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Intersections of Paraboloids and Cylinders

Use the cylinder and paraboloid equations together to identify their intersection and build correct cylindrical bounds for an enclosed solid.

Intermediate 9 min read

What You'll Learn

  • Identify the circular intersection of a coaxial paraboloid and a vertical cylinder.
  • Translate a cylinder’s equation into radial limits.
  • Determine which surface gives the upper and lower vertical bounds.
  • Set up triple integrals for solids bounded by a cylinder, a paraboloid, and a plane.
  • Check computed volumes against the geometry of the solid.

Reading the Intersection

A coaxial paraboloid and a vertical cylinder share the \(z\)-axis as their axis of symmetry. In cylindrical coordinates, a coaxial paraboloid has an equation of the form \(z=\alpha r^2+\beta\), while a vertical cylinder of radius \(R\) has equation \(r=R\). Their intersection is a circle: at every point on it, the radial coordinate is \(R\), and the height is the value of the paraboloid at that radius.

This intersection helps describe a bounded solid, but it is important to identify every boundary. A vertical cylinder alone does not close a solid above and below. A typical enclosed region also has a horizontal plane and lies between that plane and the paraboloid. The plane and paraboloid specify the vertical interval; the cylindrical wall specifies the radial extent.

Key idea: For a vertical cylinder \(r=R\) and a coaxial paraboloid \(z=\alpha r^2+\beta\), their intersection satisfies \(r=R\) and \(z=\alpha R^2+\beta\). If the solid lies inside the cylinder, its radial coordinate ranges from \(0\) to \(R\).

In cylindrical coordinates, the volume element is \(dV=r\,dz\,dr\,d\theta\). Thus, for a solid inside \(r=R\), between a lower surface \(z=L(r)\) and an upper surface \(z=U(r)\), and extending through a full revolution, the volume is

$$ V=\int_0^{2\pi}\int_0^R\int_{L(r)}^{U(r)} r\,dz\,dr\,d\theta, $$

provided \(L(r)\leq U(r)\) throughout \(0\leq r\leq R\). The factor \(r\) is part of the cylindrical volume element.

Why the Intersection Gives the Bounds

The cylinder equation and the paraboloid equation can be solved together directly. On the cylindrical wall, \(r=R\). Substituting this into \(z=\alpha r^2+\beta\) gives \(z=\alpha R^2+\beta\). The angular coordinate \(\theta\) is unrestricted for coaxial surfaces, so the intersection consists of all points with this fixed radius and height.

To describe the enclosed solid, examine a vertical line at a fixed radius \(r\). The line enters the solid at its lower surface and leaves at its upper surface. For a solid between a horizontal plane \(z=c\) and the paraboloid, the order of these surfaces must be checked across the full radial interval. Once that order is known, \(z\) runs between the two heights, \(r\) runs from the axis to the cylindrical wall, and \(\theta\) runs through the full circle.

Derivation: If the solid is inside \(r=R\), above \(z=c\), and below \(z=\alpha r^2+\beta\), its vertical bounds are valid precisely where \(c\leq\alpha r^2+\beta\). If this inequality holds for every \(r\) from \(0\) to \(R\), the integral has radial limits \(0\) and \(R\). At the wall, the intersection height is \(z=\alpha R^2+\beta\). If the two surfaces instead meet at an interior radius, that equality must be solved to locate where the vertical interval begins or ends.

This reasoning is also a useful check on the bounds: the cylinder determines the radial limit, but the intersection height indicates how the paraboloid meets the wall. The two pieces of information are related, but they are not interchangeable.

Worked Example: An Upward Paraboloid Inside a Cylinder

Worked Example: An Upward Paraboloid Inside a Cylinder

Find the volume inside the cylinder \(r=2\), above the plane \(z=0\), and below the paraboloid \(z=r^2\).

First find where the cylindrical wall and paraboloid meet. Substituting \(r=2\) into \(z=r^2\) gives

$$ z=2^2=4. $$

So the wall meets the paraboloid at height \(z=4\). At each radius \(0\leq r\leq2\), the lower height is \(0\) and the upper height is \(r^2\). The full angular range is \(0\leq\theta\leq2\pi\). Therefore

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

Integrate first with respect to \(z\):

$$ \int_0^{r^2}r\,dz =r[z]_0^{r^2} =r(r^2-0) =r^3. $$

Then integrate with respect to \(r\) and \(\theta\):

$$ \begin{aligned} V &=\int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^4}{4}\right]_0^2\,d\theta\\ &=\int_0^{2\pi}\frac{16}{4}\,d\theta\\ &=\int_0^{2\pi}4\,d\theta\\ &=4(2\pi)=8\pi. \end{aligned} $$

As a check, integrating the volume element over the stated vertical interval gives the height \(r^2\) at each radius, so the remaining radial integrand is \(r^2\cdot r=r^3\), exactly as above. The volume is \(8\pi\) cubic units.

Worked Example: A Downward Paraboloid Meets the Wall at the Base

Worked Example: A Downward Paraboloid Meets the Wall at the Base

Find the volume inside \(r=3\), above \(z=0\), and below \(z=9-r^2\).

