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Math Analysis 2

Triple integrals: cylindrical and spherical coordinates

Integration by columns and by slices, cylindrical coordinates, spherical coordinates with the volume element, volumes and moments of inertia of cylinder and sphere, distributed charges.

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The triple integral is the integral over a region of space. It is used to compute volumes, masses of bodies with variable density, charges distributed through a volume, centres of mass and moments of inertia of solids. The ideas are the same as for the double integral, with one more dimension; the real novelty is cylindrical and spherical coordinates, which make problems with symmetry simple.

Definition and reduction

    \[ \iiint_V f(x, y, z)\,dx\,dy\,dz \]

is defined with sums over small boxes, just as the double integral uses rectangles. With f = 1 it gives the volume of V; with f a mass density, the mass.

It is computed by reducing it to simpler integrals, in two ways.

By columns: if V = \{(x, y)\in D,\ h_1(x, y) \le z \le h_2(x, y)\},

    \[ \iiint_V f\,dV = \iint_D\left(\int_{h_1(x, y)}^{h_2(x, y)}f\,dz\right)dx\,dy \]

By slices: if each plane z = \text{const} cuts V in a region S(z), with z between c and d,

    \[ \iiint_V f\,dV = \int_c^d\left(\iint_{S(z)}f\,dx\,dy\right)dz \]

The second method is very handy when the slices are figures with a known area.

Example: the volume of a cone of radius R and height h. The slice at height z is a circle of radius R(1 - z/h), so

    \[ V = \int_0^h\pi R^2\left(1 - \frac zh\right)^2dz = \pi R^2\cdot\frac h3 = \frac13\pi R^2 h \]

Cylindrical coordinates

They are polar coordinates in the xy plane plus the height z:

    \[ x = r\cos\theta,\qquad y = r\sin\theta,\qquad z = z,\qquad dV = r\,dr\,d\theta\,dz \]

The Jacobian is the same as for polar coordinates, r, because z does not change. They suit cylinders, cones, solids of revolution about the z axis, and anything wrapped around an axis: a coaxial cable, a drive shaft, a solenoid.

Example: the moment of inertia of a solid cylinder of mass M, radius R and height h, about its axis:

    \[ I = \rho\int_0^h\int_0^{2\pi}\int_0^R r^2\cdot r\,dr\,d\theta\,dz = \rho\cdot h\cdot 2\pi\cdot\frac{R^4}{4} = \frac12 MR^2 \]

with \rho = M/(\pi R^2 h). The same value as the disc: the height does not matter, only how the mass is spread around the axis.

Spherical coordinates

A point is located by its distance \rho from the origin, the angle \varphi from the z axis (from 0 to \pi) and the angle \theta of its projection on the xy plane (from 0 to 2\pi), like latitude and longitude on the Earth.

Spherical coordinates of a point in space with the transformation formulas and the volume element
Point P is located by its distance \rho from the origin, the angle \varphi with the z axis and the angle \theta that its projection on the xy plane makes with the x axis.

    \[ x = \rho\sin\varphi\cos\theta,\qquad y = \rho\sin\varphi\sin\theta,\qquad z = \rho\cos\varphi \]

    \[ dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta \]

Careful: many physics texts swap the names \theta and \varphi. What counts is the definition, not the letter.

Proof of the volume element

The small “cube” of spherical coordinates has three nearly perpendicular sides: d\rho in the radial direction; \rho\,d\varphi along the meridian, an arc of radius \rho; \rho\sin\varphi\,d\theta along the parallel, an arc of radius \rho\sin\varphi, the distance from the z axis. The volume is the product of the three sides, \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta. The same result comes from the determinant of the 3×3 Jacobian.

Example: the volume of a sphere.

    \[ V = \int_0^{2\pi}\int_0^\pi\int_0^R\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta = 2\pi\cdot 2\cdot\frac{R^3}{3} = \frac43\pi R^3 \]

Example: the moment of inertia of a solid sphere about a diameter (the z axis). The squared distance from the axis is x^2 + y^2 = \rho^2\sin^2\varphi:

    \[ I = \rho_m\int_0^{2\pi}\int_0^\pi\int_0^R\rho^2\sin^2\varphi\cdot\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta = \rho_m\cdot 2\pi\cdot\frac43\cdot\frac{R^5}{5} \]

because \int_0^\pi\sin^3\varphi\,d\varphi = \frac43. With \rho_m = M/(\frac43\pi R^3) you get I = \frac25 MR^2, another value from the rigid body table, now proved.

Which system to choose

Symmetry of the problemCoordinatesVolume element
boxes, prismsCartesiandx\,dy\,dz
cylinders, cones, rotation about an axiscylindricalr\,dr\,d\theta\,dz
spheres, fields depending on the distance from a pointspherical\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

In Physics 2 this choice comes up again and again: the field of a point charge is studied in spherical coordinates, that of a wire or coaxial cable in cylindrical ones, that of a parallel-plate capacitor in Cartesian ones.

An example from electromagnetism

A sphere of radius R holds a charge whose density grows outwards, \rho_q(r) = k\,r. The total charge is

    \[ Q = \int_0^{2\pi}\int_0^\pi\int_0^R k\rho\cdot\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta = 4\pi k\frac{R^4}{4} = \pi kR^4 \]

Thanks to spherical symmetry, the integrals in \theta and \varphi always give 4\pi, the total solid angle: for any function depending only on \rho, \iiint f\,dV = 4\pi\int_0^R f(\rho)\rho^2d\rho.

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