Chapter 6 • Theory & Derivations

Unit 6: Triple Integrals in Cartesian, Cylindrical & Spherical Coordinates

Three-dimensional volume integrals: Fubini's theorem in six permutations, cylindrical and spherical coordinate frames, and general curvilinear coordinate transformations via Jacobian determinants.

§6.1 Triple Integrals over General 3D Bounded Regions

1. Definition via Riemann Triple Sums

Let $f(x, y, z)$ be defined on a bounded spatial domain $E \subset \mathbb{R}^3$. Enclose $E$ inside a rectangular box $B = [a, b] \times [c, d] \times [r, s]$.

Partition $B$ into $l \times m \times n$ sub-boxes $B_{ijk}$ of dimensions $\Delta x, \Delta y, \Delta z$ with volume $\Delta V = \Delta x \Delta y \Delta z$. The triple integral is defined as the limit:

\mathbf{\iiint_E f(x, y, z) \, dV = \lim_{l, m, n \to \infty} \sum_{i=1}^l \sum_{j=1}^m \sum_{k=1}^n f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \Delta V}

Spatial Volume: When $f(x, y, z) \equiv 1$:

\mathbf{V(E) = \iiint_E 1 \, dV}

Total Mass: If $\rho(x, y, z)$ represents density:

\mathbf{M = \iiint_E \rho(x, y, z) \, dV}

2. Iterated Triple Integrals and Fubini's Permutations

A solid region $E$ is of Type 1 if it lies between the continuous boundary surfaces $z = u_1(x, y)$ and $z = u_2(x, y)$ over a planar projection domain $D$ in the $xy$-plane:

\iiint_E f(x, y, z) \, dV = \iint_D \left( \int_{u_1(x, y)}^{u_2(x, y)} f(x, y, z) \, dz \right) dA

Depending on which coordinate is integrated first and which planar projection is chosen, there are six possible orders of integration:

dz \, dy \, dx, \quad dz \, dx \, dy, \quad dy \, dz \, dx, \quad dy \, dx \, dz, \quad dx \, dz \, dy, \quad dx \, dy \, dz

Choosing the appropriate order of integration can simplify the algebraic limits substantially and avoid splitting the domain into multiple subregions.

§6.2 Triple Integrals in Cylindrical & Spherical Coordinates

1. Cylindrical Coordinates

In cylindrical coordinates, a point $P(x, y, z)$ is represented by $(r, \theta, z)$ where:

x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = z

The volume element is obtained by multiplying the polar area element by the height $dz$:

\mathbf{dV = r \, dr \, d\theta \, dz}

Transformation Formula:

\iiint_E f(x, y, z) \, dV = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} \int_{u_1(r, \theta)}^{u_2(r, \theta)} f(r\cos\theta, r\sin\theta, z) \; r \, dz \, dr \, d\theta

When to use Cylindrical Coordinates: When the solid or integrand possesses rotational symmetry around the $z$-axis (cylinders $x^2 + y^2 \le a^2$, circular cones $z^2 = x^2 + y^2$, or paraboloids $z = x^2 + y^2$).

2. Spherical Polar Coordinates

In spherical coordinates, a point is represented by $(\rho, \theta, \phi)$ where:

$\rho \ge 0$ is the distance from the origin.
$\theta \in [0, 2\pi)$ is the azimuthal angle in the $xy$-plane.
$\phi \in [0, \pi]$ is the colatitude/polar angle measured down from the positive $z$-axis.

The Cartesian coordinates are given by:

\mathbf{x = \rho \sin\phi \cos\theta, \qquad y = \rho \sin\phi \sin\theta, \qquad z = \rho \cos\phi}

x^2 + y^2 + z^2 = \rho^2

The Spherical Volume Element:

Consider a spherical wedge bounded by $\Delta\rho, \Delta\phi, \Delta\theta$. The edges of the wedge are:

Radial edge: $\Delta\rho$
Arc length along meridian: $\rho \Delta\phi$
Arc length along parallel of latitude: $(\rho \sin\phi) \Delta\theta$

Multiplying these three mutually orthogonal infinitesimal edges yields:

\mathbf{dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta}

Transformation Formula:

\iiint_E f(x, y, z) \, dV = \int_c^d \int_\alpha^\beta \int_a^b f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi) \; \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta

When to use Spherical Coordinates: When the region involves spheres ($\rho = a$), cones ($\phi = \alpha$), or integrands involving $x^2 + y^2 + z^2 = \rho^2$.

§6.3 General Change of Variables & The Jacobian Determinant

1. General Curvilinear Coordinate Transformations

Let $T: S \to R$ be a smooth, invertible transformation mapping a region $S$ in the $uv$-plane onto a region $R$ in the $xy$-plane:

x = g(u, v), \qquad y = h(u, v)

Carl Gustav Jacobi (1841) discovered that the local area expansion factor under $T$ is given by the determinant of the transformation matrix.

