Why Conical Boundaries Call for Careful Setup
A cone may be the side of a closed solid, a boundary between two parts of a solid, or a surface that meets a plane or sphere. In each case, the Divergence Theorem turns the total outward flux through the closed boundary into an integral of divergence over the enclosed volume. The main work is then choosing bounds that describe the solid correctly and keeping the outward orientation straight.
Cylindrical coordinates are often convenient when a cone is described by a relation between height and distance from the axis. Spherical coordinates are useful when a cone cuts a sphere, because the cone becomes a constant-angle boundary. In either system, the bounds must follow from the actual region, not merely from the equation of the cone.
Writing a Cone in Cylindrical Coordinates
Take the vertical cone \(z=kr\), where \(k>0\), \(r=\sqrt{x^2+y^2}\), and \(z\) is height. The parameter \(k\) controls the slope: a larger \(k\) gives a steeper cone. For a solid above this cone and below the plane \(z=H\), with \(H>0\), the cone meets the plane where \(kr=H\), so \(r=H/k\). The cylindrical bounds are therefore \(0\leq r\leq H/k\), \(kr\leq z\leq H\), and \(0\leq\theta\leq2\pi\).
For a solid below the cone and inside the cylinder \(r=R\), the bounds instead are \(0\leq r\leq R\), \(0\leq z\leq kr\), and \(0\leq\theta\leq2\pi\). The cylindrical volume element is \(dV=r\,dz\,dr\,d\theta\). That factor \(r\) is essential; it accounts for the widening of cylindrical shells as their distance from the axis increases.
Outward Orientation on the Cone
For the cone \(z=kr\), a parametrization is \(\mathbf X(r,\theta)=(r\cos\theta,r\sin\theta,kr)\). Differentiating and taking the cross product gives a vector-area element. Its direction matters: a surface normal that points into the solid will reverse the flux sign.
To identify its orientation, define \(g(x,y,z)=z-k\sqrt{x^2+y^2}\). The cone is \(g=0\), and \(\nabla g\) points toward increasing \(g\). A solid below the cone has \(g\leq0\), so its outward direction on the cone points toward increasing \(g\). This agrees with \(\mathbf X_r\times\mathbf X_\theta\). A solid above the cone has \(g\geq0\), so its outward direction points toward decreasing \(g\), opposite to that cross product.
Worked Example: A Solid Above a Cone and Below a Plane
Worked Example: A Conical Solid Capped by a Plane
Let \(E\) be the solid \(r\leq z\leq2\), with \(0\leq\theta\leq2\pi\), and let \(\mathbf F=(x,y,z)\). Find the outward flux through the whole boundary.
The cone is \(z=r\), so the solid lies above it. At the plane \(z=2\), the cone has radius \(r=2\). The boundary consists of the conical side and the top disk. Since \(\nabla\cdot\mathbf F=1+1+1=3\), the Divergence Theorem gives \[ \iint_{\partial E}\mathbf F\cdot\mathbf n\,dS = \int_0^{2\pi}\int_0^2\int_r^2 3r\,dz\,dr\,d\theta. \]
Evaluate the integral one variable at a time: \[ \int_r^2 3r\,dz=3r(2-r), \qquad \int_0^2 3r(2-r)\,dr =3\left[r^2-\frac{r^3}{3}\right]_0^2=4. \] The remaining angular integral is \[ \int_0^{2\pi}4\,d\theta=8\pi. \] Thus the outward flux is \(8\pi\).
Check this by calculating the boundary flux pieces. The top disk has outward normal \((0,0,1)\), and there \(z=2\), so \(\mathbf F\cdot\mathbf n=2\). Its flux is \(2\cdot\pi(2)^2=8\pi\). On the cone, the solid is above, so the outward vector-area element is \((kr\cos\theta,kr\sin\theta,-r)\,dr\,d\theta\) with \(k=1\). There \(\mathbf F=(r\cos\theta,r\sin\theta,r)\), and the dot product is \(r^2-r^2=0\). The cone contributes zero. The direct boundary calculation also gives \(8\pi\).
Worked Example: A Solid Below a Cone and Inside a Cylinder
Worked Example: A Cone Over a Circular Base
Let \(E\) be bounded by \(z=0\), the cone \(z=2r\), and the cylinder \(r=1\). For \(\mathbf F=(0,0,z^2)\), find the outward flux through the closed boundary.
The region lies below the cone, so its cylindrical bounds are \(0\leq r\leq1\), \(0\leq z\leq2r\), and \(0\leq\theta\leq2\pi\). The divergence is \[ \nabla\cdot\mathbf F = \frac{\partial 0}{\partial x} +\frac{\partial 0}{\partial y} +\frac{\partial z^2}{\partial z} =2z. \] Therefore, \[ \iint_{\partial E}\mathbf F\cdot\mathbf n\,dS = \int_0^{2\pi}\int_0^1\int_0^{2r}2z\,r\,dz\,dr\,d\theta. \]
First integrate with respect to \(z\): \[ \int_0^{2r}2zr\,dz =r[z^2]_0^{2r} =4r^3. \] Then \[ \int_0^1 4r^3\,dr=[r^4]_0^1=1, \qquad \int_0^{2\pi}1\,d\theta=2\pi. \] The outward flux is \(2\pi\).
