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The Divergence Theorem · Tutorial 973 of 999

Worked Problems on Mixed Closed Boundaries

Learn to set up and check flux integrals for closed solids whose boundaries combine planes, cylinders, cones, and curved graphs.

Advanced 11 min read

What You'll Learn

  • Identify the closed solid and its outward-oriented boundary components.
  • Check when the Divergence Theorem applies before replacing surface flux with a volume integral.
  • Use cylindrical coordinates and the correct volume element for rotationally described solids.
  • Split volume integrals where competing boundary surfaces intersect.
  • Verify flux calculations by evaluating the boundary pieces directly.
  • Recognize and correct common errors in choosing upper bounds and radial intervals.

Mixed Boundaries Require a Careful Description of the Solid

A closed solid may have several different kinds of boundary at once: a plane at its base, a cylinder at its side, and a cone or curved graph at its top. For flux problems, the Divergence Theorem can replace the sum of fluxes across all those boundary pieces with one volume integral. The theorem does not remove the need to describe the solid accurately. In particular, if two candidate top surfaces exchange roles, the volume integral must be split where they meet.

The practical sequence is to identify the solid, determine which surface bounds it in each part of the coordinate domain, select coordinates that match its geometry, and then integrate the divergence over the resulting region. Directly checking the boundary flux can expose a misplaced bound or an incorrect normal direction.

Theorem: Let \(E\) be a bounded solid with a piecewise smooth closed boundary \(\partial E\), oriented by its outward unit normal \(\mathbf{n}\). If the vector field \(\mathbf{F}\) has continuous first partial derivatives on an open set containing \(E\), then
$$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_E \nabla\cdot\mathbf{F}\,dV. $$
Here, if \(\mathbf{F}=(P,Q,R)\), its divergence is \(\nabla\cdot\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\). The surface must be closed, and the field must be regular throughout the solid; an interior singularity prevents direct use of this form of the theorem.

Choosing Coordinates and Setting Bounds

Cylindrical coordinates are often convenient when the solid contains circular cylinders, cones, or surfaces described by \(z\) as a function of the radius. They use \(x=r\cos\theta\), \(y=r\sin\theta\), and the volume element \(dV=r\,dz\,dr\,d\theta\). The factor \(r\) is essential.

To see where this volume element comes from, consider the coordinate map \((r,\theta,z)\mapsto(r\cos\theta,r\sin\theta,z)\). Its Jacobian determinant is

$$ \det \begin{pmatrix} \cos\theta & -r\sin\theta & 0\\ \sin\theta & r\cos\theta & 0\\ 0&0&1 \end{pmatrix} = r\cos^2\theta+r\sin^2\theta = r. $$

Thus a small coordinate volume has ordinary volume \(r\,dz\,dr\,d\theta\). When one surface provides the upper \(z\)-bound for some radii and another provides it for the rest, locate their intersection first. Each resulting radial interval gets its own correct bound.

Worked Example: A Paraboloid, a Plane, and a Cylinder

Worked Example: A Paraboloid, a Plane, and a Cylinder

Find the outward flux of \(\mathbf{F}=(x,y,2z)\) through the closed boundary of the solid \(E\) lying above \(z=0\), below \(z=4-x^2-y^2\), and inside \(x^2+y^2=1\).

Step 1: Check the theorem and compute the divergence. The boundary consists of the bottom plane, the cylindrical side, and the part of the paraboloid over the disk \(r\leq 1\). It is closed and piecewise smooth, and the components of \(\mathbf{F}\) are polynomials, so the theorem applies. The divergence is

$$ \nabla\cdot\mathbf{F}=1+1+2=4. $$

Step 2: Describe the region in cylindrical coordinates. The radial range is \(0\leq r\leq 1\), the angular range is \(0\leq\theta\leq 2\pi\), and \(0\leq z\leq 4-r^2\). Therefore,

$$ \begin{aligned} \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS &=\int_0^{2\pi}\int_0^1\int_0^{4-r^2}4r\,dz\,dr\,d\theta\\ &=2\pi\int_0^1 4(4-r^2)r\,dr\\ &=8\pi\left[2r^2-\frac{r^4}{4}\right]_0^1\\ &=8\pi\left(2-\frac14\right) =14\pi. \end{aligned} $$

Check by direct surface flux. On the bottom, \(z=0\), so \(\mathbf{F}\cdot(0,0,-1)=0\). On the cylindrical side \(r=1\), the outward radial component of the field is \(1\); the side runs from \(z=0\) to \(z=3\), giving flux \(1\cdot(2\pi)(3)=6\pi\). On the paraboloid, use the upward-oriented graph normal \((-f_x,-f_y,1)=(2x,2y,1)\), where \(f=4-x^2-y^2\). There, \(\mathbf{F}\cdot(2x,2y,1)=2x^2+2y^2+2(4-r^2)=8\), so the flux is \(8\cdot\pi=8\pi\) over the unit disk. The total is \(0+6\pi+8\pi=14\pi\), agreeing with the volume integral.

Piecewise Bounds When a Cone and Plane Meet

A piecewise integral is necessary when a top boundary is the lower of two candidate surfaces and which one is lower changes with radius. The surface intersection determines the split; it is not an arbitrary convenience. The two pieces cover the solid without overlapping volume, since their shared interface has zero volume.

Worked Example: A Cone, a Plane, and a Cylindrical Wall

Find the outward flux of \(\mathbf{F}=(xz,yz,z^2)\) through the closed solid inside \(r=3\), above \(z=0\), and below both \(z=r\) and \(z=2\).

