Triple Integrals | Sahithyan's Notes

Let $G$ be a bounded solid region and $f$ a real-valued function defined on $G$ . Partition a rectangular box containing $G$ into sub-boxes by planes parallel to the coordinate planes, and choose a sample point $(x_k^*, y_k^*, z_k^*)$ in each resulting subregion of volume $\Delta V_k$ . The triple integral of $f$ over $G$ is

\iiint_G f\,\text{d}V = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*, y_k^*, z_k^*)\,\Delta V_k

provided the limit exists.

Properties

Suppose $f$ and $g$ integrable over $G$ , and $c \in \mathbb{R}$ :

$\displaystyle \iiint_G 1\,\text{d}V$ : volume of $G$
$\displaystyle\iiint_G f\,\text{d}V$ : mass of $G$ with $f$ density
$\displaystyle\iiint_G cf\,\text{d}V = c\iiint_G f\,\text{d}V$
$\displaystyle\iiint_G (f+g)\,\text{d}V = \iiint_G f\,\text{d}V + \iiint_G g\,\text{d}V$
If $G = G_1 \cup G_2$ with $G_1$ , $G_2$ having disjoint interiors,
$\displaystyle\iiint_G f\,\text{d}V = \iiint_{G_1} f\,\text{d}V + \iiint_{G_2} f\,\text{d}V$

Fubini’s Theorem

Let $G$ be the rectangular box $a \le x \le b$ , $c \le y \le d$ , $k \le z \le l$ .

If $f$ is continuous on $G$ :

\iiint_G f(x,y,z)\,\text{d}V = \int_a^b \int_c^d \int_k^l f(x,y,z)\,\text{d}z\,\text{d}y\,\text{d}x

The box is cut into thin slabs perpendicular to the $x$ -axis. Each slab is cut into columns parallel to the $z$ -axis. The inner integral adds up $f$ along 1 column. The middle integral adds up the columns across 1 slab. The outer integral adds up the slabs from $x = a$ to $x = b$ .

Integrating f along a column gives its total. Integrating across the columns of a slab gives the slab total. Integrating across the slabs gives the integral over the box.

If $f(x,y,z) = p(x)\,q(y)\,r(z)$ is a product of single-variable continuous functions, the iterated integral splits into 3 single integrals. Each inner integral pulls the factors that don’t depend on its variable outside, and the limits are constants because $G$ is a box.

\iiint_G p(x)\,q(y)\,r(z)\,\text{d}V = \int_a^b p(x)\,\text{d}x \cdot \int_c^d q(y)\,\text{d}y \cdot \int_k^l r(z)\,\text{d}z

Regions of Integration

A solid region $G$ is a simple $xy$ -solid region iff:

Bounded above by $z = v(x,y)$ and below by $z = u(x,y)$
$u$ , $v$ continuous on the projection $R$ of $G$ onto the $xy$ -plane
$u(x,y) \le v(x,y)$ for all $(x,y) \in R$

A solid bounded above by z = v(x, y) and below by z = u(x, y) over its projection R onto the xy-plane. Each vertical line through R enters at u and leaves at v. Drag to rotate.

For $f$ continuous on $G$ ,

\iiint_G f(x,y,z)\,\text{d}V = \iint_R \left( \int_{u(x,y)}^{v(x,y)} f(x,y,z)\,\text{d}z \right) \text{d}A

The projection $R$ can be:

Type I, or vertically simple
$R = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}$
Type II, or horizontally simple
$R = \{(x,y) : c \le y \le d,\ h_1(y) \le x \le h_2(y)\}$

If $R$ is Type I,

\iiint_G f(x,y,z)\,\text{d}V = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{u(x,y)}^{v(x,y)} f(x,y,z)\,\text{d}z\,\text{d}y\,\text{d}x

A Type I region R . Each slice at a fixed x runs in y from g₁(x) to g₂(x). The solid above the slice forms a cross-section. Drag to rotate.

