Triple integrals and three-dimensional coordinates

Accumulation in space

For a continuous density $\rho$ on a suitable bounded solid $E$, its mass is $M=\iiint_E\rho\,dV$. For constant density one, the integral is the volume. Iterated integration slices the solid successively, provided the bounds actually describe $E$; Fubini's theorem applies under standard continuity and regularity assumptions.

Worked example

A tetrahedral solid

Let $E=\{x,y,z\ge0:x+y+z\le1\}$. The bounds $0\le x\le1$, $0\le y\le1-x$, $0\le z\le1-x-y$ give

$$\operatorname{Vol}(E)=\int_0^1\int_0^{1-x}\int_0^{1-x-y}1\,dz\,dy\,dx=\frac16.$$

The inner upper bound is the plane expressed as $z=1-x-y$; projecting $E$ onto the $xy$-plane gives the triangular outer bounds.

Cylindrical and spherical coordinates

Definition

Coordinate conventions and volume elements

Cylindrical coordinates use $x=r\cos\theta$, $y=r\sin\theta$, $z=z$ with $r\ge0$ and $dV=r\,dr\,d\theta\,dz$. Spherical coordinates here use

$$x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi,$$

where $\rho\ge0$, $0\le\phi\le\pi$ is the angle from the positive $z$-axis and $\theta$ is azimuth. The volume element is $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$. Other sources may exchange names of the two angles; check definitions before using a formula.

Proof

Why the spherical factor is $\rho^2\sin\phi$

Differentiate the position vector with respect to $\rho,\phi,\theta$. The absolute value of their scalar triple product is $\rho^2\sin\phi$ for $0\le\phi\le\pi$. The three-dimensional change-of-variables theorem then supplies the volume element; coordinate singularities on the axis and at the origin are negligible in standard solid integrations.

Worked example

A sphere above a cone

Let $E$ be inside $x^2+y^2+z^2\le4$ and above the cone $z=\sqrt{x^2+y^2}$. In spherical coordinates this is $0\le\rho\le2$, $0\le\phi\le\pi/4$, $0\le\theta\le2\pi$. Hence

$$\operatorname{Vol}(E)=\int_0^{2\pi}\int_0^{\pi/4}\int_0^2\rho^2\sin\phi\,d\rho\,d\phi\,d\theta =\frac{16\pi}{3}\left(1-\frac{\sqrt2}{2}\right).$$

The cone fixes an angle, and the sphere fixes a radius. Those geometric observations determine bounds before any integration.

Worked example

Cylindrical shells

For the solid $x^2+y^2\le1$, $0\le z\le2-x^2-y^2$, use $0\le r\le1$, $0\le\theta\le2\pi$, $0\le z\le2-r^2$:

$$\operatorname{Vol}(E)=\int_0^{2\pi}\int_0^1\int_0^{2-r^2}r\,dz\,dr\,d\theta=\frac{3\pi}{2}.$$

2. Derive spherical scale before using radial symmetry

Use coordinates

$$T(r,\phi,\theta)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),$$

where $\phi$ is measured from the positive $z$ axis, $0\le\phi\le\pi$, and $\theta$ is azimuth over one full turn. The derivative vectors have lengths $1,r,r\sin\phi$ and are mutually perpendicular with the corresponding positive orientation on the interior. Their scalar triple product is $r^2\sin\phi$, giving

$$dV=r^2\sin\phi\,dr\,d\phi\,d\theta.$$

To verify the determinant explicitly, the derivative columns in the order $(r,\phi,\theta)$ are

$$T_r=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$ $$T_\phi=(r\cos\phi\cos\theta,r\cos\phi\sin\theta,-r\sin\phi),\qquad T_\theta=(-r\sin\phi\sin\theta,r\sin\phi\cos\theta,0).$$

Their triple product is $r^2\sin\phi>0$ when $r>0$ and $0<\phi<\pi$. On the open parameter domain $(0,\infty)\times(0,\pi)\times(-\pi,\pi)$ the map is one-to-one onto its slit image, with smooth inverse $r=\sqrt{x^2+y^2+z^2}$, $\phi=\arccos(z/r)$ and the usual smooth slit-plane azimuth. Thus the proved finite-dimensional change-of-variables theorem applies to every compact truncated parameter box inside that domain.

The omitted target regions also need estimates independent of the formula we are proving. The ball of radius $\varepsilon$ is in a cube of volume $8\varepsilon^3$. In a radius-$R$ ball, the two polar caps with angle at most $\delta<\pi/2$ from the axis lie in the box $|x|,|y|\le R\sin\delta$, $|z|\le R$, of volume $8R^3\sin^2\delta$. An azimuth seam of total angular width $2\eta$, with $0<\eta<\pi/2$, lies in $|x|,|z|\le R$, $|y|\le R\sin\eta$, a box of volume $8R^3\sin\eta$. All three bounds vanish with their cutoff. Multiplying by a bound for the continuous target integrand controls its omitted integral, while the parameter estimates below control the other side. This completes the nonsingular-exhaustion argument without using spherical volume to justify its own formula.

The origin, poles and angular seam are coordinate-boundary exceptions. For bounded continuous integrands on a ball, use nonsingular truncated parameter regions and pass to their limit; the omitted parameter integral bounds tend to zero. For a bounded integrand, the radial factor near zero is bounded by a constant times $\int_0^\varepsilon r^2dr=\varepsilon^3/3$, while a polar cap has factor $\int_0^\varepsilon\sin\phi\,d\phi=1-\cos\varepsilon\to0$ and a shrinking azimuth seam has vanishing angle width. The formula is not a license to ignore an arbitrary unbounded singularity at those locations.

Choosing coordinates

Cartesian coordinates suit plane-bounded solids; cylindrical coordinates suit circular cross-sections around an axis; spherical coordinates suit spheres and cones about the origin. Transform the integrand as well as the bounds and include the Jacobian. Angles must cover the intended solid exactly once except for negligible coordinate seams. Triple integration computes signed accumulation if the integrand changes sign; geometric volume uses integrand $1$.

Practice

1. Find the volume of the box $0\le x\le2$, $0\le y\le3$, $0\le z\le4$.
2. Find the volume of the cylinder $x^2+y^2\le4$, $0\le z\le5$ using cylindrical coordinates.
3. Find the volume of the ball $x^2+y^2+z^2\le R^2$ using spherical coordinates.
4. Rewrite $x^2+y^2+z^2$ and $z$ in the spherical convention of this unit.

Full solutions

1. Integrate $1$ over the box: $2\cdot3\cdot4=24$.
2. $\int_0^{2\pi}\int_0^2\int_0^5r\,dz\,dr\,d\theta=20\pi$.
3. For $R\ge0$, $\int_0^{2\pi}\int_0^\pi\int_0^R\rho^2\sin\phi\,d\rho\,d\phi\,d\theta=4\pi R^3/3$.
4. They are $\rho^2$ and $\rho\cos\phi$, respectively.

Synthesis

A triple integral is accumulated density over volume. Choose coordinates adapted to the solid, express every boundary and the integrand in those coordinates, and include the appropriate volume factor.