For a direct check, the bottom disk contributes zero because \(z=0\) there. On the cylindrical side, the normal is horizontal and \(\mathbf F\) is vertical, so the flux is zero. On the cone, the solid is below, so the outward vector-area element is \((-2r\cos\theta,-2r\sin\theta,r)\,dr\,d\theta\). The field there is \((0,0,4r^2)\), and their dot product is \(4r^3\). Hence the cone flux is \[ \int_0^{2\pi}\int_0^1 4r^3\,dr\,d\theta =2\pi. \] The direct calculation agrees with the volume integral.
When a Cone Meets a Sphere
Spherical coordinates describe points using their distance \(\rho\) from the origin, polar angle \(\phi\) measured down from the positive \(z\)-axis, and azimuthal angle \(\theta\). They satisfy \(z=\rho\cos\phi\) and \(r=\rho\sin\phi\). Substituting these relations into \(z=kr\) gives \[ \rho\cos\phi=k\rho\sin\phi. \] Away from the origin, division by \(\rho\sin\phi\) yields \(\cot\phi=k\), or \(\tan\phi=1/k\). Thus the cone corresponds to the constant angle \(\alpha=\arctan(1/k)\). Points above the cone have \(0\leq\phi\leq\alpha\). The spherical volume element is \(\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\).
Worked Example: A Spherical Solid Above a Cone
Worked Example: A Cone Cutting a Sphere
Let \(E\) be inside the sphere \(\rho\leq a\) and above the cone \(z=\sqrt{3}\,r\), where \(a>0\). For \(\mathbf F=(x,y,z)\), find the outward flux through the entire boundary.
Here \(k=\sqrt{3}\), so the cone angle is \(\alpha=\arctan(1/\sqrt{3})=\pi/6\). The bounds are \(0\leq\rho\leq a\), \(0\leq\phi\leq\pi/6\), and \(0\leq\theta\leq2\pi\). Since \(\nabla\cdot\mathbf F=3\), \[ \iint_{\partial E}\mathbf F\cdot\mathbf n\,dS = \int_0^{2\pi}\int_0^{\pi/6}\int_0^a 3\rho^2\sin\phi\,d\rho\,d\phi\,d\theta. \]
The radial integral is \[ \int_0^a3\rho^2\,d\rho=a^3. \] The polar integral is \[ \int_0^{\pi/6}\sin\phi\,d\phi =[-\cos\phi]_0^{\pi/6} =1-\frac{\sqrt{3}}{2}. \] The azimuthal integral contributes \(2\pi\). Therefore the flux is \[ 2\pi a^3\left(1-\frac{\sqrt{3}}{2}\right). \]
Check the boundary pieces. On the spherical cap \(\rho=a\), the outward normal is radial and \(\mathbf F\cdot\mathbf n=a\). The surface element is \(a^2\sin\phi\,d\phi\,d\theta\), so its flux is \[ \int_0^{2\pi}\int_0^{\pi/6}a^3\sin\phi\,d\phi\,d\theta = 2\pi a^3\left(1-\frac{\sqrt{3}}{2}\right). \] On the conical side, \(\mathbf F\) is tangent to the cone: its dot product with the cone’s normal is zero. Thus the cone contributes no flux, and the direct check matches the volume integral.
A Sign Error to Avoid
Suppose the second example is evaluated with the cone’s vector-area element reversed. The incorrectly chosen element would be \((2r\cos\theta,2r\sin\theta,-r)\,dr\,d\theta\). Its dot product with \((0,0,4r^2)\) is \(-4r^3\), giving cone flux \(-2\pi\). That sign is wrong: the solid lies below the cone, so the outward direction points toward increasing \(z-2r\), not decreasing it. The correct element is \((-2r\cos\theta,-2r\sin\theta,r)\,dr\,d\theta\), which gives \(+2\pi\), consistent with the positive divergence integral.
More generally, a negative flux through one boundary piece is not automatically an error: the field may point inward through that piece. But reversing the normal changes the sign of that piece’s flux. Check the side of the cone containing the solid before assigning an orientation.
Choosing Coordinates and Checking the Setup
Use cylindrical coordinates when the cone is naturally expressed as \(z=kr\), especially when planes or cylinders give simple radial and height limits. Use spherical coordinates when the solid is cut by a sphere and the conical boundary passes through the origin, since the cone then becomes a fixed polar angle.
Before integrating, verify that the limits describe the intended solid. For example, a region above \(z=kr\) and below \(z=H\) must have \(0\leq r\leq H/k\); using a larger radial bound would include points where the lower height \(kr\) exceeds the cap \(H\). Also identify every piece of the closed boundary and orient each piece outward. The Divergence Theorem gives the sum of fluxes over all those pieces, not the flux through the cone alone.
Check Your Understanding
Use the definitions, bounds, and orientation rules above to answer these questions.
- For the solid above \(z=2r\) and below \(z=6\), what is the maximum cylindrical radius?
- For a solid below \(z=kr\), which direction does the outward normal point relative to the cone?
- What polar-angle bound describes points above \(z=kr\) in spherical coordinates?
- Why must a cylindrical volume integral include a factor of \(r\)?
- For a closed solid bounded by a cone, a plane, and a cylinder, does the Divergence Theorem give flux through the cone alone or through the entire boundary?