Sphere-Cone Intersections

Reading a Sphere and a Cone Together

A cone with its vertex at the origin is especially simple in spherical coordinates: its side is described by a fixed angle. A sphere centered at the origin is equally simple, because its boundary has a fixed distance from the origin. When both surfaces bound a region, the main task is to identify which coordinates each surface constrains and where the surfaces meet.

A sphere whose center is not the origin requires more care. Its radial boundary can depend on the angle, so the cone’s angular constraint and the sphere’s radial constraint must be compared. The intersection tells us where the two boundary surfaces meet and provides a useful check on the limits.

Coordinate convention: We use \(\rho\) for distance from the origin, \(\phi\) for the angle measured down from the positive \(z\)-axis, and \(\theta\) for the azimuthal angle in the \(xy\)-plane. Thus \(0\leq\phi\leq\pi\) and \(0\leq\theta<2\pi\), and

x=\rho\sin\phi\cos\theta,\qquad y=\rho\sin\phi\sin\theta,\qquad z=\rho\cos\phi.

Which Coordinate Does Each Surface Constrain?

A cone with vertex at the origin and axis along the positive \(z\)-axis can be written as \(\phi=\alpha\), where \(0<\alpha<\pi/2\). Its interior, the cone around the positive \(z\)-axis, corresponds to \(0\leq\phi\leq\alpha\). The cone itself fixes an angle; it does not, by itself, set an upper bound on \(\rho\).

A sphere centered at the origin with radius \(R\) has equation \(\rho=R\). Its interior gives \(0\leq\rho\leq R\), independent of angle. Together, such a sphere and cone constrain different coordinates: the cone restricts \(\phi\), and the sphere restricts \(\rho\).

For comparison, consider a sphere centered at \((0,0,a)\) with radius \(a\), where \(a>0\). Its equation is \(x^2+y^2+(z-a)^2=a^2\). In spherical coordinates, this becomes a radial boundary that depends on \(\phi\). The cone and sphere can then meet at a circle whose location must be found using both constraints.

Definition: A sphere-cone intersection is the set of points satisfying the equations of both surfaces. For a region bounded by those surfaces, use the cone to identify the allowed angular range and the sphere to identify the allowed radial range. Check the intersection to see where those boundary descriptions agree.

Deriving the Radial Boundary for a Shifted Sphere

For the sphere centered at \((0,0,a)\) with radius \(a\), substitute \(x^2+y^2=\rho^2\sin^2\phi\) and \(z=\rho\cos\phi\) into its Cartesian equation. The algebra shows why the radial limit is not generally constant.

\begin{aligned} x^2+y^2+(z-a)^2&=a^2,\\ \rho^2\sin^2\phi+(\rho\cos\phi-a)^2&=a^2,\\ \rho^2\sin^2\phi+\rho^2\cos^2\phi-2a\rho\cos\phi+a^2&=a^2,\\ \rho^2-2a\rho\cos\phi&=0,\\ \rho(\rho-2a\cos\phi)&=0. \end{aligned}

The factor \(\rho=0\) is the sphere’s point at the origin. Away from that point, the spherical boundary is \(\rho=2a\cos\phi\). To find the sphere’s interior, use the inequality \(x^2+y^2+(z-a)^2\leq a^2\). The same substitution gives \(\rho(\rho-2a\cos\phi)\leq0\). Since \(\rho\geq0\), nonzero points in the sphere must satisfy \(0<\rho\leq2a\cos\phi\). In particular, \(\cos\phi\) must be nonnegative, so this sphere lies in the angular range \(0\leq\phi\leq\pi/2\), apart from its boundary point at the origin.

Now intersect this sphere with the cone \(\phi=\alpha\). Substituting the cone angle into the sphere’s radial boundary gives the distance of the intersection circle from the origin. Its height and its distance from the \(z\)-axis follow from the spherical-coordinate formulas.

\rho=2a\cos\alpha,\qquad z=2a\cos^2\alpha,\qquad \sqrt{x^2+y^2}=2a\sin\alpha\cos\alpha.

A Reliable Method for Choosing the Limits

When the sphere and cone form the boundary of a region, first decide which side of each surface is included. Then translate those choices into angular and radial inequalities. The following steps summarize the reasoning for the common case of a sphere centered on the positive \(z\)-axis.

Translate the cone.
Write the cone as \(\phi=\alpha\), then decide whether the region is inside it, \(0\leq\phi\leq\alpha\), or on the other side.

Translate the sphere.
Use the sphere equation to find its radial boundary. For a sphere centered at \((0,0,a)\) with radius \(a\), the interior has \(0\leq\rho\leq2a\cos\phi\) wherever \(\cos\phi\geq0\).

Find the meeting location.
Set \(\phi=\alpha\) in the sphere’s radial boundary. This identifies the circle at which the cone and sphere intersect.

Assemble and check the limits.
Let \(\theta\) run through the azimuths included by the region, and verify that the radial and angular bounds describe the intended side of both surfaces.

This process is not a rule to choose the same angular interval in every problem. A cone separates directions into different angular ranges, and a region may lie inside or outside it. The sphere determines how far each allowed ray continues. The selected interval must match the actual solid, not merely the equations of its surfaces.

Worked Examples

Worked Example: A Cone Inside a Sphere Centered at the Origin

Find spherical-coordinate limits for the region inside the sphere \(x^2+y^2+z^2=16\) and inside the cone \(\phi=\pi/3\), with the cone opening around the positive \(z\)-axis.

