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Surface Area and Scalar Integrals · Tutorial 857 of 999

Surface Integrals on Spherical Patches

Learn to set spherical parameter bounds and use the area element to calculate areas and scalar surface integrals on spherical patches.

Advanced 10 min read

What You'll Learn

  • Parametrize a sphere using the polar angle and azimuthal angle
  • Derive the spherical area element from the cross product of the tangent vectors
  • Set bounds for spherical caps, bands, and sectors
  • Calculate areas using spherical parameter integrals
  • Evaluate scalar surface integrals over spherical patches
  • Recognize common angle and area-element errors

Why Spherical Patches Are Useful

A sphere has the same radius in every direction, so spherical coordinates often make its surface integrals simpler than a description in Cartesian coordinates. A spherical cap, for example, is described by a single bound on the angle measured from the positive \(z\)-axis. A band is described by two such angle bounds, while a sector can also restrict the azimuthal angle.

The essential ingredient is the spherical area element. It is not simply the product of the angle differentials: the surface area represented by a small parameter rectangle depends on its position on the sphere. Once that factor is included, areas and scalar surface integrals reduce to ordinary double integrals.

Spherical Parametrization and Angle Convention

Let \(R>0\) be the radius of a sphere centered at the origin. We use \(\phi\) for the polar angle measured down from the positive \(z\)-axis, and \(\theta\) for the azimuthal angle in the \(xy\)-plane, measured from the positive \(x\)-axis. Thus \(0\leq\phi\leq\pi\), and a full revolution uses \(0\leq\theta\leq2\pi\).

Definition: A spherical parametrization of the sphere of radius \(R\) is $$ \mathbf{r}(\phi,\theta) = \bigl(R\sin\phi\cos\theta,\ R\sin\phi\sin\theta,\ R\cos\phi\bigr). $$ Here \(\phi\) is the polar angle and \(\theta\) is the azimuthal angle.

The convention matters. In this parametrization, \(\phi=0\) is the north pole and \(\phi=\pi/2\) is the equator. The \(z\)-coordinate is \(R\cos\phi\), while the distance from the \(z\)-axis is \(R\sin\phi\).

Deriving the Spherical Area Element

For a parametrized surface, the area element is the magnitude of the cross product of its parameter tangent vectors, multiplied by the parameter differentials. Differentiate \(\mathbf{r}\) with respect to each parameter:

$$ \mathbf{r}_{\phi} = \bigl(R\cos\phi\cos\theta,\ R\cos\phi\sin\theta,\ -R\sin\phi\bigr), \qquad \mathbf{r}_{\theta} = \bigl(-R\sin\phi\sin\theta,\ R\sin\phi\cos\theta,\ 0\bigr). $$

The cross product is

$$ \mathbf{r}_{\phi}\times\mathbf{r}_{\theta} = \bigl(R^2\sin^2\phi\cos\theta,\ R^2\sin^2\phi\sin\theta,\ R^2\sin\phi\cos\phi\bigr). $$

Its magnitude follows by squaring the components and using \(\sin^2\theta+\cos^2\theta=1\):

$$ \begin{aligned} \left\lVert\mathbf{r}_{\phi}\times\mathbf{r}_{\theta}\right\rVert &=R^2\sqrt{\sin^4\phi(\cos^2\theta+\sin^2\theta)+\sin^2\phi\cos^2\phi}\\ &=R^2\sqrt{\sin^2\phi(\sin^2\phi+\cos^2\phi)}\\ &=R^2\sin\phi. \end{aligned} $$

The last equality uses \(0\leq\phi\leq\pi\), so \(\sin\phi\geq0\). Therefore, the surface area element is \(dS=R^2\sin\phi\,d\phi\,d\theta\). The factor \(\sin\phi\) expresses how the width of a line of constant \(\phi\) shrinks toward either pole.

Result: On a sphere of radius \(R\), the spherical area element is $$ dS=R^2\sin\phi\,d\phi\,d\theta. $$ For a scalar function \(f\) defined on the sphere, its scalar surface integral over a patch \(S\) is $$ \iint_S f\,dS = \int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} f\bigl(\mathbf{r}(\phi,\theta)\bigr)R^2\sin\phi\,d\phi\,d\theta, $$ when the patch is represented by the rectangular parameter bounds \(\phi_1\leq\phi\leq\phi_2\) and \(\theta_1\leq\theta\leq\theta_2\).

