Triple integral of a sphere in cylindrical coordinates is a fundamental technique in multivariable calculus that simplifies the evaluation of volumes, masses, and other physical quantities when the region of integration exhibits rotational symmetry about the z‑axis. But by converting Cartesian coordinates ((x,y,z)) to cylindrical coordinates ((r,\theta,z)), the integrand and the limits often become far easier to handle, especially for solids bounded by spheres, cylinders, or paraboloids. This article walks through the theory, the step‑by‑step setup, and worked examples that illustrate why cylindrical coordinates are the preferred choice for many sphere‑related problems Simple as that..
1. Why Use Cylindrical Coordinates for a Sphere?
A sphere centered at the origin with radius (R) is described in Cartesian coordinates by the equation
[ x^{2}+y^{2}+z^{2}=R^{2}. ]
When we switch to cylindrical coordinates, we use the relations
[ x = r\cos\theta,\qquad y = r\sin\theta,\qquad z = z, ]
with the Jacobian determinant (|J| = r). Substituting these into the sphere’s equation gives
[ r^{2}+z^{2}=R^{2}\quad\Longrightarrow\quad z = \pm\sqrt{R^{2}-r^{2}}. ]
Notice that the dependence on (\theta) disappears; the sphere is axisymmetric. But consequently, the integration over (\theta) simply contributes a factor of (2\pi), reducing a three‑dimensional problem to a two‑dimensional one in the ((r,z))‑plane. This simplification is the main reason cylindrical coordinates are favored for evaluating triple integrals over spheres Easy to understand, harder to ignore..
2. Setting Up the Triple Integral
The general form of a triple integral in cylindrical coordinates is
[ \iiint_{V} f(x,y,z),dV = \int_{\theta=\theta_{1}}^{\theta_{2}} \int_{r=r_{1}(\theta)}^{r_{2}(\theta)} \int_{z=z_{1}(r,\theta)}^{z_{2}(r,\theta)} f\bigl(r\cos\theta,,r\sin\theta,,z\bigr); r;dz,dr,d\theta . ]
For a sphere of radius (R) centered at the origin, the limits become:
| Variable | Lower limit | Upper limit |
|---|---|---|
| (\theta) | (0) | (2\pi) |
| (r) | (0) | (R) (the projection of the sphere onto the xy‑plane is a disk of radius (R)) |
| (z) | (-\sqrt{R^{2}-r^{2}}) | (+\sqrt{R^{2}-r^{2}}) |
Thus the integral reads
[ \boxed{ \displaystyle \int_{0}^{2\pi}\int_{0}^{R}\int_{-\sqrt{R^{2}-r^{2}}}^{;\sqrt{R^{2}-r^{2}}} f(r\cos\theta,,r\sin\theta,,z); r;dz,dr,d\theta }. ]
If the integrand (f) depends only on (r) and (z) (or is constant), the (\theta)‑integration can be performed immediately, yielding a factor (2\pi).
3. Example 1: Volume of a Sphere
To find the volume (V) of the sphere, set (f\equiv 1). The integral simplifies to
[ V = \int_{0}^{2\pi}\int_{0}^{R}\int_{-\sqrt{R^{2}-r^{2}}}^{;\sqrt{R^{2}-r^{2}}} r;dz,dr,d\theta . ]
Step‑by‑step evaluation
-
Integrate with respect to (z):
[ \int_{-\sqrt{R^{2}-r^{2}}}^{;\sqrt{R^{2}-r^{2}}} dz = 2\sqrt{R^{2}-r^{2}} . ]
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Insert the result:
[ V = \int_{0}^{2\pi}\int_{0}^{R} 2r\sqrt{R^{2}-r^{2}};dr,d\theta . ]
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Integrate with respect to (r):
Use the substitution (u = R^{2}-r^{2}), (du = -2r,dr) Still holds up..[ \int_{0}^{R} 2r\sqrt{R^{2}-r^{2}};dr = \int_{u=R^{2}}^{0} -\sqrt{u};du = \int_{0}^{R^{2}} \sqrt{u};du = \frac{2}{3} \bigl(R^{2}\bigr)^{3/2} = \frac{2}{3}R^{3}. ]
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Integrate with respect to (\theta):
[ V = \int_{0}^{2\pi} \frac{2}{3}R^{3};d\theta = \frac{2}{3}R^{3},(2\pi) = \frac{4}{3}\pi R^{3}. ]
Hence we recover the familiar volume formula (V=\frac{4}{3}\pi R^{3}) Not complicated — just consistent..
4. Example 2: Mass of a Sphere with Variable Density
Suppose the density (\rho) (mass per unit volume) varies with the distance from the z‑axis only: (\rho(r,z)=k,r^{2}), where (k) is a constant. The mass (M) is
[ M = \iiint_{V} \rho(r,z),dV = \int_{0}^{2\pi}\int_{0}^{R}\int_{-\sqrt{R^{2}-r^{2}}}^{;\sqrt{R^{2}-r^{2}}} k,r^{2}; r;dz,dr,d\theta = k\int_{0}^{2\pi}\int_{0}^{R}\int_{-\sqrt{R^{2}-r^{2}}}^{;\sqrt{R^{2}-r^{2}}} r^{3};dz,dr,d\theta . ]
Carrying out the (z)‑integration gives (2\sqrt{R^{2}-r^{2}}) as before:
[ M = 2k\int_{0}^{2\pi}\int_{0}^{R} r^{3}\sqrt{R^{2}-r^{2}};dr,d\theta . ]
The (\theta)‑integral yields a factor (2\pi):
[ M = 4\pi k \int_{0}^{R} r^{3}\sqrt{R^{2}-r^{2}};dr . ]
Use the substitution (u = R^{2}-r^{2}) ((du = -2r,dr), (r^{2}=R^{2}-u)):
[ \begin{aligned} \int_{0}^{R} r^{3}\sqrt{R^{2}-r^{2}};dr &= \int_{u=R^{2}}^{0} (R^{2}-u)\sqrt{u},\Bigl(-\frac{du}{2}\Bigr) \ &= \frac{1}{2}\int_{0}^{R^{
