Parametrizing Intersections of Two Surfaces

One Parameter, Two Surface Conditions

A space curve formed by two surfaces must satisfy both surface equations at the same time. A parametrization is useful only if its three coordinates share one parameter and every resulting point lies on both surfaces. If the goal is to describe the entire intersection, every point of that intersection must also be reached.

A practical strategy is to begin with a parametrization of one surface, then use the other surface’s equation to determine the remaining coordinates or restrict the parameter. For a cylinder, an angular parameter often supplies two coordinates immediately. For a sphere cut by a plane, the intersection can be described as a circle in that plane.

Definition: A parametrization of the intersection of two surfaces \(S_1\) and \(S_2\) is a vector-valued function \(\mathbf{r}(t)=(x(t),y(t),z(t))\) such that \(\mathbf{r}(t)\in S_1\cap S_2\) for every allowed \(t\). It represents the whole intersection if every point of \(S_1\cap S_2\) equals \(\mathbf{r}(t)\) for at least one allowed parameter value.

These are two separate requirements. Substituting into the two surface equations checks that the curve is contained in the intersection. A separate coverage argument checks that the parametrization does not omit part of the intersection.

Using a Cylinder’s Angular Parameter

For the circular cylinder \(x^2+y^2=a^2\), the angular parametrization \(x=a\cos t,\ y=a\sin t\) satisfies the cylinder equation because \(\cos^2t+\sin^2t=1\). If the second surface is a plane, substitute those expressions into the plane equation to find the third coordinate or the allowed parameter values.

Worked Example: A Plane Cutting a Circular Cylinder

Find a parametrization of the intersection of the cylinder \(x^2+y^2=9\) and the plane \(z=x+y\).

The cylinder has radius \(3\), so use its angle \(t\) to set \(x=3\cos t\) and \(y=3\sin t\). The plane equation then determines \(z\):

z=x+y=3\cos t+3\sin t.

Thus a parametrization is

\mathbf{r}(t)=\bigl(3\cos t,\ 3\sin t,\ 3\cos t+3\sin t\bigr), \qquad 0\le t<2\pi.

Check the cylinder equation:

x^2+y^2 =(3\cos t)^2+(3\sin t)^2 =9\cos^2t+9\sin^2t =9.

Check the plane equation:

x+y=3\cos t+3\sin t=z.

Both equations hold for every allowed \(t\). To check coverage, take any point \((x,y,z)\) on the intersection. Since \(x^2+y^2=9\), the pair \((x/3,y/3)\) lies on the unit circle, so there is an angle \(t\) with \(x=3\cos t\) and \(y=3\sin t\). The plane then forces \(z=x+y\), exactly the third coordinate above. Therefore the parametrization covers the whole intersection.

The interval \(0\le t<2\pi\) covers the cylinder’s circular cross-section once. Allowing all real \(t\) would describe the same curve repeatedly because sine and cosine have period \(2\pi\).

Sphere Intersections with a Plane

A plane cutting a sphere can produce no points, one point, or a circle. The outcome depends on the plane’s distance from the sphere’s center. The following result gives both the condition and a parametrization.

Theorem: Let a sphere have center \(\mathbf{C}\) and radius \(R>0\). Let a plane be written as \(\mathbf{n}\cdot\mathbf{x}=c\), where \(\mathbf{n}\) is a unit normal vector. Set \(d=c-\mathbf{n}\cdot\mathbf{C}\). If \(|d|<R\), the intersection is a circle of radius \(\rho=\sqrt{R^2-d^2}\), centered at \(\mathbf{C}+d\mathbf{n}\). Choose perpendicular unit vectors \(\mathbf{u}\) and \(\mathbf{v}\) that are both perpendicular to \(\mathbf{n}\). Then the full circle is parametrized by \(\mathbf{r}(t)=\mathbf{C}+d\mathbf{n}+\rho(\mathbf{u}\cos t+\mathbf{v}\sin t)\), \(0\le t<2\pi\). If \(|d|=R\), the intersection is the single point \(\mathbf{C}+d\mathbf{n}\). If \(|d|>R\), the surfaces do not intersect.