The scalar function must be evaluated at the parametrized point. For example, if \(f(x,y,z)=z\), then \(f(\mathbf{r}(\phi,\theta))=R\cos\phi\). A single value at a pole may have multiple azimuthal parameter descriptions, but the poles have zero surface area, so this duplication does not alter the integral.

Bounds for Caps, Bands, and Sectors

A north polar cap with angular radius \(\alpha\), where \(0\leq\alpha\leq\pi\), uses \(0\leq\phi\leq\alpha\) and a full azimuthal turn. A band between polar angles \(\alpha\) and \(\beta\), with \(0\leq\alpha\leq\beta\leq\pi\), uses \(\alpha\leq\phi\leq\beta\) and a full turn. A sector restricts the azimuth as well as the polar angle; for example, \(\theta_1\leq\theta\leq\theta_2\). A single sector description should not cover the same surface points more than once, apart from boundary duplications.

Integrating \(dS\) gives the corresponding areas. The cap formula follows from integrating \(\sin\phi\) from \(0\) to \(\alpha\); the band formula uses the same antiderivative with different bounds.

$$ \begin{aligned} A_{\text{cap}} &=\int_0^{2\pi}\int_0^\alpha R^2\sin\phi\,d\phi\,d\theta =2\pi R^2(1-\cos\alpha),\\ A_{\text{band}} &=\int_0^{2\pi}\int_\alpha^\beta R^2\sin\phi\,d\phi\,d\theta =2\pi R^2(\cos\alpha-\cos\beta),\\ A_{\text{sector}} &=\int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2} R^2\sin\phi\,d\phi\,d\theta =R^2(\theta_2-\theta_1)(\cos\phi_1-\cos\phi_2). \end{aligned} $$

These formulas assume the stated bounds and a sphere centered at the origin. Angles are in radians. The sector expression applies when \(\theta_2\geq\theta_1\) and the interval describes the intended azimuthal sweep without unintended repeated coverage.

Worked Examples

Worked Example: Area of a North Polar Cap

Find the area of the cap on a sphere of radius \(3\) cut off by the polar angle \(\phi=\pi/3\).

The cap has \(0\leq\phi\leq\pi/3\) and \(0\leq\theta\leq2\pi\). Using the spherical area element gives

$$ \begin{aligned} A &=\int_0^{2\pi}\int_0^{\pi/3}9\sin\phi\,d\phi\,d\theta\\ &=\int_0^{2\pi}9\left[-\cos\phi\right]_0^{\pi/3}\,d\theta\\ &=\int_0^{2\pi}9\left(1-\frac12\right)\,d\theta\\ &=\int_0^{2\pi}\frac92\,d\theta =9\pi. \end{aligned} $$

Check with the cap formula: \(2\pi(3^2)(1-\cos(\pi/3))=18\pi(1-\frac12)=9\pi\), agreeing with the direct integral.

Worked Example: Scalar Integral of \(z^2\) over a Spherical Band

Evaluate \(\iint_S z^2\,dS\) on the unit sphere over the band \(\pi/3\leq\phi\leq2\pi/3\), with a full azimuthal turn.

On the unit sphere, \(z=\cos\phi\), so \(z^2=\cos^2\phi\), and \(dS=\sin\phi\,d\phi\,d\theta\). Thus

$$ \begin{aligned} \iint_S z^2\,dS &=\int_0^{2\pi}\int_{\pi/3}^{2\pi/3}\cos^2\phi\sin\phi\,d\phi\,d\theta\\ &=\int_0^{2\pi}\left[-\frac{\cos^3\phi}{3}\right]_{\pi/3}^{2\pi/3}\,d\theta\\ &=\int_0^{2\pi}\frac{1}{3}\left(\frac18-\left(-\frac18\right)\right)\,d\theta\\ &=\int_0^{2\pi}\frac{1}{12}\,d\theta =\frac{\pi}{6}. \end{aligned} $$

For the check, the antiderivative differentiates as \(\frac{d}{d\phi}(-\cos^3\phi/3)=\cos^2\phi\sin\phi\). The bounds give \(1/12\) per unit of azimuth, and multiplying by the full azimuthal width \(2\pi\) yields \(\pi/6\).