Step 1: Find where the top boundary changes. The cone and plane meet when \(r=2\). For \(0\leq r\leq 2\), the cone \(z=r\) is lower; for \(2\leq r\leq 3\), the plane \(z=2\) is lower. The top bound is therefore \(r\) on the first interval and \(2\) on the second.

Step 2: Apply the theorem. The boundary is closed and piecewise smooth, and the polynomial field is continuously differentiable throughout the solid. Its divergence is \(\nabla\cdot\mathbf{F}=z+z+2z=4z\). Hence

$$ \begin{aligned} \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS &=\int_0^{2\pi}\left( \int_0^2\int_0^r 4z\,r\,dz\,dr +\int_2^3\int_0^2 4z\,r\,dz\,dr \right)d\theta\\ &=4\pi\left(\int_0^2 r^3\,dr+\int_2^3 4r\,dr\right)\\ &=4\pi\left(4+10\right) =56\pi. \end{aligned} $$

Check by boundary pieces. The bottom has zero flux because \(\mathbf{F}=0\) at \(z=0\). On the cone \(z=r\), an upward-oriented graph normal is \((-\cos\theta,-\sin\theta,1)\), with projected area element \(r\,dr\,d\theta\). The field there is \((r^2\cos\theta,r^2\sin\theta,r^2)\), whose dot product with that normal is \(-r^2+r^2=0\). The plane \(z=2\) covers the annulus \(2\leq r\leq3\); its flux is \(4\) times the annulus area, or \(4\pi(9-4)=20\pi\). On the cylinder \(r=3\), the outward radial field component is \(3z\), and \(dS=3\,d\theta\,dz\) for \(0\leq z\leq2\). Its flux is \(2\pi\int_0^2 9z\,dz=36\pi\). The total boundary flux is \(0+20\pi+36\pi=56\pi\), as required.

Worked Example: A Cone and a Plane as Competing Top Surfaces

Worked Example: A Cone and a Plane as Competing Top Surfaces

Find the outward flux of \(\mathbf{F}=(x,y,z)\) through the solid inside \(r=3\), above \(z=0\), and below both \(z=2r\) and \(z=4\).

The cone and plane meet at \(2r=4\), or \(r=2\). Thus the top is \(2r\) for \(0\leq r\leq2\), then \(4\) for \(2\leq r\leq3\). The boundary is closed and piecewise smooth, and the field is continuously differentiable throughout the solid. Since \(\nabla\cdot\mathbf{F}=3\), the flux is

$$ \begin{aligned} \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS &=3\int_0^{2\pi}\left( \int_0^2\int_0^{2r}r\,dz\,dr +\int_2^3\int_0^4 r\,dz\,dr \right)d\theta\\ &=6\pi\left(\int_0^2 2r^2\,dr+\int_2^3 4r\,dr\right)\\ &=6\pi\left(\frac{16}{3}+10\right) =92\pi. \end{aligned} $$

Check by boundary pieces. The bottom contributes zero because its normal is vertical and the field has \(z\)-component zero there. On the cone, the upward graph normal is \((-2\cos\theta,-2\sin\theta,1)\); the field on \(z=2r\) is \((r\cos\theta,r\sin\theta,2r)\), and their dot product is \(-2r+2r=0\). The plane \(z=4\) covers \(2\leq r\leq3\); its flux is \(4\) times the annulus area, giving \(4\pi(9-4)=20\pi\). On the cylindrical wall \(r=3\), the outward radial field component is \(3\), the surface element is \(3\,d\theta\,dz\), and \(0\leq z\leq4\). Its flux is \(3\cdot3\cdot(2\pi)\cdot4=72\pi\). The total is \(20\pi+72\pi=92\pi\), matching the theorem.

Common Mistake: Using One Top Bound Everywhere

In the cone-and-plane example with top \(z=\min(r,2)\), using \(z=2\) for every radius up to \(3\) ignores the cone over \(0\leq r<2\). That incorrect setup gives

$$ \int_0^{2\pi}\int_0^3\int_0^2 4z\,r\,dz\,dr\,d\theta =4\pi\int_0^3 4r\,dr =72\pi. $$

The correct flux is \(56\pi\). The wrong integral overcounts the region where the cone lies below the plane. The correction is to solve \(r=2\), then use \(z=r\) up to that radius and \(z=2\) afterward. Always test which surface is actually lower on each side of an intersection.

Why the Setup Matters

Mixed-boundary problems reward a geometric check before any integration begins. Draw or describe the solid, identify the intersections of candidate boundary surfaces, and test a point in each resulting interval to determine which surface is active. Then verify that the region is closed and that the field has no singularity inside it before applying the Divergence Theorem. A direct boundary calculation, when manageable, provides a strong independent check of the resulting flux.

Check Your Understanding

Use these questions to test your setup and interpretation of mixed-boundary flux problems.

  1. For a solid with top surfaces \(z=r\) and \(z=3\), at what radius do the candidate upper bounds exchange roles?
  2. Why does a cylindrical-coordinate volume integral include a factor of \(r\)?
  3. What conditions on the boundary and vector field are needed for the Divergence Theorem in the form stated here?
  4. If a cone is below a plane for small radii and above it for larger radii, what information determines where the radial integral must be split?
  5. In a direct flux check, why must the normal on each boundary piece point outward from the solid?

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