Why the Sphere-Plane Formula Works

First, the center of the intersection circle must lie in the plane. Moving from \(\mathbf{C}\) a signed distance \(d\) in the unit-normal direction gives \(\mathbf{C}+d\mathbf{n}\), and

\mathbf{n}\cdot(\mathbf{C}+d\mathbf{n}) =\mathbf{n}\cdot\mathbf{C}+d =c.

Any point in the plane can be written as \(\mathbf{x}=\mathbf{C}+d\mathbf{n}+\mathbf{w}\), where \(\mathbf{w}\) lies parallel to the plane and hence \(\mathbf{w}\cdot\mathbf{n}=0\). Since the sphere equation is \(\|\mathbf{x}-\mathbf{C}\|^2=R^2\), perpendicularity gives

R^2 =\|d\mathbf{n}+\mathbf{w}\|^2 =d^2+\|\mathbf{w}\|^2.

Therefore \(\|\mathbf{w}\|^2=R^2-d^2\). If \(|d|<R\), the possible vectors \(\mathbf{w}\) form a circle of radius \(\rho=\sqrt{R^2-d^2}\) in the plane. The perpendicular unit vectors \(\mathbf{u}\) and \(\mathbf{v}\) give coordinates in that plane, and \(\mathbf{w}=\rho(\mathbf{u}\cos t+\mathbf{v}\sin t)\) traces that circle. If \(|d|=R\), then \(\|\mathbf{w}\|=0\), leaving only the center point of the circle. If \(|d|>R\), the equation would require \(\|\mathbf{w}\|^2<0\), which is impossible.

Worked Example: A Horizontal Plane Cutting a Sphere

Parametrize the intersection of the sphere \(x^2+y^2+z^2=25\) and the plane \(z=3\).

Substitute \(z=3\) into the sphere equation:

x^2+y^2+3^2=25, \qquad x^2+y^2=16.

The intersection is a circle of radius \(4\) in the plane \(z=3\). Use the circular parameter for its \(x\)- and \(y\)-coordinates, keeping \(z\) fixed:

\mathbf{r}(t)=\bigl(4\cos t,\ 4\sin t,\ 3\bigr), \qquad 0\le t<2\pi.

Substitution checks the sphere:

x^2+y^2+z^2 =16\cos^2t+16\sin^2t+9 =16+9 =25.

The plane check is immediate from the third coordinate: \(z=3\). For coverage, every point of the intersection has \(z=3\) and satisfies \(x^2+y^2=16\), so its first two coordinates lie on the circle parametrized by \(x=4\cos t,\ y=4\sin t\). Thus every intersection point occurs.

An Oblique Plane and a Sphere

The same circle construction works when the plane is not parallel to a coordinate plane. For a concrete example, consider the sphere \(x^2+y^2+z^2=25\) and the plane \(x+z=6\). The sphere has center \(\mathbf{C}=(0,0,0)\) and radius \(5\). A unit normal to the plane is \(\mathbf{n}=(1/\sqrt{2},0,1/\sqrt{2})\), and the plane equation becomes \(\mathbf{n}\cdot\mathbf{x}=3\sqrt{2}\). Thus \(d=3\sqrt{2}\), and the circle radius is

\rho=\sqrt{5^2-(3\sqrt{2})^2} =\sqrt{25-18} =\sqrt{7}.

