Workshop: A Cylinder Intersecting a Paraboloid

Introduction

A vertical cylinder and a coaxial paraboloid are both rotationally symmetric about the $z$-axis. That symmetry makes cylindrical coordinates a natural choice for finding volumes bounded by them. The main challenge is choosing the radial interval correctly: a horizontal plane may meet the paraboloid inside the cylinder, at its side, or beyond it.

We will use the intersection radius to determine which boundaries actually limit the solid. A finite volume also needs enough boundaries to enclose it. For the regions in this tutorial, those boundaries include a horizontal plane or a base plane as well as the cylinder and paraboloid.

Paraboloid, Cylinder, and Intersection Radius

Consider a coaxial upward-opening paraboloid with equation $z=z_0+kr^2$, where $k>0$, and a vertical cylinder of radius $R>0$, whose equation in cylindrical coordinates is $r=R$. The interior of the cylinder has $0\leq r\leq R$. A horizontal plane has equation $z=H$.

Definition: The intersection radius $r_*$ is the nonnegative radial distance at which the horizontal plane $z=H$ meets the paraboloid $z=z_0+kr^2$. When $H\geq z_0$, it is $$ r_*=\sqrt{\frac{H-z_0}{k}}. $$ If $H<z_0$, the plane lies below the paraboloid's vertex, so there is no real intersection radius.

The formula follows by setting the two heights equal: $H=z_0+kr^2$, so $r^2=(H-z_0)/k$. Since $r$ is a nonnegative distance, we take the nonnegative square root. Comparing $r_*$ with $R$ tells us whether the intersection lies inside the cylinder, on its side, or outside it.

Comparison	Geometric consequence
$r_*<R$	The plane and paraboloid meet inside the cylinder; the intersection can restrict the radial bounds.
$r_*=R$	The plane and paraboloid meet at the cylinder's side.
$r_*>R$	The intersection is outside the cylinder; the cylinder side limits the radial bounds first.

In cylindrical coordinates, the volume element is $dV=r\,dz\,dr\,d\theta$, with $0\leq\theta\leq2\pi$ for a full rotation. If the lower and upper heights are $z_{\mathrm{low}}(r)$ and $z_{\mathrm{high}}(r)$, the volume integral is therefore

V=\int_0^{2\pi}\int_{r_{\min}}^{r_{\max}} \int_{z_{\mathrm{low}}(r)}^{z_{\mathrm{high}}(r)} r\,dz\,dr\,d\theta.

Integrating first with respect to $z$ replaces the vertical interval by its height, $z_{\mathrm{high}}(r)-z_{\mathrm{low}}(r)$. The radial bounds must include only radii where the lower height does not exceed the upper height.

Choosing the Bounds

For a solid inside the cylinder, above the plane $z=z_0$, and below the paraboloid $z=z_0+kr^2$, the radial interval is the full cylinder. Its bounds are $0\leq r\leq R$, $0\leq\theta\leq2\pi$, and $z_0\leq z\leq z_0+kr^2$.

V=\int_0^{2\pi}\int_0^R \int_{z_0}^{z_0+kr^2}r\,dz\,dr\,d\theta.

For a region inside the cylinder, above the paraboloid, and below the plane $z=H$, the paraboloid must lie no higher than the plane. This requires $z_0+kr^2\leq H$, or $r\leq r_*$. Thus, if $H\geq z_0$, the radial interval is $0\leq r\leq\min(R,r_*)$.

V=\int_0^{2\pi}\int_0^{\min(R,r_*)} \int_{z_0+kr^2}^{H}r\,dz\,dr\,d\theta.

These bounds describe two different arrangements. A solid below the paraboloid and above its vertex plane extends across the cylinder; a solid above the paraboloid and below a horizontal plane exists only where the paraboloid is below that plane. If instead the region is above the plane $z=H$ and below the paraboloid, it requires $z_0+kr^2\geq H$. When $z_0\leq H$ and $r_*\leq R$, its radial bounds are $r_*\leq r\leq R$, and its vertical bounds are $H\leq z\leq z_0+kr^2$. This last arrangement illustrates why the intersection radius can be a lower, rather than an upper, radial bound.

Why the Intersection Radius Gives the Correct Bounds

Result: For $k>0$, the paraboloid $z=z_0+kr^2$ lies below the plane $z=H$ exactly when $r\leq r_*$, provided $H\geq z_0$. Inside a cylinder of radius $R$, the radii available to a region between the paraboloid and plane are therefore $0\leq r\leq\min(R,r_*)$.

To justify the result, compare the two heights at an arbitrary radius $r$. The paraboloid lies below the plane precisely when

z_0+kr^2\leq H.

Subtract $z_0$ and divide by $k>0$, which preserves the inequality. This gives $r^2\leq(H-z_0)/k$. For $r\geq0$, that is equivalent to $r\leq\sqrt{(H-z_0)/k}=r_*$. The cylinder also requires $r\leq R$, so both restrictions together give $r\leq\min(R,r_*)$. This argument also explains why there are no such radii if $H<z_0$: even the paraboloid's lowest point lies above the plane.

Worked Volumes

Worked Example: Below a Paraboloid Inside a Cylinder

Find the volume inside the cylinder $r\leq2$, above the plane $z=0$, and below the paraboloid $z=3r^2$.

The cylinder supplies $0\leq r\leq2$, and the solid makes one full turn, so $0\leq\theta\leq2\pi$. At each radius, the vertical bounds are $0\leq z\leq3r^2$. Therefore

V=\int_0^{2\pi}\int_0^2\int_0^{3r^2}r\,dz\,dr\,d\theta.

Integrating with respect to $z$ gives the height $3r^2$:

V=\int_0^{2\pi}\int_0^2 3r^3\,dr\,d\theta.

