Triple Integrals over Cone-Sphere Regions

Reading a Cone-Sphere Region in Spherical Coordinates

A solid bounded by a cone and a sphere often has a particularly clear description in spherical coordinates. The cone controls the range of polar angles, while the sphere controls how far each ray extends. The important step is to derive both bounds from the surfaces and verify where the surfaces meet.

Use spherical coordinates \((\rho,\phi,\theta)\), with \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\). Here \(\rho\geq 0\) is distance from the origin, \(\phi\) is the angle down from the positive \(z\)-axis, and \(\theta\) is the azimuthal angle in the \(xy\)-plane. Write \(r=\sqrt{x^2+y^2}\) for the horizontal distance from the \(z\)-axis; thus \(r=\rho\sin\phi\), while \(z=\rho\cos\phi\). The volume element is \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\).

Deriving the Cone and Sphere Bounds

Consider the upper nappe of the cone \(z=kr\), where \(k>0\). At points on this cone with \(\rho>0\), substitution of \(z=\rho\cos\phi\) and \(r=\rho\sin\phi\) gives \(\cos\phi=k\sin\phi\). Therefore the cone has constant polar angle \(\alpha=\arctan(1/k)\). The part of space above the cone, toward the positive \(z\)-axis, has \(0\leq\phi\leq\alpha\).

Now consider a sphere centered at \((0,0,c)\) with radius \(R\). Its equation is \(r^2+(z-c)^2=R^2\). Since \(r^2+z^2=\rho^2\), its spherical equation is \(\rho^2-2c\rho\cos\phi+c^2-R^2=0\). In general, solving this quadratic gives candidate radial intersections; one must choose the roots that actually describe the desired solid.

A particularly useful case is a sphere passing through the origin, with center \((0,0,R)\) and radius \(R\). Its equation simplifies to \(r^2+(z-R)^2=R^2\), or \(r^2+z^2=2Rz\). Substitution gives \(\rho^2=2R\rho\cos\phi\), so the sphere meets a ray at \(\rho=0\) and at \(\rho=2R\cos\phi\). For the portion inside the sphere along that ray, \(0\leq\rho\leq 2R\cos\phi\), provided \(\cos\phi\geq0\).

Worked Example: Volume Inside a Sphere and Above a Cone

Why the Intersection Determines the Limits

The radial limit is not an independent guess. Fix an angle \(\phi\), and imagine moving outward along the ray from the origin with that direction. If the ray passes through the solid, it starts at \(\rho=0\) and leaves the sphere where the sphere equation is satisfied. At the edge of the angular range, the ray also lies on the cone. Therefore, the last point on that ray satisfies both surface equations.

For a sphere through the origin and the cone \(z=kr\), the cone angle obeys \(\tan\alpha=1/k\). At that angle, the nonzero sphere intersection is \(\rho=2R\cos\alpha\). The corresponding horizontal distance and height are \(r=\rho\sin\alpha\) and \(z=\rho\cos\alpha\). Since \(\cos\alpha=k\sin\alpha\), these satisfy \(z=kr\), as required. This calculation is a direct check that the radial and angular boundaries meet on both surfaces.

Worked Example: A Steeper Cone and a Larger Sphere

Integrating a Function over a Cone-Sphere Solid

Once the bounds are correct, a triple integral uses the same region with the chosen integrand. For example, the height \(z\) becomes \(\rho\cos\phi\). The volume element still contributes \(\rho^2\sin\phi\), so the full integrand for integrating \(z\) is \(\rho^3\cos\phi\sin\phi\). Keeping the Jacobian factor \(\rho^2\sin\phi\) is essential: it is part of \(dV\), not part of the function being integrated.

A Common Bound-Setting Mistake

A frequent error is to mistake the angle \(\phi\) for the angle measured up from the \(xy\)-plane. With the convention used here, \(\phi\) is measured down from the positive \(z\)-axis. For the cone \(z=r\), an incorrect setup might claim that the cone has \(\phi=\pi/4\) but that the region above it has \(\pi/4\leq\phi\leq\pi/2\). That describes directions below the cone, not above it.

For a concrete check, a direction on the positive \(z\)-axis has \(\phi=0\). It is above the cone, so the allowed angular interval must include \(0\). The corrected interval is \(0\leq\phi\leq\pi/4\). A second error is to treat the intersection at \(\rho=\sqrt2,\phi=\pi/4\) as if both \(r\) and \(z\) were \(\sqrt2/2\). In fact, \(r=\rho\sin\phi=1\) and \(z=\rho\cos\phi=1\). Substituting these actual coordinates confirms both \(z=r\) and \(r^2+(z-1)^2=1\).

Setting Up the Region Reliably

Why These Regions Are Useful

Cone-sphere regions are a focused setting for coordinating angular and radial bounds. A cone passing through the origin has a constant polar angle, while a sphere centered on the axis can have a radial boundary that varies with that angle. Their intersection ties the two descriptions together and provides a natural check on the setup.

For a full rotationally symmetric solid, the azimuthal interval is \(0\leq\theta\leq2\pi\). If the solid is restricted to a wedge, that interval must be changed to match the region; it should not be assumed to be a full revolution automatically. In every case, the bounds must describe the solid before the integrand is evaluated.