Because the sphere is centered at the origin, its equation becomes \(\rho^2=16\). Since \(\rho\geq0\), its boundary is \(\rho=4\), and the interior has \(0\leq\rho\leq4\). Being inside the cone gives \(0\leq\phi\leq\pi/3\). The region includes all azimuthal directions, so \(0\leq\theta<2\pi\). Therefore, one set of limits is

0\leq\rho\leq4,\qquad 0\leq\phi\leq\frac{\pi}{3},\qquad 0\leq\theta<2\pi.

The cone and sphere meet where \(\rho=4\) and \(\phi=\pi/3\). On that circle, the cylindrical radius is \(4\sin(\pi/3)=2\sqrt{3}\), and the height is \(4\cos(\pi/3)=2\). Checking a point on the circle gives \(x^2+y^2+z^2=(2\sqrt{3})^2+2^2=12+4=16\); also, its angle from the positive \(z\)-axis satisfies \(\tan\phi=(2\sqrt{3})/2=\sqrt{3}\), so \(\phi=\pi/3\). Both surface equations hold.

Worked Example: Inside a Cone and a Shifted Sphere

Find limits for the region inside the sphere \(x^2+y^2+z^2=2z\) and inside the cone \(\phi=\pi/4\), with the cone opening around the positive \(z\)-axis.

First rewrite the sphere as \(x^2+y^2+(z-1)^2=1\), so it is centered at \((0,0,1)\) and has radius \(1\). The radial boundary derived above, with \(a=1\), is \(\rho=2\cos\phi\). The sphere’s interior is \(0\leq\rho\leq2\cos\phi\) for \(0\leq\phi\leq\pi/2\). The cone further restricts the angle to \(0\leq\phi\leq\pi/4\). The full azimuthal range is allowed. Thus the limits are

0\leq\rho\leq2\cos\phi,\qquad 0\leq\phi\leq\frac{\pi}{4},\qquad 0\leq\theta<2\pi.

At the intersection, \(\phi=\pi/4\), so \(\rho=2\cos(\pi/4)=\sqrt{2}\). Its height is \(\rho\cos\phi=\sqrt{2}(\sqrt{2}/2)=1\), and its distance from the \(z\)-axis is \(\rho\sin\phi=1\). Check the sphere equation: \(x^2+y^2+z^2=1+1=2=2z\). Check the cone: the cylindrical radius and height are both \(1\), so the angle from the positive \(z\)-axis is \(\pi/4\). The intersection circle satisfies both boundaries.

Worked Example: The Part of the Shifted Sphere Outside the Cone

For the same sphere \(x^2+y^2+z^2=2z\) and cone \(\phi=\pi/4\), find limits for the part of the sphere outside the cone, on the side with \(z\geq0\).

The sphere’s interior still gives \(0\leq\rho\leq2\cos\phi\). The sphere has no nonzero points with \(\phi>\pi/2\), since there \(\cos\phi<0\) and the radial upper bound would be negative. Being outside the cone on the specified side means \(\pi/4\leq\phi\leq\pi/2\). The region includes all azimuths, so its limits are

0\leq\rho\leq2\cos\phi,\qquad \frac{\pi}{4}\leq\phi\leq\frac{\pi}{2},\qquad 0\leq\theta<2\pi.

At the lower angular boundary \(\phi=\pi/4\), the radial limit is \(\sqrt{2}\), as in the previous example. At \(\phi=\pi/2\), the radial limit is \(2\cos(\pi/2)=0\), so the region tapers to the origin in the horizontal direction. A point on the cone-sphere intersection has cylindrical radius \(1\) and height \(1\); substituting gives \(1^2+1^2=2(1)\), and equal radius and height confirm that it lies on the cone. The chosen angular interval takes the other side of the cone from the previous example while retaining the sphere’s radial boundary.

Why the Intersection Check Matters

A frequent mistake is to assume that every sphere has a constant radial boundary. That is true only for a sphere centered at the origin. For the sphere \(x^2+y^2+z^2=2z\), replacing its boundary by \(\rho=\sqrt{2}\), for example, would confuse the equation \(\rho^2=2\rho\cos\phi\) with \(\rho^2=2\). The actual nonzero boundary is \(\rho=2\cos\phi\).

The error can be exposed directly at \(\phi=\pi/3\). The correct radial boundary gives \(\rho=2\cos(\pi/3)=1\). The incorrect constant-radius guess gives \(\rho=\sqrt{2}\). Substituting that guess into the sphere equation in spherical form would require \(\rho^2=2\rho\cos\phi\), but its two sides are \(2\) and \(2(\sqrt{2})(1/2)=\sqrt{2}\), respectively. Since \(2\neq\sqrt{2}\), that point is not on the sphere. Comparing the constraints and checking their intersection prevents this kind of incorrect limit.

Key takeaway: A cone with vertex at the origin supplies an angular boundary. A sphere supplies a radial boundary, which is constant only when the sphere is centered at the origin. Set the cone angle in the sphere boundary to locate their intersection, and verify that your limits describe the intended side of each surface.

Check Your Understanding

Use the coordinate conventions and reasoning developed above to answer these questions.

For a sphere centered at the origin with radius \(R\), which spherical coordinate is fixed on the sphere? Which coordinate does a cone \(\phi=\alpha\) fix?
Convert \(x^2+y^2+(z-a)^2=a^2\) into spherical coordinates and solve for its nonzero radial boundary.
For that shifted sphere and the cone \(\phi=\alpha\), where is their intersection circle in terms of \(\rho\), \(z\), and its distance from the \(z\)-axis?
For the sphere \(x^2+y^2+z^2=2z\), what angular range describes the part inside the sphere and outside the cone \(\phi=\pi/6\), on the side \(z\geq0\)?
Why is a constant radial upper bound generally incorrect for a sphere centered away from the origin?

What You'll Learn