Spherical Surface Patches and Their Parameters

Why Angular Parameters Fit a Sphere

A sphere is naturally described by directions from its center. Fixing the distance from the center while allowing the direction to vary traces its surface. Two angles specify that direction: one measures the angle down from the positive vertical axis, and the other measures rotation around that axis.

This gives a compact way to describe not just an entire sphere, but also a hemisphere, cap, band, or wedge-shaped portion. The parameter ranges determine which patch is covered. Differentiating the parametrization then gives tangent directions, while a cross product gives a normal direction.

Definition: The sphere of radius \(R>0\) centered at \(\mathbf{c}=(a,b,d)\) is the set of points whose distance from \(\mathbf{c}\) is \(R\). A spherical patch is a portion of this sphere described by restricting angular parameters to specified intervals.

Polar Angle and Longitude

Use \(\phi\) for the polar angle, measured from the positive \(z\)-axis, and \(\theta\) for the longitude, measured around that axis from the positive \(x\)-axis. For the sphere centered at the origin, the parametrization is

\mathbf{r}(\phi,\theta) = \bigl(R\sin\phi\cos\theta,\; R\sin\phi\sin\theta,\; R\cos\phi\bigr), \qquad 0\leq\phi\leq\pi,\quad 0\leq\theta\leq 2\pi.

For a sphere centered at \(\mathbf{c}\), add \(\mathbf{c}\) to the right-hand side. The distance from the center is \(R\), since

(R\sin\phi\cos\theta)^2 +(R\sin\phi\sin\theta)^2 +(R\cos\phi)^2 = R^2\bigl(\sin^2\phi+\cos^2\phi\bigr) = R^2.

The parameter \(\phi\) selects height: \(\phi=0\) is the north pole, \(\phi=\pi/2\) is the equator, and \(\phi=\pi\) is the south pole. Longitude selects a vertical half-plane through the \(z\)-axis. A full turn can be covered by \(0\leq\theta\leq2\pi\); the two endpoints describe the same meridian, so the parametrization repeats that seam.

Choosing Ranges for Common Patches

A parameter domain is a set of allowed angle pairs \((\phi,\theta)\). Restricting that domain selects the desired patch. For instance, the upper hemisphere uses \(0\leq\phi\leq\pi/2\), while the lower hemisphere uses \(\pi/2\leq\phi\leq\pi\). A spherical cap around the north pole can be specified by \(0\leq\phi\leq\alpha\), where \(0<\alpha<\pi\).

A longitude interval selects a wedge between meridians. For example, \(0\leq\theta\leq\pi/2\) selects the part between the positive \(x\)- and positive \(y\)-directions. Restricting both angles gives a band or sector. These descriptions concern the curved spherical surface, not the solid ball inside it.

Patch	One suitable parameter range
Upper hemisphere	\(0\leq\phi\leq\pi/2,\ 0\leq\theta\leq2\pi\)
Lower hemisphere	\(\pi/2\leq\phi\leq\pi,\ 0\leq\theta\leq2\pi\)
North-pole cap	\(0\leq\phi\leq\alpha,\ 0\leq\theta\leq2\pi\)
Angular wedge in upper hemisphere	\(0\leq\phi\leq\pi/2,\ \theta_0\leq\theta\leq\theta_1\)
Spherical band	\(\phi_0\leq\phi\leq\phi_1,\ 0\leq\theta\leq2\pi\)

Endpoints should match the geometric description. If a wedge spans less than a full turn, choose an interval \([\theta_0,\theta_1]\) describing that wedge without accidentally including other longitudes. The boundary values of the angle ranges trace the patch's boundary curves, with the understanding that coordinate edges at the poles or at the repeated longitude seam may collapse or coincide.

Latitude-Longitude Form

Some descriptions use latitude \(\lambda\), measured up from the equatorial plane, instead of polar angle. The two conventions are related by \(\lambda=\pi/2-\phi\). Thus \(-\pi/2\leq\lambda\leq\pi/2\), and the same sphere can be written

\mathbf{r}(\lambda,\theta) = \bigl(R\cos\lambda\cos\theta,\; R\cos\lambda\sin\theta,\; R\sin\lambda\bigr).

Positive latitude is north of the equator; negative latitude is south. The upper hemisphere has \(0\leq\lambda\leq\pi/2\), and the lower hemisphere has \(-\pi/2\leq\lambda\leq0\). A cap whose lowest latitude is \(\lambda_0\) uses \(\lambda_0\leq\lambda\leq\pi/2\), together with a full longitude range if the cap goes all the way around the axis.