Worked Example: Integral of \(x\) over a Spherical Sector

Evaluate \(\iint_S x\,dS\) on the sphere of radius \(2\), with \(0\leq\phi\leq\pi/2\) and \(0\leq\theta\leq\pi/3\).

The parametrization gives \(x=2\sin\phi\cos\theta\), and the area element is \(4\sin\phi\,d\phi\,d\theta\). Therefore

$$ \begin{aligned} \iint_S x\,dS &=\int_0^{\pi/3}\int_0^{\pi/2} 8\sin^2\phi\cos\theta\,d\phi\,d\theta\\ &=8\left(\int_0^{\pi/3}\cos\theta\,d\theta\right) \left(\int_0^{\pi/2}\sin^2\phi\,d\phi\right)\\ &=8\left[\sin\theta\right]_0^{\pi/3} \left[\frac{\phi}{2}-\frac{\sin(2\phi)}{4}\right]_0^{\pi/2}\\ &=8\left(\frac{\sqrt3}{2}\right)\left(\frac{\pi}{4}\right) =\sqrt3\pi. \end{aligned} $$

Check: the azimuthal factor is \(\sin(\pi/3)-\sin(0)=\sqrt3/2\), and the polar factor is \(\pi/4\). Their product times \(8\) is \(8(\sqrt3/2)(\pi/4)=\sqrt3\pi\), as calculated.

A Common Error: Omitting the Sine Factor

A frequent mistake is to use \(R^2\,d\phi\,d\theta\) for the area element. For the cap in the first example, that incorrect choice would give

$$ \int_0^{2\pi}\int_0^{\pi/3}9\,d\phi\,d\theta =9\left(\frac{\pi}{3}\right)(2\pi) =6\pi^2. $$

This is wrong because equal increments in \(\theta\) do not represent the same surface distance at every polar angle. Near a pole, a line of constant \(\phi\) has small circumference; its radius is \(R\sin\phi\), not \(R\). The correct cross-product magnitude is \(R^2\sin\phi\), so the cap area is \(9\pi\), not \(6\pi^2\). A useful check is that the factor \(\sin\phi\) makes the area contribution shrink to zero at either pole.

Using the Method Reliably

1
Identify the angular region.
Translate the geometric description into bounds for \(\phi\) and \(\theta\), using \(\phi\) as the angle from the positive \(z\)-axis.
2
Substitute into the scalar function.
Replace \(x,y,z\) by \(R\sin\phi\cos\theta\), \(R\sin\phi\sin\theta\), and \(R\cos\phi\), respectively.
3
Include the area factor.
Multiply the substituted function by \(R^2\sin\phi\), or use that factor alone when calculating area.
4
Integrate over the parameter bounds.
Evaluate the resulting double integral and check the bounds, trigonometric signs, and any symmetry suggested by the patch.

Keep the radius factor visible: \(R^2\) belongs to the area element, and additional powers of \(R\) may appear when the integrand itself contains \(x\), \(y\), or \(z\). Also distinguish a polar angle from a latitude. In this convention, latitude-like descriptions must be converted to \(\phi\); for example, a point at the equator has \(\phi=\pi/2\), not \(\phi=0\).

Takeaway: On a sphere of radius \(R\), use \(dS=R^2\sin\phi\,d\phi\,d\theta\). Correct bounds encode whether the patch is a cap, band, or sector; substituting the scalar function and integrating then gives the surface integral.

Check Your Understanding

For each question, identify the parameter bounds before attempting the integration.

  1. What are the spherical bounds for a north polar cap ending at \(\phi=\pi/4\), and what area element should be used?
  2. Find the area of the band on a sphere of radius \(2\) given by \(\pi/3\leq\phi\leq\pi/2\), with a full azimuthal turn.
  3. On the unit sphere, how would you express \(z\) in terms of the spherical parameters before integrating \(z\,dS\)?
  4. For a sector with \(\phi_1\leq\phi\leq\phi_2\) and \(\theta_1\leq\theta\leq\theta_2\), which factor in the area element accounts for the shrinking azimuthal circles near the poles?
  5. Why does the incorrect area element \(R^2\,d\phi\,d\theta\) fail near a pole?

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