Cap

In the examples below, use the cap inside the sphere of radius \(2\) and above the plane \(z=1\). Its bounds are \(0\le\theta\le2\pi\), \(0\le\phi\le\pi/3\), and \(1/\cos\phi\le\rho\le2\).

Worked Example: Integrating the Height \(z\)

Evaluate \(\iiint_D z\,dV\). Since \(z=\rho\cos\phi\), multiplication by the volume element gives \(\rho^3\cos\phi\sin\phi\). Thus

\begin{aligned} \iiint_D z\,dV &=\int_0^{2\pi}\int_0^{\pi/3}\int_{1/\cos\phi}^{2} \rho^3\cos\phi\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\frac{\pi}{2}\int_0^{\pi/3} \cos\phi\sin\phi\left(16-\frac{1}{\cos^4\phi}\right)d\phi\\ &=\frac{\pi}{2}\left( 16\int_0^{\pi/3}\cos\phi\sin\phi\,d\phi -\int_0^{\pi/3}\frac{\sin\phi}{\cos^3\phi}\,d\phi \right). \end{aligned}

The first antiderivative is \(\sin^2\phi/2\), so its definite integral is \(3/8\). The second antiderivative is \(1/(2\cos^2\phi)\), so its definite integral is \(2-1/2=3/2\). Therefore

\iiint_D z\,dV =\frac{\pi}{2}\left(16\cdot\frac38-\frac32\right) =\frac{\pi}{2}\left(6-\frac32\right) =\frac{9\pi}{4}.

For verification, differentiating \(\sin^2\phi/2\) gives \(\sin\phi\cos\phi\), and differentiating \(1/(2\cos^2\phi)\) gives \(\sin\phi/\cos^3\phi\). Substitution of the bounds gives the displayed integrals and hence the stated value.

Worked Example: Integrating \(x^2\)

Now evaluate \(\iiint_D x^2\,dV\). Here \(x^2=\rho^2\sin^2\phi\cos^2\theta\), so the full integrand, including the volume element, is \(\rho^4\sin^3\phi\cos^2\theta\). Since \(\int_0^{2\pi}\cos^2\theta\,d\theta=\pi\),

\begin{aligned} \iiint_D x^2\,dV &=\frac{\pi}{5}\int_0^{\pi/3} \sin^3\phi\left(32-\frac{1}{\cos^5\phi}\right)d\phi\\ &=\frac{\pi}{5}\left( 32\int_0^{\pi/3}\sin^3\phi\,d\phi -\int_0^{\pi/3}\frac{\sin^3\phi}{\cos^5\phi}\,d\phi \right). \end{aligned}

For the first integral, an antiderivative is \(-\cos\phi+\cos^3\phi/3\). Its values at \(\pi/3\) and \(0\) differ by \(5/24\). For the second, let \(u=\cos\phi\); then the integral becomes \(\int_{1/2}^{1}(u^{-5}-u^{-3})\,du\). An antiderivative is \(-u^{-4}/4+u^{-2}/2\); its values at \(1\) and \(1/2\) differ by \(1/4-(-2)=9/4\). Therefore

\iiint_D x^2\,dV =\frac{\pi}{5}\left(32\cdot\frac{5}{24}-\frac94\right) =\frac{\pi}{5}\left(\frac{80-27}{12}\right) =\frac{53\pi}{60}.

Differentiating the stated antiderivatives recovers the two angular integrands; the substitution \(u=\cos\phi\) has \(du=-\sin\phi\,d\phi\), which gives the transformed second integral with the displayed limits. Both contributions are positive overall after subtraction, as expected for the integral of \(x^2\).

Worked Example: An Integral That Vanishes by Azimuthal Symmetry

Evaluate \(\iiint_D x\,dV\). Since \(x=\rho\sin\phi\cos\theta\), the integrand with the volume element is \(\rho^3\sin^2\phi\cos\theta\). The bounds separate the azimuthal integral:

\begin{aligned} \iiint_D x\,dV &=\int_0^{2\pi}\cos\theta\,d\theta \int_0^{\pi/3}\sin^2\phi\,d\phi \int_{1/\cos\phi}^{2}\rho^3\,d\rho\\ &=\left[\sin\theta\right]_0^{2\pi} \int_0^{\pi/3}\sin^2\phi \left[ \frac{\rho^4}{4}\right]_{1/\cos\phi}^{2} \,d\phi\\ &=0. \end{aligned}

The radial and polar integrals are finite, while the azimuthal factor is \(\sin(2\pi)-\sin(0)=0\). Thus the result is zero. The cap is unchanged by reflection across the \(yz\)-plane, whereas \(x\) changes sign under that reflection; this provides an independent symmetry check.

A Common Bounds Error

A frequent mistake is to use \(\rho\ge h\) as the lower bound for the cap above \(z=h\). For example, at \(\phi=\pi/3\), the plane condition requires \(\rho\cos(\pi/3)\ge1\), hence \(\rho\ge2\), not \(\rho\ge1\). The correct lower bound is \(1/\cos\phi\). At this angle it meets the sphere at \(\rho=2\), which is why \(\phi=\pi/3\) is the largest allowed polar angle.

Another error is to transform the function but omit the volume factor. For \(x^2\), the spherical expression for the function is \(\rho^2\sin^2\phi\cos^2\theta\), but the full integrand is \(\rho^4\sin^3\phi\cos^2\theta\) after multiplication by \(\rho^2\sin\phi\). Both the transformed function and the volume element are necessary.

When This Setup Applies

The bounds above describe the cap inside a centered sphere and above a horizontal plane \(z=h\), with \(0<h<R\). If the plane is below the origin, or if the cutting plane has a different orientation, these same limits need not describe the region; the angle ranges and radial bounds must be derived from the actual inequalities. For the stated cap, coordinate-dependent functions are handled by substitution, and symmetries in \(\theta\) can simplify or eliminate parts of the integral.

Key takeaway: For a nonradial integral over an upper spherical cap, transform every coordinate in the function, use \(\rho\ge h/\cos\phi\) for the plane, restrict \(\phi\) so this bound stays inside the sphere, and include \(\rho^2\sin\phi\).

Check Your Understanding

Use the cap inside a ball of radius \(R\) above the plane \(z=h\), where \(0<h<R\).

Write the spherical-coordinate bounds for this cap and explain why the upper bound for \(\phi\) is \(\arccos(h/R)\).
What is the complete spherical integrand, including the volume element, when the Cartesian integrand is \(y\)?
For the cap of radius \(2\) above \(z=1\), why does the radial lower bound equal \(2\) at \(\phi=\pi/3\)?
Explain why the integral of \(x\) over this cap is zero.
Identify the error in using \(\rho\ge h\) as the radial lower bound for the plane \(z=h\).