Worked Example: An Upper-Hemisphere Wedge

Describe the part of the sphere of radius \(2\) centered at the origin that lies in the upper hemisphere and between longitudes \(0\) and \(\pi/2\). Use polar angle and longitude.

The upper hemisphere requires \(0\leq\phi\leq\pi/2\), and the wedge requires \(0\leq\theta\leq\pi/2\). Therefore the parametrization and domain are

\mathbf{r}(\phi,\theta) =(2\sin\phi\cos\theta,\;2\sin\phi\sin\theta,\;2\cos\phi), \qquad 0\leq\phi\leq\frac{\pi}{2},\quad 0\leq\theta\leq\frac{\pi}{2}.

For a point with \(\phi=\pi/3\) and \(\theta=\pi/4\), substitute the exact values \(\sin(\pi/3)=\sqrt{3}/2\), \(\cos(\pi/3)=1/2\), and \(\cos(\pi/4)=\sin(\pi/4)=\sqrt{2}/2\):

\mathbf{r}\left(\frac{\pi}{3},\frac{\pi}{4}\right) = \left( 2\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}, 2\cdot\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}, 2\cdot\frac{1}{2} \right) = \left(\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2},1\right).

Check that the point lies on the stated sphere and in the upper hemisphere. Its squared distance from the origin is \(6/4+6/4+1=4\), so its distance is \(2\); its \(z\)-coordinate is \(1>0\), as required.

Tangent Directions and the Cross Product

For the polar-angle parametrization, differentiate with respect to each parameter while holding the other fixed:

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

These are tangent vectors because varying one parameter while fixing the other traces a curve on the sphere. Their cross product gives a vector perpendicular to both tangent directions. Direct calculation yields

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

The vector in parentheses is the unit radial direction from the sphere's center to the point. Thus, away from the poles, this parameter order gives an outward-pointing normal. Reversing the cross-product order changes its sign, because \(\mathbf{r}_{\theta}\times\mathbf{r}_{\phi}=-(\mathbf{r}_{\phi}\times\mathbf{r}_{\theta})\).

Why the cross product is normal: The cross product of two vectors is perpendicular to each of them. Since \(\mathbf{r}_{\phi}\) and \(\mathbf{r}_{\theta}\) are tangent to the parameter curves, their cross product is perpendicular to the tangent directions and hence normal to the surface. In this case the displayed formula also identifies its direction: it is a positive multiple of the outward radial unit vector whenever \(0<\phi<\pi\).

The magnitude of the cross product is \(R^2\sin\phi\). It is positive for \(0<\phi<\pi\), so the two parameter directions are independent there. At either pole, \(\sin\phi=0\); every longitude describes the same pole, and the longitude tangent vector vanishes. This is a limitation of these angular coordinates at the poles, not a claim that the sphere itself has no tangent plane there.

Worked Example: Tangents and an Outward Normal

Find the two coordinate tangent vectors and an outward normal direction at \(\phi=\pi/3,\theta=\pi/4\) on the sphere of radius \(2\). The point was found above to be \((\sqrt{6}/2,\sqrt{6}/2,1)\).

Substitute \(R=2\), \(\cos\phi=1/2\), \(\sin\phi=\sqrt{3}/2\), and \(\cos\theta=\sin\theta=\sqrt{2}/2\) into the derivative formulas:

\mathbf{r}_{\phi} = \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},-\sqrt{3}\right), \qquad \mathbf{r}_{\theta} = \left(-\frac{\sqrt{6}}{2},\frac{\sqrt{6}}{2},0\right).

The cross product is

\mathbf{r}_{\phi}\times\mathbf{r}_{\theta} = \left(\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2},\sqrt{3}\right).

As a check, the radial vector at the point is \((\sqrt{6}/2,\sqrt{6}/2,1)\). The cross product equals \(\sqrt{3}\) times this radial vector, so it points outward and is perpendicular to the radius. Its magnitude is \(2\sqrt{3}\), which agrees with \(R^2\sin\phi=4(\sqrt{3}/2)=2\sqrt{3}\).

Computing with Latitude and Checking the Direction

Latitude coordinates give a different pair of parameter derivatives. For the centered sphere,

\mathbf{r}_{\lambda} = (-R\sin\lambda\cos\theta,\;-R\sin\lambda\sin\theta,\;R\cos\lambda), \qquad \mathbf{r}_{\theta} = (-R\cos\lambda\sin\theta,\;R\cos\lambda\cos\theta,\;0).

