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 $0\le r\le1$, $0\le\theta\le2\pi$, $0\le z\le2-r^2$:

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

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.