An antiderivative of $3r^3$ is $\frac{3}{4}r^4$, since its derivative is $3r^3$. Evaluating at the radial bounds gives $\frac{3}{4}(2^4)-\frac{3}{4}(0^4)=12$. The angular integral contributes $2\pi$, so

V=2\pi(12)=24\pi.

Check: the height of the paraboloid at the cylinder wall is $3(2^2)=12$, and the volume integral uses the nonnegative height $3r^2$ over every radius from $0$ to $2$. Differentiating the radial antiderivative returns $3r^3$, the integrand obtained after integrating in $z$.

Worked Example: The Intersection Lies Inside the Cylinder

Find the volume inside $r\leq3$, above the paraboloid $z=r^2$, and below the plane $z=4$.

Set the paraboloid height equal to the plane height:

r^2=4,\qquad r_*=2.

Because $2<3$, the intersection occurs inside the cylinder. At radii greater than $2$, the paraboloid is higher than the plane, so there is no vertical interval above the paraboloid and below the plane. The full bounds are $0\leq\theta\leq2\pi$, $0\leq r\leq2$, and $r^2\leq z\leq4$.

V=\int_0^{2\pi}\int_0^2\int_{r^2}^4 r\,dz\,dr\,d\theta =\int_0^{2\pi}\int_0^2(4-r^2)r\,dr\,d\theta.

Expand the radial integrand: $(4-r^2)r=4r-r^3$. An antiderivative is $2r^2-\frac{r^4}{4}$, whose derivative is $4r-r^3$. Thus

V=2\pi\left[2r^2-\frac{r^4}{4}\right]_0^2 =2\pi\left(8-4\right)=8\pi.

Check: at $r=2$, the lower and upper heights agree because $r^2=4$; at $r=0$, the vertical height is $4$. The radial antiderivative differentiates to the required integrand $4r-r^3$, and the resulting volume is positive.

Worked Example: The Intersection Lies Outside the Cylinder

Find the volume inside $r\leq2$, above the paraboloid $z=2r^2$, and below the plane $z=10$.

Solving for the intersection gives

2r^2=10,\qquad r_*^2=5,\qquad r_*=\sqrt{5}.

Since $\sqrt{5}>2$, the intersection is outside the cylinder. The paraboloid remains below the plane throughout the cylinder: at the largest allowed radius, its height is $2(2^2)=8<10$. Consequently, every radius from $0$ to $2$ is included. The complete bounds are $0\leq\theta\leq2\pi$, $0\leq r\leq2$, and $2r^2\leq z\leq10$.

V=\int_0^{2\pi}\int_0^2\int_{2r^2}^{10}r\,dz\,dr\,d\theta =\int_0^{2\pi}\int_0^2(10-2r^2)r\,dr\,d\theta.

The radial integrand is $10r-2r^3$. Its antiderivative is $5r^2-\frac{r^4}{2}$, because differentiating gives $10r-2r^3$. Therefore

V=2\pi\left[5r^2-\frac{r^4}{2}\right]_0^2 =2\pi(20-8)=24\pi.

Check: at $r=2$, the vertical height is $10-8=2>0$, so the entire cylinder cross-section is feasible. At $r=0$, the height is $10$. The derivative of the radial antiderivative is the integrand, and the final volume is positive.

A Common Intersection-Radius Mistake

When a plane at height $4$ meets the paraboloid $z=r^2$, the intersection equation is $r^2=4$. A common error is to set the radius equal to the height, writing $r_*=4$, rather than solving for $r$. For a cylinder of radius $3$, that incorrect choice would lead someone to integrate the proposed height difference over the entire cylinder:

2\pi\int_0^3(4-r^2)r\,dr =2\pi\left[2r^2-\frac{r^4}{4}\right]_0^3 =2\pi\left(18-\frac{81}{4}\right) =-\frac{9\pi}{2}.

The negative result signals a problem: for $2<r\leq3$, $r^2>4$, so the supposed lower height is above the upper height. The correct intersection is $r_*=\sqrt{4}=2$. Integrating only where the region exists gives $2\pi\int_0^2(4-r^2)r\,dr=8\pi$, as computed above.

Why the Case Comparison Matters

Finding $r_*$ is not the end of setting up the volume. It must be compared with the cylinder radius. If $r_*<R$, the plane-paraboloid intersection cuts off the region before the cylinder wall. If $r_*\geq R$, the cylinder wall is reached first, so the entire cylinder's radial range may be included, provided the vertical bounds remain in the correct order.

In each case, check the vertical interval at the radial limits. For a region above the paraboloid and below a plane, verify that $H-(z_0+kr^2)\geq0$ throughout the proposed radial interval. This simple check catches incorrect bounds before the integration begins.

Check Your Understanding

For each question, identify how the intersection radius affects the bounds before attempting any integration.

For $z=2r^2$ and $z=18$, find the intersection radius. If the cylinder has radius $4$, is the intersection inside, on, or outside its side?
State the full cylindrical bounds for the region inside $r\leq5$, above $z=r^2$, and below $z=9$.
For $z=1+3r^2$ and $z=13$, find the intersection radius and describe the radial bounds for the region between these surfaces inside a cylinder of radius $3$.
Explain why a region above $z=r^2$ and below $z=4$ cannot include radii greater than $2$.
What inequality should be checked to confirm that a proposed radial interval gives nonnegative vertical height between a paraboloid and a horizontal plane?

Comparison	Geometric consequence
\(r_*<R\)	The plane and paraboloid meet inside the cylinder; the intersection can restrict the radial bounds.
\(r_*=R\)	The plane and paraboloid meet at the cylinder's side.
\(r_*>R\)	The intersection is outside the cylinder; the cylinder side limits the radial bounds first.