Here the outward order is \(\mathbf{r}_{\theta}\times\mathbf{r}_{\lambda}\), not \(\mathbf{r}_{\lambda}\times\mathbf{r}_{\theta}\). In fact, \(\mathbf{r}_{\lambda}\times\mathbf{r}_{\theta}\) points inward for \(-\pi/2<\lambda<\pi/2\). Checking the direction against the radial vector prevents an orientation sign error.

Worked Example: A Latitude Patch on a Shifted Sphere

Consider the sphere of radius \(2\) centered at \(\mathbf{c}=(1,-1,2)\), and a patch between latitudes \(\pi/6\) and \(\pi/2\), with longitude from \(0\) to \(\pi/2\). Parametrize it and find tangent and outward normal directions at \(\lambda=\pi/6,\theta=0\).

The requested parameter domain is \(\pi/6\leq\lambda\leq\pi/2\), \(0\leq\theta\leq\pi/2\). Add the center to the latitude parametrization:

\mathbf{r}(\lambda,\theta) = (1+2\cos\lambda\cos\theta,\;-1+2\cos\lambda\sin\theta,\;2+2\sin\lambda).

At the specified parameter values,

\mathbf{r}\left(\frac{\pi}{6},0\right) =(1+\sqrt{3},-1,3).

The tangent vectors are obtained by differentiating the nonconstant part; translating the sphere does not change its tangent directions. Substitution gives

\mathbf{r}_{\lambda}\left(\frac{\pi}{6},0\right)=(-1,0,\sqrt{3}), \qquad \mathbf{r}_{\theta}\left(\frac{\pi}{6},0\right)=(0,\sqrt{3},0).

Taking the outward order of the cross product gives

\mathbf{r}_{\theta}\times\mathbf{r}_{\lambda} = (0,\sqrt{3},0)\times(-1,0,\sqrt{3}) = (3,0,\sqrt{3}).

The vector from the center to the point is \((\sqrt{3},0,1)\), and \((3,0,\sqrt{3})=\sqrt{3}(\sqrt{3},0,1)\). Therefore the normal points outward. Its length is \(2\sqrt{3}\), matching \(R^2\cos\lambda=4(\sqrt{3}/2)=2\sqrt{3}\), the magnitude for this latitude-coordinate cross product.

A Common Angle-Convention Mistake

A frequent error is to use the latitude formula but put \(\sin\lambda\) in the horizontal coordinates. For latitude measured from the equator, the horizontal radius is \(R\cos\lambda\), and the vertical coordinate is \(R\sin\lambda\). The incorrect expression \( (R\sin\lambda\cos\theta,R\sin\lambda\sin\theta,R\cos\lambda) \) actually uses the polar-angle pattern, not the latitude convention.

For example, at latitude \(\lambda=0\), a point on the equator should have height zero and horizontal distance \(R\) from the axis. The incorrect expression gives horizontal distance \(0\) and height \(R\), placing the point at the north pole instead. The correction is to use \(R\cos\lambda\) horizontally and \(R\sin\lambda\) vertically—or to convert consistently using \(\phi=\pi/2-\lambda\).

Using the Parametrization Reliably

When describing a spherical patch, first identify whether the angles are polar angle or latitude. Then select intervals that match the requested cap, hemisphere, band, or longitude wedge. After differentiating, state the parameter order used for the cross product and check whether the resulting normal points inward or outward by comparing it with the vector from the center to the surface point.

Fix the sphere.
Specify its center and positive radius, then write the matching angular parametrization.

Restrict the angles.
Translate the geometric description into intervals for latitude or polar angle and longitude.

Differentiate and inspect.
Compute the parameter tangent vectors, take their cross product in the stated order, and check its direction against the outward radial vector.

These checks are useful because a correct formula with an incorrect angle convention or parameter range describes the wrong patch. Likewise, a valid normal vector can point inward if the cross-product order is reversed. Keeping the geometry, parameter ranges, and orientation together makes the parametrization interpretable and verifiable.

Check Your Understanding

Answer these questions using the parameter conventions in this tutorial.

What polar-angle and longitude ranges describe the lower hemisphere?
How is latitude \(\lambda\) related to polar angle \(\phi\), and what is the latitude parametrization of a centered sphere?
For the polar-angle parametrization, why does \(\mathbf{r}_{\phi}\times\mathbf{r}_{\theta}\) point outward away from the poles?
What happens to the longitude tangent vector at either pole, and what does that imply about these coordinates there?
For a latitude parametrization, which order of the two parameter tangent vectors gives an outward normal?