Paraboloid-Bounded Solids

From a Curved Boundary to a Volume Integral

A paraboloid and a plane can enclose a finite solid whose boundary has both a curved piece and a flat piece. Computing the flux across each piece separately can require careful normal vectors and surface integrals. When the vector field is smooth throughout the solid, the Divergence Theorem offers another route: compute the divergence, then integrate it over the volume.

The main geometric task is to describe the solid correctly. For a circular paraboloid such as \(z=x^2+y^2\), cylindrical coordinates make the surface equation \(z=r^2\). The intersection with a horizontal plane then gives a simple radial bound. When the plane is tilted, its intersection can instead give an angle-dependent radial bound.

Core strategy: Identify which surface is above the other, find their intersection to bound the projection in the \(xy\)-plane, and integrate the divergence over the resulting solid using cylindrical coordinates.

The Divergence Theorem and Cylindrical Bounds

For a vector field \(\mathbf{F}=\langle P,Q,R\rangle\), its divergence is \(\nabla\cdot\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\). If \(E\) is a bounded solid with a piecewise smooth closed boundary \(\partial E\), and \(\mathbf{F}\) has continuous first partial derivatives on an open region containing \(E\), the Divergence Theorem states that the flux with outward orientation is

\iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_E \nabla\cdot\mathbf{F}\,dV.

Here \(\mathbf{n}\) is the outward unit normal. A paraboloid joined to a plane along their intersection forms a closed, piecewise smooth boundary in the examples below. The theorem directly gives the net outward flux through that entire boundary; it does not require computing the flux through each piece separately.

In cylindrical coordinates, \(x=r\cos\theta\), \(y=r\sin\theta\), and the volume element is \(dV=r\,dz\,dr\,d\theta\). The factor \(r\) is essential. It is the absolute value of the Jacobian determinant of the coordinate map \((r,\theta,z)\mapsto(x,y,z)\): the determinant has absolute value \(r\), so \(dV=r\,dr\,d\theta\,dz\). The order of differentials can be arranged to match the chosen integration bounds.

For a solid between \(z=g(r,\theta)\) and \(z=h(r,\theta)\), with \(g\leq h\) over its projection \(D\), the volume integral is therefore

\iiint_E \nabla\cdot\mathbf{F}\,dV = \int_{\theta=a}^{b}\int_{r=u(\theta)}^{v(\theta)} \int_{z=g(r,\theta)}^{h(r,\theta)} (\nabla\cdot\mathbf{F})\,r\,dz\,dr\,d\theta.

Why the Bounds Have This Form

A point in the solid must lie between the two boundary surfaces. For the common case of a paraboloid below a plane, this means the vertical coordinate lies from the paraboloid's height to the plane's height. The allowed \((r,\theta)\) values are those for which the lower height does not exceed the upper height. Their boundary is found by setting the two heights equal: that equation describes where the surfaces meet.

Write both surfaces in cylindrical coordinates.
Use \(x^2+y^2=r^2\); a circular paraboloid \(z=x^2+y^2\) becomes \(z=r^2\).

Find the projection.
Set the surface heights equal to determine the radial and angular region in which the solid exists.

Integrate through the solid.
Put the lower surface first and the upper surface second in the \(z\)-bounds, and include the factor \(r\).

Interpret the result.
The volume integral of divergence equals the net outward flux through the complete closed boundary.

This setup follows directly from the inequalities defining the solid. For example, if \(r^2\leq z\leq c\), then \(r^2\leq c\), so \(0\leq r\leq\sqrt{c}\). The angular range is a full revolution when the intersection is a circle centered on the \(z\)-axis. These bounds describe the solid without separately parameterizing the curved and planar boundary pieces.

Worked Examples

Worked Example: A Paraboloid Capped by a Horizontal Plane

Find the outward flux of \(\mathbf{F}=\langle x,y,z\rangle\) through the closed surface formed by \(z=x^2+y^2\) and \(z=4\).

First calculate the divergence:

\nabla\cdot\mathbf{F} = \frac{\partial x}{\partial x} +\frac{\partial y}{\partial y} +\frac{\partial z}{\partial z} =1+1+1=3.

The surfaces meet when \(x^2+y^2=4\), or \(r^2=4\), so \(0\leq r\leq2\) and \(0\leq\theta\leq2\pi\). Within that disk, the paraboloid is below the plane. Thus \(r^2\leq z\leq4\). The field is smooth everywhere, and the boundary is closed and piecewise smooth, so the Divergence Theorem applies.

\begin{aligned} \text{Flux} &=\int_0^{2\pi}\int_0^2\int_{r^2}^4 3r\,dz\,dr\,d\theta\\ &=2\pi\int_0^2 3r(4-r^2)\,dr\\ &=6\pi\int_0^2(4r-r^3)\,dr\\ &=6\pi\left[2r^2-\frac{r^4}{4}\right]_0^2\\ &=6\pi(8-4)=24\pi. \end{aligned}

Check: The same bounds give the solid's volume as \(2\pi\int_0^2 r(4-r^2)\,dr=8\pi\). Since the divergence is constantly \(3\), its volume integral is \(3(8\pi)=24\pi\), agreeing with the computed flux.

Worked Example: A Larger Horizontal Cap

Find the outward flux of \(\mathbf{F}=\langle 2x,2y,z\rangle\) through the surface formed by \(z=x^2+y^2\) and \(z=9\).

The divergence is constant:

\nabla\cdot\mathbf{F} = \frac{\partial (2x)}{\partial x} +\frac{\partial (2y)}{\partial y} +\frac{\partial z}{\partial z} =2+2+1=5.

