Regions Between Spheres And Cylinders

Describing a Sphere–Cylinder Intersection

A sphere and a coaxial cylinder impose different kinds of distance conditions. A sphere centered at the origin limits distance from the origin, while a cylinder whose axis is the $z$-axis limits distance from that axis. Combining the two conditions describes a familiar solid: the part of the ball that remains after removing its central cylindrical portion.

Writing both conditions as inequalities makes the geometry precise. It also reveals which surfaces form the boundary, whether the intersection exists, and how to express the same region in cylindrical or spherical coordinates.

Definition: Let $R>0$ be the radius of a sphere centered at the origin, and let $a>0$ be the radius of a cylinder whose axis is the $z$-axis. The region inside the sphere and outside the cylinder consists of points satisfying both the sphere condition and the cylinder condition.

We use $r$ for cylindrical radius, so $r=\sqrt{x^2+y^2}$, and $\rho$ for spherical radius, so $\rho=\sqrt{x^2+y^2+z^2}$. The spherical angle $\phi$ is measured from the positive $z$-axis. Thus $r=\rho\sin\phi$, $z=\rho\cos\phi$, and $0\leq\phi\leq\pi$.

The Two Conditions in Cylindrical Coordinates

Being inside or on the sphere of radius $R$ means that the distance from the origin is at most $R$. In cylindrical coordinates, the squared distance is $r^2+z^2$. Being outside or on the cylinder of radius $a$ means that the distance from the $z$-axis is at least $a$, or $r\geq a$. Therefore the region is described by

r^2+z^2\leq R^2, \qquad r\geq a.

The sphere condition alone gives $0\leq r\leq R$. Imposing $r\geq a$ leaves points only if $a\leq R$. If $0<a<R$, the allowed cylindrical radii begin at $a$, and the sphere limits the height through $z^2\leq R^2-r^2$. Equivalently, one may use height first: for a fixed $z$, the sphere allows $r\leq\sqrt{R^2-z^2}$, which must be at least $a$.

-\sqrt{R^2-a^2}\leq z\leq \sqrt{R^2-a^2}, \qquad a\leq r\leq\sqrt{R^2-z^2}, \qquad 0\leq\theta<2\pi.

These bounds describe the same solid without gaps or extra points. The height range follows by requiring that the largest allowable radius $\sqrt{R^2-z^2}$ is at least $a$. Squaring that condition gives $R^2-z^2\geq a^2$, hence $z^2\leq R^2-a^2$. For each such height, the cylindrical radius runs from the cylinder to the sphere.

Deriving the Spherical Bounds

In spherical coordinates, being inside the sphere is simply $\rho\leq R$. The cylinder condition must still measure distance from the $z$-axis, not distance from the origin. Since $r=\rho\sin\phi$, being outside the cylinder becomes $\rho\sin\phi\geq a$. Because $\sin\phi\geq0$ for $0\leq\phi\leq\pi$, this is equivalent to $\rho\geq a/\sin\phi$.

For a given angle $\phi$, there are points satisfying both radial inequalities only when $a/\sin\phi\leq R$, or $\sin\phi\geq a/R$. When $0<a<R$, define $\alpha=\arcsin(a/R)$, where $0<\alpha<\pi/2$. On the range $0\leq\phi\leq\pi$, the angles satisfying this sine inequality are $\alpha\leq\phi\leq\pi-\alpha$.

Theorem: If $0<a<R$, the region inside the sphere of radius $R$ and outside the coaxial cylinder of radius $a$ has spherical bounds $$ \alpha\leq\phi\leq\pi-\alpha, \qquad \frac{a}{\sin\phi}\leq\rho\leq R, \qquad 0\leq\theta<2\pi, \qquad \alpha=\arcsin\left(\frac{a}{R}\right). $$ These bounds use the spherical convention in which $\phi$ is measured from the positive $z$-axis.

The lower radial bound varies with angle: at angles near the positive or negative $z$-axis, $\sin\phi$ is small, so a point must be farther from the origin to reach the cylinder’s exterior. The angular limits mark where the sphere and cylinder meet. At either limiting angle, the lower radial bound equals the sphere radius, leaving no radial interval at that angle.

Worked Examples

Worked Example: A Sphere of Radius 5 and a Cylinder of Radius 3

Describe the portion inside the sphere $x^2+y^2+z^2=25$ and outside the cylinder $x^2+y^2=9$. First, the sphere gives $r^2+z^2\leq25$, and the cylinder exterior gives $r\geq3$. For the height limits, require $25-z^2\geq9$: this gives $z^2\leq16$, hence $-4\leq z\leq4$. For each such $z$, the sphere gives $r\leq\sqrt{25-z^2}$, so the cylindrical bounds are

-4\leq z\leq4, \qquad 3\leq r\leq\sqrt{25-z^2}, \qquad 0\leq\theta<2\pi.

For spherical bounds, $a/R=3/5$, so $\alpha=\arcsin(3/5)$. The condition $\rho\sin\phi\geq3$, together with $\rho\leq5$, gives

\arcsin(3/5)\leq\phi\leq\pi-\arcsin(3/5), \qquad \frac{3}{\sin\phi}\leq\rho\leq5, \qquad 0\leq\theta<2\pi.