The center of the circle is \(\mathbf{C}+d\mathbf{n}=(3,0,3)\). Two perpendicular unit vectors parallel to the plane are \(\mathbf{u}=(1/\sqrt{2},0,-1/\sqrt{2})\) and \(\mathbf{v}=(0,1,0)\). Applying the circle parametrization gives:

Worked Example: A Sphere Cut by an Oblique Plane

Using the center, radius, and plane directions above, obtain the coordinate functions by expanding the circle formula:

\begin{aligned} \mathbf{r}(t) &=(3,0,3)+\sqrt{7}\left( \left(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}\right)\cos t +(0,1,0)\sin t\right)\\ &=\left(3+\sqrt{\frac{7}{2}}\cos t,\ \sqrt{7}\sin t,\ 3-\sqrt{\frac{7}{2}}\cos t\right), \qquad 0\le t<2\pi. \end{aligned}

For the plane equation, the first and third coordinates add to \(6\):

x+z =3+\sqrt{\frac{7}{2}}\cos t +3-\sqrt{\frac{7}{2}}\cos t =6.

For the sphere equation, write \(a=\sqrt{7/2}\cos t\). Then \(x=3+a\), \(z=3-a\), and \(y=\sqrt{7}\sin t\). Hence

\begin{aligned} x^2+y^2+z^2 &=(3+a)^2+7\sin^2t+(3-a)^2\\ &=18+2a^2+7\sin^2t\\ &=18+7\cos^2t+7\sin^2t\\ &=25. \end{aligned}

The parametrization therefore lies on both surfaces. The two perpendicular unit directions span the plane, and \((\cos t,\sin t)\) traces the unit circle, so the formula reaches every point of the intersection circle.

A Common Error: Satisfying Only One Surface

Suppose the cylinder and plane from the first example are being parametrized. A tempting but incorrect choice is

\mathbf{r}(t)=\bigl(3\cos t,\ 3\sin t,\ 3\bigr).

This curve does lie on the cylinder, since \(x^2+y^2=9\). But it does not generally lie on the plane \(z=x+y\). At \(t=0\), for instance, the proposed point is \((3,0,3)\), while \(x+y=3\); this value happens to satisfy the plane. At \(t=\pi/2\), the point is \((0,3,3)\), while \(x+y=3\), also satisfying it. But at \(t=\pi/4\), \(x+y=3\sqrt{2}\), which is not \(z=3\). The error is treating the plane coordinate as constant instead of substituting the cylinder coordinates into the plane equation.

The correction is to use \(z=x+y=3\cos t+3\sin t\). Then both surface equations hold for every parameter value. A reliable check is to substitute the proposed coordinates into each defining equation independently.

Building and Checking an Intersection Parametrization

For simple intersections, the reasoning can be organized into a short sequence:

Choose a useful parameter.
For a circular cylinder, its angle often supplies two coordinates. For a sphere cut by a plane, identify the circle in the plane.

Enforce the second surface equation.
Substitute the chosen coordinate functions into the other surface equation, rather than assigning the remaining coordinate independently.

Check both equations.
Substitute all three coordinate functions into each surface equation and simplify.

Check coverage.
Explain why every point satisfying both equations corresponds to some allowed parameter value.

The coverage step matters because satisfying the equations establishes only that the parametrized curve is contained in the intersection. In the cylinder example, the angle covers every point on the circular cross-section, while the plane fixes the last coordinate. In the sphere-plane examples, the two in-plane directions cover the entire intersection circle.

Intersections can also be empty, consist of a single point, or have more complicated shapes. The sphere-plane theorem identifies those possibilities for that particular pair of surfaces; for other surface equations, substitution and geometric inspection determine which parameter values are possible. In every case, a valid parametrization must use one common parameter and satisfy both surface conditions.

Check Your Understanding

For each question, distinguish between checking that a parametrization lies in an intersection and checking that it covers the whole intersection.

For \(x^2+y^2=16\) and \(z=2x\), use the cylinder angle to write all three coordinate functions and verify both surface equations.
A sphere of radius \(6\) is cut by a plane at distance \(2\) from its center. What is the radius of the intersection circle?
For a sphere of radius \(4\), what do the cases \(|d|<4\), \(|d|=4\), and \(|d|>4\) mean for its intersection with a plane?
Why does verifying both surface equations for every \(t\) not, by itself, prove that a parametrization represents the entire intersection?
In the oblique-plane example, which two unit vectors provide directions within the plane, and why must they be perpendicular to the plane’s normal?