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Solids Between Coaxial Paraboloids

Find the radial intersection, choose the correct vertical bounds, and calculate the enclosed volume between coaxial paraboloids.

Intermediate 9 min read

What You'll Learn

  • Express coaxial paraboloids using the radial coordinate.
  • Solve for the radius of their circular intersection.
  • Determine the upper and lower surfaces by comparing their heights at the axis.
  • Set up cylindrical bounds for the enclosed solid.
  • Evaluate and check volumes between paraboloids.

Introduction

Two coaxial paraboloids share the same vertical axis, so their heights depend only on the distance from that axis. When they intersect along a circle, they can enclose a finite solid above the disk inside that circle. To find its volume, first locate the circle and decide which paraboloid supplies the upper vertical bound. Then integrate the vertical height across the circular projection.

The main care is not the integration itself. It is setting up the region correctly. The surfaces may have different curvatures and different heights at the axis, so a reliable setup must use both the intersection equation and a comparison of the surfaces.

Writing the Surfaces and Finding Their Intersection

Use cylindrical coordinates \((r,\theta,z)\), where \(r\) is the distance to the common axis. A paraboloid with this axis has an equation of the form \(z=ar^2+c\), where \(a\ne 0\). The coefficient \(a\) determines how the height changes with radius, while \(c\) is the height where the surface meets the axis.

Definition: Two coaxial paraboloids can be written as \(z_1=a_1r^2+c_1\) and \(z_2=a_2r^2+c_2\), with the same \(r\)-coordinate measured from their common axis.

At an intersection, both surfaces have the same \(z\)-value. Thus their radial intersection satisfies \(a_1r^2+c_1=a_2r^2+c_2\). If \(a_1\ne a_2\), solving gives the candidate squared radius

$$ R^2=\frac{c_2-c_1}{a_1-a_2}. $$

A finite circular intersection requires \(R^2>0\). The radius is \(R=\sqrt{R^2}\). If the coefficients \(a_1\) and \(a_2\) are equal, the surfaces either never meet (if \(c_1\ne c_2\)) or coincide (if \(c_1=c_2\)); neither case gives a single intersection circle enclosing the solid considered here.

Once \(R\) is known, compare the two heights at \(r=0\), the axis. Those heights are \(c_1\) and \(c_2\). Whichever is larger supplies the upper \(z\)-bound throughout the disk \(0\le r\le R\). To see why, subtract the surface heights: their difference is linear in \(r^2\), and it is zero at \(r=R\). It cannot change sign strictly inside the disk without having another zero there.

Key setup: For the finite solid between two coaxial paraboloids meeting at radius \(R\), use \(0\le\theta\le 2\pi\), \(0\le r\le R\), and \(z\) between the lower and upper surface. The cylindrical volume element is \(r\,dz\,dr\,d\theta\).

Worked Example: A Bowl and a Downward-Opening Paraboloid

Worked Example: A Bowl and a Downward-Opening Paraboloid

Find the volume enclosed by \(z=r^2+1\) and \(z=7-r^2\).

First find the radial intersection:

$$ r^2+1=7-r^2 \quad\Longrightarrow\quad 2r^2=6 \quad\Longrightarrow\quad r^2=3. $$

Therefore \(R=\sqrt{3}\). At the axis, the surfaces have heights \(1\) and \(7\), so \(7-r^2\) is the upper surface and \(r^2+1\) is the lower surface over the disk. The volume integral is

$$ V=\int_0^{2\pi}\int_0^{\sqrt{3}}\int_{r^2+1}^{7-r^2} r\,dz\,dr\,d\theta. $$

Integrating first with respect to \(z\) gives the height difference:

$$ V=2\pi\int_0^{\sqrt{3}}r\bigl((7-r^2)-(r^2+1)\bigr)\,dr =2\pi\int_0^{\sqrt{3}}(6r-2r^3)\,dr. $$

An antiderivative is \(3r^2-\frac12r^4\). Hence

$$ V=2\pi\left[3r^2-\frac12r^4\right]_0^{\sqrt{3}} =2\pi\left(9-\frac92\right) =9\pi. $$

Check: the vertical gap is \(6-2r^2\), which is \(6\) at the axis and \(0\) at \(r=\sqrt{3}\), as required. The general volume formula derived below gives \(\frac{\pi}{2}(6)(3)=9\pi\), agreeing with the integral.

Why the Surface Order Stays Fixed Inside the Circle

It is useful to justify the choice of upper and lower surface rather than infer it from a sketch. Let the upper surface at the axis be \(z_U=a_Ur^2+c_U\) and the lower surface there be \(z_L=a_Lr^2+c_L\). Their height difference is

$$ H(r)=z_U-z_L=(c_U-c_L)+(a_U-a_L)r^2. $$

The surfaces intersect at \(r=R\), so \(H(R)=0\). Since \(H\) is a linear function of \(r^2\), this condition rewrites it as

$$ H(r)=(c_U-c_L)\left(1-\frac{r^2}{R^2}\right). $$

If the upper surface at the axis is strictly higher, then \(c_U-c_L>0\). For every \(0\le r<R\), the factor \(1-r^2/R^2\) is also positive. Thus \(H(r)>0\) throughout the interior of the disk, and the chosen upper and lower bounds are correct. They meet only at the boundary \(r=R\).

The same expression also yields a useful volume formula. Integrating the gap over the disk, with the full angular range, gives

$$ \begin{aligned} V &=2\pi\int_0^R H(r)r\,dr\\ &=2\pi(c_U-c_L)\int_0^R \left(1-\frac{r^2}{R^2}\right)r\,dr\\ &=2\pi(c_U-c_L)\left(\frac{R^2}{2}-\frac{R^2}{4}\right)\\ &=\frac{\pi}{2}(c_U-c_L)R^2. \end{aligned} $$

This is not a replacement for finding the correct bounds: it depends on a circular projection and on two surfaces of the stated form that meet at its outer radius. It is a convenient check once those conditions are satisfied.