The intersection satisfies \(r^2=9\), giving \(0\leq r\leq3\) and \(0\leq\theta\leq2\pi\). The vertical bounds are \(r^2\leq z\leq9\). The field is smooth on and inside this bounded, closed surface, so the theorem applies.

\begin{aligned} \text{Flux} &=\int_0^{2\pi}\int_0^3\int_{r^2}^9 5r\,dz\,dr\,d\theta\\ &=2\pi\int_0^3 5r(9-r^2)\,dr\\ &=10\pi\int_0^3(9r-r^3)\,dr\\ &=10\pi\left[\frac{9r^2}{2}-\frac{r^4}{4}\right]_0^3\\ &=10\pi\left(\frac{81}{2}-\frac{81}{4}\right)\\ &=10\pi\cdot\frac{81}{4} =\frac{405\pi}{2}. \end{aligned}

Check: The volume is \(2\pi\int_0^3r(9-r^2)\,dr=2\pi(81/4)=81\pi/2\). Multiplying by the constant divergence \(5\) gives \(5(81\pi/2)=405\pi/2\), as required.

Worked Example: A Tilted Plane and an Off-Center Projection

Find the outward flux of \(\mathbf{F}=\langle x,y,z\rangle\) through the closed surface formed by the paraboloid \(z=x^2+y^2\) and the plane \(z=2x\).

Again, \(\nabla\cdot\mathbf{F}=3\). In cylindrical coordinates, the surfaces are \(z=r^2\) and \(z=2r\cos\theta\). Their intersection satisfies \(r^2=2r\cos\theta\). Besides \(r=0\), the intersection curve is \(r=2\cos\theta\). Since \(r\geq0\), this requires \(-\pi/2\leq\theta\leq\pi/2\). The plane lies above the paraboloid for \(0\leq r\leq2\cos\theta\), so the solid has \(r^2\leq z\leq2r\cos\theta\). This angle-dependent bound describes a disk centered at \((1,0)\) in the \(xy\)-plane.

\begin{aligned} \text{Flux} &=3\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta} \int_{r^2}^{2r\cos\theta} r\,dz\,dr\,d\theta\\ &=3\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta} (2r^2\cos\theta-r^3)\,dr\,d\theta\\ &=3\int_{-\pi/2}^{\pi/2} \left[\frac{2r^3\cos\theta}{3}-\frac{r^4}{4}\right]_0^{2\cos\theta} \,d\theta\\ &=3\int_{-\pi/2}^{\pi/2} \left(\frac{16}{3}\cos^4\theta-4\cos^4\theta\right)\,d\theta\\ &=4\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta. \end{aligned}

Use \(\cos^4\theta=(3+4\cos2\theta+\cos4\theta)/8\). The sine terms vanish at the endpoints, so

\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta = \left[\frac{3\theta}{8}+\frac{\sin2\theta}{4}+\frac{\sin4\theta}{32}\right]_{-\pi/2}^{\pi/2} =\frac{3\pi}{8}.

Therefore the flux is \(4(3\pi/8)=3\pi/2\). Check: Removing the divergence factor \(3\) from the volume integral shows that the solid's volume is \(\pi/2\). Multiplying this volume by the constant divergence gives \(3(\pi/2)=3\pi/2\), confirming the result.

A Common Mistake: Omitting the Factor \(r\)

The cylindrical volume element is \(r\,dz\,dr\,d\theta\), not \(dz\,dr\,d\theta\). The factor accounts for the fact that a small angular change covers a wider arc at larger radius. Forgetting it changes the volume integral and therefore the flux.

For the first example, omitting \(r\) would lead to the incorrect calculation

\int_0^{2\pi}\int_0^2\int_{r^2}^4 3\,dz\,dr\,d\theta = 2\pi\int_0^2 3(4-r^2)\,dr = 2\pi[12r-r^3]_0^2 = 32\pi.

The one-dimensional integral here is \(24-8=16\), so the displayed value really is \(2\pi(16)=32\pi\); it is not the correct flux. The correct setup includes the Jacobian factor:

\int_0^{2\pi}\int_0^2\int_{r^2}^4 3r\,dz\,dr\,d\theta=24\pi.

This error is especially easy to miss because the omitted-factor integral still has plausible bounds and can be evaluated correctly. Always write the cylindrical volume element into the setup before integrating.

What the Method Accomplishes

Paraboloid-bounded solids are useful examples of how geometry and integration work together. The surfaces determine the limits, while the vector field determines the integrand through its divergence. A constant divergence makes the calculation a constant times the volume; a nonconstant divergence remains inside the volume integral. In either case, the outward orientation is built into the Divergence Theorem, provided the boundary is taken with its outward normal.

Before evaluating an integral, check that the solid is bounded, that the two surfaces meet along the proposed projection boundary, that the lower height is no greater than the upper height throughout the projection, and that the field satisfies the theorem's smoothness requirement. Then verify that the cylindrical Jacobian \(r\) appears exactly once.

Check Your Understanding

For each question, focus on the geometry and the setup before carrying out any lengthy calculation.

For the solid between \(z=x^2+y^2\) and \(z=16\), what are the cylindrical bounds for \(r\), \(\theta\), and \(z\)?
Why does a cylindrical volume integral include a factor of \(r\)?
For the surfaces \(z=r^2\) and \(z=6\), what equation gives the radius of their intersection?
For \(z=r^2\) and \(z=2r\cos\theta\), what angular interval permits nonnegative radial bounds?
What smoothness and boundary conditions are needed to use the Divergence Theorem for these solids?

What You'll Learn