If $R$ is Type II,

\iiint_G f(x,y,z)\,\text{d}V = \int_c^d \int_{h_1(y)}^{h_2(y)} \int_{u(x,y)}^{v(x,y)} f(x,y,z)\,\text{d}z\,\text{d}x\,\text{d}y

The region is similar to type 1 but the axes switched.

Change of Variables

Let $T$ be a transformation from $uvw$ -space to $xyz$ -space mapping a region $S$ onto a region $R$ . If:

$T$ is one-to-one on $S$
$T$ has continuous first partial derivatives
The Jacobian $\dfrac{\partial(x,y,z)}{\partial(u,v,w)}$ is non-zero and does not change sign on $S$

then

\iiint_R f(x,y,z)\,\text{d}V = \iiint_S f(x(u,v,w), y(u,v,w), z(u,v,w)) \left| \frac{\partial(x,y,z)}{\partial(u,v,w)} \right| \text{d}V

Cylindrical Coordinates

In cylindrical coordinates, the volume element is

\text{d}V = r\,\text{d}z\,\text{d}r\,\text{d}\theta

Here $r$ is the Jacobian $\dfrac{\partial(x,y,z)}{\partial(r,\theta,z)}$ of the change of variables.

Let $G$ have upper surface $z = v(r,\theta)$ and lower surface $z = u(r,\theta)$ , with projection onto the $xy$ -plane a simple polar region $R'$ in the $r\theta$ -plane. For $f$ continuous on $G$ ,

\iiint_G f(x,y,z)\,\text{d}V = \iint_{R'} \left( \int_{u(r,\theta)}^{v(r,\theta)} f(r\cos\theta, r\sin\theta, z)\,\text{d}z \right) r\,\text{d}A

If for each fixed $\theta \in [\alpha, \beta]$ , $r$ varies from $r_1(\theta)$ to $r_2(\theta)$ ,

\iiint_G f(x,y,z)\,\text{d}V = \int_\alpha^\beta \int_{r_1(\theta)}^{r_2(\theta)} \int_{u(r,\theta)}^{v(r,\theta)} f(r\cos\theta, r\sin\theta, z)\,r\,\text{d}z\,\text{d}r\,\text{d}\theta

Spherical Coordinates

In spherical coordinates, the volume element is

\text{d}V = \rho^2 \sin\phi\,\text{d}\rho\,\text{d}\phi\,\text{d}\theta

Here $\rho^2\sin\phi$ is the Jacobian $\dfrac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)}$ of the change of variables.

For $f$ continuous on $G$ , with appropriate limits (see common spherical regions),

\iiint_G f(x,y,z)\,\text{d}V = \iiint f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi)\,\rho^2\sin\phi\,\text{d}\rho\,\text{d}\phi\,\text{d}\theta

Worked Examples

Box

Evaluate $\iiint_G xyz\,\text{d}V$ over $G: 0 \le x \le 1,\ 0 \le y \le 2,\ 0 \le z \le 3$ .

The integrand $xyz$ is a product of single-variable functions and $G$ is a box, so the integral separates into 3 single integrals.

\int_0^1 x\,\text{d}x \cdot \int_0^2 y\,\text{d}y \cdot \int_0^3 z\,\text{d}z = \frac{1}{2} \cdot 2 \cdot \frac{9}{2} = \frac{9}{2}

Tetrahedron

Find the volume of the solid bounded by the planes $x = 0$ , $y = 0$ , $z = 0$ and $x + y + z = 1$ .

The solid lies between $z = 0$ and $z = 1 - x - y$ . The projection onto the $xy$ -plane is the triangle $0 \le x \le 1$ , $0 \le y \le 1 - x$ .

V = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} \text{d}z\,\text{d}y\,\text{d}x

Integrate with respect to $z$ .

V = \int_0^1 \int_0^{1-x} (1 - x - y)\,\text{d}y\,\text{d}x

Integrate with respect to $y$ .

V = \int_0^1 \left[ (1-x)y - \frac{y^2}{2} \right]_0^{1-x} \text{d}x = \int_0^1 \frac{(1-x)^2}{2}\,\text{d}x

Integrate with respect to $x$ .