At the cylindrical wall, the paraboloid has height

$$ z=9-3^2=0. $$

Thus the paraboloid meets the plane \(z=0\) exactly at the wall. For \(0\leq r\leq3\), \(9-r^2\geq0\), so the paraboloid is the upper surface and the plane is the lower surface. The volume is

$$ V=\int_0^{2\pi}\int_0^3\int_0^{9-r^2}r\,dz\,dr\,d\theta. $$

Evaluate the vertical integral:

$$ \int_0^{9-r^2}r\,dz=r(9-r^2). $$

Then

$$ \begin{aligned} V &=2\pi\int_0^3(9-r^2)r\,dr\\ &=2\pi\int_0^3(9r-r^3)\,dr\\ &=2\pi\left[\frac{9r^2}{2}-\frac{r^4}{4}\right]_0^3\\ &=2\pi\left(\frac{81}{2}-\frac{81}{4}\right)\\ &=2\pi\left(\frac{81}{4}\right) =\frac{81\pi}{2}. \end{aligned} $$

At \(r=0\), the paraboloid reaches height \(9\); at \(r=3\), its height is \(0\). This confirms that the vertical thickness decreases to zero at the wall, consistent with the intersection calculation. The volume is \(\frac{81\pi}{2}\) cubic units.

Worked Example: A Paraboloid Above a Plane

Worked Example: A Paraboloid Above a Plane

Find the volume inside \(r=1\), between the paraboloid \(z=2+r^2\) and the plane \(z=2\).

On the cylindrical wall, \(r=1\), so the paraboloid’s height is

$$ z=2+1^2=3. $$

The lower surface is \(z=2\), and the upper surface is \(z=2+r^2\). Their difference is \(r^2\), which is nonnegative for every \(r\) in the interval \(0\leq r\leq1\). Hence the bounds are in the correct order, and

$$ V=\int_0^{2\pi}\int_0^1\int_2^{2+r^2}r\,dz\,dr\,d\theta. $$

Integrating vertically gives

$$ \int_2^{2+r^2}r\,dz =r[(2+r^2)-2] =r^3. $$

Therefore

$$ \begin{aligned} V &=2\pi\int_0^1r^3\,dr\\ &=2\pi\left[\frac{r^4}{4}\right]_0^1\\ &=2\pi\left(\frac14-0\right) =\frac{\pi}{2}. \end{aligned} $$

The vertical thickness is \(r^2\), which is zero at the axis and \(1\) at the wall. Multiplying that thickness by the cylindrical factor \(r\) gives the radial integrand \(r^3\), confirming the setup. The volume is \(\frac{\pi}{2}\) cubic units.

A Common Mistake: Confusing Intersection Height and Radius

For the paraboloid \(z=r^2\) and cylinder \(r=2\), the intersection is at \(r=2,\ z=4\). A common error is to treat \(z=4\) as the upper radial limit, for example by writing \(0\leq r\leq4\). That mixes the height coordinate with the radial coordinate and describes points beyond the cylinder.

$$ \text{Wrong: }0\leq r\leq4 \qquad\qquad \text{Correct: }0\leq r\leq2,\quad 0\leq z\leq r^2. $$

The correction follows directly from the cylinder equation: the wall is \(r=2\), so the radial coordinate cannot exceed \(2\). The value \(z=4\) is the height where the paraboloid reaches that wall. Another error is to omit the factor \(r\) from \(dV\); the cylindrical volume element is \(r\,dz\,dr\,d\theta\), not \(dz\,dr\,d\theta\).

Using the Geometry as a Check

Before evaluating a triple integral, check that the proposed bounds actually describe the stated solid. For a coaxial paraboloid and a vertical cylinder, the following checks are particularly useful:

  • Substitute the cylinder radius into the paraboloid equation to find the height of their intersection.
  • Check which surface is upper and which is lower over the entire radial interval.
  • Confirm that an inside-cylinder region uses radial limits from \(0\) to the cylinder radius.
  • Use \(0\leq\theta\leq2\pi\) only when the solid goes all the way around the axis.
  • Include the factor \(r\) in the cylindrical volume element.

These checks distinguish three roles: the cylinder fixes the radial boundary, the paraboloid provides a curved vertical boundary, and a plane or another surface supplies any remaining boundary needed to enclose the solid. The intersection tells how those surfaces meet; it does not replace the need to specify the complete solid.

Check Your Understanding

Use the surface equations and the geometry to answer each question.

  1. Where does the paraboloid \(z=2r^2\) meet the cylinder \(r=3\)? State both the radius and height.
  2. For the solid inside \(r=2\), above \(z=0\), and below \(z=8-r^2\), what are the \(r\)- and \(z\)-limits?
  3. For the solid inside \(r=1\), between \(z=1\) and \(z=1+3r^2\), which surface is upper? What is the volume element?
  4. Explain why the intersection height of a paraboloid and cylinder is not itself a radial limit.
  5. Set up, but do not evaluate, the volume integral inside \(r=2\), above \(z=1\), and below \(z=1+r^2\).

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