Worked Example: Two Paraboloids Opening in the Same Direction

Worked Example: Two Paraboloids Opening in the Same Direction

Find the volume between \(z=3r^2+2\) and \(z=r^2+6\), using the finite region enclosed by their intersection.

Set their heights equal to find the intersection:

$$ 3r^2+2=r^2+6 \quad\Longrightarrow\quad 2r^2=4 \quad\Longrightarrow\quad r^2=2. $$

Thus \(R=\sqrt{2}\). At \(r=0\), the first surface has height \(2\), and the second has height \(6\). Therefore \(r^2+6\) is the upper surface on the disk. Set up and evaluate the integral:

$$ \begin{aligned} V &=\int_0^{2\pi}\int_0^{\sqrt{2}}\int_{3r^2+2}^{r^2+6} r\,dz\,dr\,d\theta\\ &=2\pi\int_0^{\sqrt{2}}r\bigl((r^2+6)-(3r^2+2)\bigr)\,dr\\ &=2\pi\int_0^{\sqrt{2}}(4r-2r^3)\,dr\\ &=2\pi\left[2r^2-\frac12r^4\right]_0^{\sqrt{2}}\\ &=2\pi(4-2)=4\pi. \end{aligned} $$

Check: the gap \(4-2r^2\) equals \(4\) on the axis and \(0\) at \(r=\sqrt{2}\). The formula gives \(\frac{\pi}{2}(4)(2)=4\pi\), matching the calculated volume.

Worked Example: The Steeper Surface Is Higher at the Axis

Worked Example: The Steeper Surface Is Higher at the Axis

Find the volume enclosed by \(z=2r^2+9\) and \(z=5r^2+3\).

The intersection radius follows from

$$ 2r^2+9=5r^2+3 \quad\Longrightarrow\quad 3r^2=6 \quad\Longrightarrow\quad r^2=2. $$

Thus \(R=\sqrt{2}\). At the axis, the first surface is at height \(9\) and the second is at height \(3\). The first is the upper surface, despite having the smaller coefficient of \(r^2\). The bounds and integral are

$$ \begin{aligned} V &=\int_0^{2\pi}\int_0^{\sqrt{2}}\int_{5r^2+3}^{2r^2+9} r\,dz\,dr\,d\theta\\ &=2\pi\int_0^{\sqrt{2}}r\bigl((2r^2+9)-(5r^2+3)\bigr)\,dr\\ &=2\pi\int_0^{\sqrt{2}}(6r-3r^3)\,dr\\ &=2\pi\left[3r^2-\frac34r^4\right]_0^{\sqrt{2}}\\ &=2\pi(6-3)=6\pi. \end{aligned} $$

Check: the upper-minus-lower gap is \(6-3r^2\); it is positive for \(0\le r<\sqrt{2}\) and zero at the intersection. The general formula gives \(\frac{\pi}{2}(6)(2)=6\pi\), confirming the result.

A Common Mistake: Reversing the Vertical Bounds

In the first example, reversing the surfaces would use \(r^2+1\) as the upper bound and \(7-r^2\) as the lower bound. That produces a negative answer:

$$ 2\pi\int_0^{\sqrt{3}}r\bigl((r^2+1)-(7-r^2)\bigr)\,dr =2\pi\left[\frac12r^4-3r^2\right]_0^{\sqrt{3}} =-9\pi. $$

The negative sign is a warning: the supposed upper surface is actually lower over the interior of the disk. Correct the setup by comparing the surface heights at the axis, then write the larger one as the upper \(z\)-bound. The volume must be nonnegative.

A different setup error is to solve for \(r^2\) but then use that value as \(R\). For example, \(r^2=3\) means \(R=\sqrt{3}\), not \(R=3\). The radial integration limit is a radius; the intersection equation first gives its square.

When a Finite Enclosed Solid Exists

The intersection equation also tests whether the stated surfaces have the circular boundary needed for this finite solid. If it gives \(R^2<0\), there is no real radial intersection. If it gives \(R^2=0\), the surfaces meet only on the axis, not along a circle of positive radius. If the coefficients of \(r^2\) agree, the surfaces do not have a unique circular intersection. In these cases, the finite disk-bounded setup above does not apply as stated.

When \(R^2>0\), the surface comparison at \(r=0\) determines the vertical order on the disk. The intersection circle closes the solid at the rim, and the projection is the full disk because the surfaces share an axis. This is why the angular bounds run through \(2\pi\), rather than covering only a sector.

Takeaway: Solve the surface equation for \(R^2\), require a positive value, and compare the surfaces at the axis. Use that comparison to order the vertical bounds over \(0\le r\le R\), then integrate with the cylindrical factor \(r\).

Check Your Understanding

For each question, identify the intersection radius and the correct surface order before evaluating any integral.

  1. For \(z=2r^2+1\) and \(z=8-r^2\), find \(R\) and identify the upper surface over the disk.
  2. For \(z=4r^2+3\) and \(z=r^2+9\), write the cylindrical triple integral for the enclosed volume.
  3. Why is comparing the surface heights at \(r=0\) enough to determine the vertical order inside the intersection circle?
  4. What does a negative value of \(R^2\) tell you about the finite disk-bounded setup?
  5. For surfaces whose axis heights differ by \(5\) and whose intersection has \(R^2=4\), use the derived formula to find the volume.

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