V = \left[ -\frac{(1-x)^3}{6} \right]_0^1 = \frac{1}{6}

Paraboloid

Find the volume of the solid between the plane $z = 0$ and the paraboloid $z = 4 - x^2 - y^2$ .

The paraboloid meets $z = 0$ on the circle $x^2 + y^2 = 4$ , so the projection of the solid onto the $xy$ -plane is a disk. In Cartesian coordinates, the disk gives limits with square roots.

V = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_0^{4 - x^2 - y^2} \text{d}z\,\text{d}y\,\text{d}x

The surface and the disk both depend on $x$ and $y$ only through $x^2 + y^2$ , which is $r^2$ in cylindrical coordinates. Cylindrical coordinates turn the disk into the rectangle $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ and the height into $z = 4 - r^2$ , so every limit is simple. The volume element is $\text{d}V = r\,\text{d}z\,\text{d}r\,\text{d}\theta$ .

V = \int_0^{2\pi} \int_0^2 \int_0^{4 - r^2} r\,\text{d}z\,\text{d}r\,\text{d}\theta

Integrate with respect to $z$ .

V = \int_0^{2\pi} \int_0^2 (4r - r^3)\,\text{d}r\,\text{d}\theta

Integrate with respect to $r$ .

V = \int_0^{2\pi} \left[ 2r^2 - \frac{r^4}{4} \right]_0^2 \text{d}\theta = \int_0^{2\pi} 4\,\text{d}\theta = 8\pi

Ball

Evaluate $\iiint_G (x^2 + y^2 + z^2)\,\text{d}V$ over the unit ball $x^2 + y^2 + z^2 \le 1$ .

In Cartesian coordinates, the ball gives limits with nested square roots.

\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} (x^2 + y^2 + z^2)\,\text{d}z\,\text{d}y\,\text{d}x

The region and the integrand both depend on $x$ , $y$ and $z$ only through $x^2 + y^2 + z^2 = \rho^2$ , the squared distance from the origin. Spherical coordinates turn the ball into the box $0 \le \rho \le 1$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ and the integrand into $\rho^2$ , so every limit is constant. The volume element is $\text{d}V = \rho^2 \sin\phi\,\text{d}\rho\,\text{d}\phi\,\text{d}\theta$ .

\iiint_G (x^2 + y^2 + z^2)\,\text{d}V = \int_0^{2\pi} \int_0^\pi \int_0^1 \rho^2 \cdot \rho^2 \sin\phi\,\text{d}\rho\,\text{d}\phi\,\text{d}\theta

The integral separates.

\int_0^{2\pi} \text{d}\theta \cdot \int_0^\pi \sin\phi\,\text{d}\phi \cdot \int_0^1 \rho^4\,\text{d}\rho = 2\pi \cdot 2 \cdot \frac{1}{5} = \frac{4\pi}{5}

Cone

Find the volume of the solid enclosed laterally by the cone $\phi = \frac{\pi}{4}$ and above by the plane $z = 1$ .

From the last region in common spherical regions with $a = 1$ , the limits are $0 \le \rho \le \sec\phi$ , $0 \le \phi \le \frac{\pi}{4}$ , $0 \le \theta \le 2\pi$ .

V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{\sec\phi} \rho^2 \sin\phi\,\text{d}\rho\,\text{d}\phi\,\text{d}\theta

Integrate with respect to $\rho$ .

V = \int_0^{2\pi} \int_0^{\pi/4} \frac{\sec^3\phi \sin\phi}{3}\,\text{d}\phi\,\text{d}\theta = \frac{2\pi}{3} \int_0^{\pi/4} \tan\phi \sec^2\phi\,\text{d}\phi

Integrate with respect to $\phi$ .

V = \frac{2\pi}{3} \left[ \frac{\tan^2\phi}{2} \right]_0^{\pi/4} = \frac{2\pi}{3} \cdot \frac{1}{2} = \frac{\pi}{3}