Check a point on the sphere–cylinder intersection: at $r=3$ and $z=4$, $r^2+z^2=9+16=25$, so it lies on the sphere, and $r=3$ puts it on the cylinder. In spherical coordinates, $\sin\phi=3/5$ and $\rho=5$ give $\rho\sin\phi=5(3/5)=3$, confirming the cylinder condition as well.

Worked Example: Finding the Intersection Circle

Suppose a sphere of radius $13$ is centered at the origin and a coaxial cylinder has radius $5$. To find where their surfaces meet, impose both equalities: $r=5$ and $r^2+z^2=169$. Substitution gives $25+z^2=169$, so $z^2=144$ and $z=\pm12$. The surfaces therefore meet in two horizontal circles, one at height $12$ and one at height $-12$, each of cylindrical radius $5$.

The solid itself has the cylindrical description

-12\leq z\leq12, \qquad 5\leq r\leq\sqrt{169-z^2}, \qquad 0\leq\theta<2\pi.

For a check, take a point on the upper intersection circle. It has $r=5$, $z=12$, and any azimuth $\theta$. Its squared distance from the origin is $5^2+12^2=25+144=169=13^2$, and its distance from the $z$-axis is $5$. Thus it lies on both surfaces, as required.

Worked Example: A Sphere Centered Away from the Origin on the Axis

A sphere centered at $(0,0,2)$ with radius $5$ has equation $r^2+(z-2)^2=25$. Intersect its interior with the exterior of the coaxial cylinder $r=3$. The inequalities are $r^2+(z-2)^2\leq25$ and $r\geq3$. For the height limits, the largest available $r$ must be at least $3$, so $25-(z-2)^2\geq9$. Therefore $(z-2)^2\leq16$, giving $-2\leq z\leq6$. The bounds are

-2\leq z\leq6, \qquad 3\leq r\leq\sqrt{25-(z-2)^2}, \qquad 0\leq\theta<2\pi.

At the upper intersection circle, $z=6$ and $r=3$. Substitution into the sphere equation gives $3^2+(6-2)^2=9+16=25$, and $r=3$ also satisfies the cylinder boundary equation. This check confirms both the height endpoint and the intersection of the surfaces. The spherical radial bound $\rho\leq R$ from a sphere centered at the origin does not apply here: the sphere’s distance condition is measured from $(0,0,2)$, not from the origin.

Boundary Surfaces and Existence

For $0<a<R$, the region is bounded by two surfaces. The outer boundary is the sphere $r^2+z^2=R^2$, and the inner boundary is the cylinder $r=a$. The sphere supplies the upper radial limit in cylindrical coordinates; the cylinder supplies the lower radial limit. Their intersection consists of the two circles found by setting $r=a$ in the sphere equation.

The size comparison between $a$ and $R$ determines the kind of intersection. If $a<R$, the inequalities leave a three-dimensional region with positive volume. If $a=R$, they allow only $r=R,\ z=0$: the circle where the cylinder touches the equator of the sphere. There is no three-dimensional interior. If $a>R$, every point in the ball has $r\leq R<a$, so no point can be outside the cylinder and the intersection is empty.

Geometric takeaway: For a sphere centered at the origin and a coaxial cylinder, the intersection is determined by comparing their radii. When the cylinder radius is smaller, the sphere and cylinder bound a solid; when the radii are equal, only their common circle remains; when the cylinder is larger, there is no intersection.

A Common Coordinate Mistake

A frequent error is to translate “outside the cylinder” into $\rho\geq a$ in spherical coordinates. That condition describes points at least $a$ units from the origin; it describes a spherical shell boundary, not a cylinder. A cylinder measures distance from its axis, so the correct spherical inequality is $\rho\sin\phi\geq a$.

For example, take $\rho=4$, $\phi=\pi/6$, and cylinder radius $a=3$. The incorrect test gives $\rho=4\geq3$, and would classify the point as outside. But its cylindrical radius is $r=\rho\sin\phi=4(1/2)=2$, which is less than $3$. The point is inside the cylinder. The correction is to test $r=\rho\sin\phi$ directly; only if $\rho\sin\phi\geq3$ is the point outside or on this cylinder.

Keeping the two distances distinct is the key to describing these regions correctly: $\rho$ measures distance from the origin, while $r$ measures distance from the $z$-axis. The sphere bounds the first, and the coaxial cylinder bounds the second.

Check Your Understanding

Use the inequalities and boundary interpretations developed above to answer these questions.

Write cylindrical-coordinate inequalities for the region inside a sphere of radius $10$ centered at the origin and outside a coaxial cylinder of radius $6$.
For that same region, find the range of $z$ and the two heights where the sphere and cylinder meet.
State the spherical-coordinate condition for being outside a coaxial cylinder of radius $a$, and explain what distance it measures.
For a sphere of radius $R$ and a coaxial cylinder of radius $a$, describe what happens when $a=R$ and when $a>R$.
Identify the two boundary surfaces of the solid when $0<a<R$, and state which one gives the lower cylindrical-radius bound.