Definition (The Jacobian in 2D): The Jacobian of $x$ and $y$ with respect to $u$ and $v$ is:

\mathbf{J(u, v) = \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}

Change of Variables Theorem in 2D:

\mathbf{\iint_R f(x, y) \, dA = \iint_S f(g(u, v), h(u, v)) \left| \frac{\partial(x, y)}{\partial(u, v)} \right| \, du \, dv}

2. The 3D Jacobian and Space Transformations

For a transformation from $uvw$-space to $xyz$-space: $x = g(u, v, w), y = h(u, v, w), z = k(u, v, w)$:

\mathbf{\frac{\partial(x, y, z)}{\partial(u, v, w)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{vmatrix}}

Change of Variables in 3D:

\mathbf{\iiint_R f(x, y, z) \, dV = \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| \, du \, dv \, dw}

Verification for Spherical Coordinates:

\frac{\partial(x, y, z)}{\partial(\rho, \phi, \theta)} = \begin{vmatrix} \sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta \\ \sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta \\ \cos\phi & -\rho\sin\phi & 0 \end{vmatrix} = \rho^2 \sin\phi \quad \blacksquare

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Solved Problem Example 6.1: Volume of a Bounded Tetrahedron via Iterated Triple Integral

Find the volume of the solid tetrahedron $E$ bounded by the coordinate planes $x = 0, y = 0, z = 0$ and the plane:

$$x + 2y + 3z = 6$$

Step 1: Determine the Bounds of the Solid From $x + 2y + 3z = 6$:

Upper boundary for $z$: $z = \frac{6 - x - 2y}{3}$
Lower boundary for $z$: $z = 0$

Setting $z = 0$ gives the projection region $D$ in the $xy$-plane:

x + 2y = 6 \implies y = \frac{6 - x}{2}

In the first quadrant ($x \ge 0, y \ge 0$):

$x$ varies from $0$ to $6$
For a fixed $x$, $y$ varies from $0$ to $\frac{6 - x}{2}$

Thus:

E = \left\{ (x, y, z) : 0 \le x \le 6, \; 0 \le y \le \frac{6 - x}{2}, \; 0 \le z \le \frac{6 - x - 2y}{3} \right\}

Step 2: Set up the Iterated Triple Integral

V = \int_0^6 \int_0^{(6-x)/2} \int_0^{(6-x-2y)/3} 1 \, dz \, dy \, dx

Step 3: Evaluate the Inner Integral ($z$)

\int_0^{(6-x-2y)/3} dz = \frac{6 - x - 2y}{3}

Step 4: Evaluate the Middle Integral ($y$)

\int_0^{(6-x)/2} \frac{6 - x - 2y}{3} \, dy = \frac{1}{3} \left[ (6 - x)y - y^2 \right]_{y=0}^{y=(6-x)/2}

= \frac{1}{3} \left[ (6 - x)\frac{6 - x}{2} - \left(\frac{6 - x}{2}\right)^2 \right] = \frac{1}{3} \left[ \frac{(6 - x)^2}{2} - \frac{(6 - x)^2}{4} \right] = \frac{1}{3} \frac{(6 - x)^2}{4} = \frac{(6 - x)^2}{12}

Step 5: Evaluate the Outer Integral ($x$)

V = \int_0^6 \frac{(6 - x)^2}{12} \, dx

Let $u = 6 - x \implies du = -dx$:

V = \frac{1}{12} \left[ -\frac{(6 - x)^3}{3} \right]_0^6 = \frac{1}{36} [ 0 - (-6^3) ] = \frac{216}{36} = \mathbf{6} \text{ cubic units}

(Check via geometry: $V = \frac{1}{6} a b c = \frac{1}{6}(6)(3)(2) = 6$. $\checkmark$)

Solved Problem Example 6.2: Volume of an Ice-Cream Cone Solid in Spherical Coordinates

Find the volume of the solid $E$ that lies within the sphere $x^2 + y^2 + z^2 \le a^2$ and inside the upper cone $z \ge \sqrt{x^2 + y^2}$ ($a > 0$).

Step 1: Translate the Boundaries into Spherical Coordinates

1. Sphere:

x^2 + y^2 + z^2 = a^2 \implies \rho^2 = a^2 \implies \rho = a

Thus, $0 \le \rho \le a$.

2. Cone:

In spherical coordinates:

z = \rho\cos\phi, \qquad \sqrt{x^2 + y^2} = \rho\sin\phi

The cone equation becomes:

\rho\cos\phi = \rho\sin\phi \implies \tan\phi = 1 \implies \phi = \frac{\pi}{4}

The solid lies inside the cone, so $0 \le \phi \le \frac{\pi}{4}$.

3. Azimuthal Angle:

The solid has complete rotational symmetry around the $z$-axis:

0 \le \theta \le 2\pi

Step 2: Set up the Spherical Triple Integral Recall $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$:

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

Step 3: Evaluate the Factored Iterated Integrals Because the limits of integration are all constants and the integrand factors completely:

V = \left( \int_0^{2\pi} d\theta \right) \left( \int_0^{\pi/4} \sin\phi \, d\phi \right) \left( \int_0^a \rho^2 \, d\rho \right)

Evaluate each single integral:

$\int_0^{2\pi} d\theta = 2\pi$
$\int_0^{\pi/4} \sin\phi \, d\phi = [-\cos\phi]_0^{\pi/4} = -\frac{\sqrt{2}}{2} - (-1) = 1 - \frac{\sqrt{2}}{2}$
$\int_0^a \rho^2 \, d\rho = \frac{a^3}{3}$

Multiply:

V = 2\pi \left( 1 - \frac{\sqrt{2}}{2} \right) \left( \frac{a^3}{3} \right) = \mathbf{\frac{2\pi a^3}{3} \left( 1 - \frac{\sqrt{2}}{2} \right)} = \frac{\pi a^3}{3}(2 - \sqrt{2}) \approx 0.6134 a^3

Solved Problem Example 6.3: Volume of the Steinmetz Solid (Intersection of Two Cylinders)

Find the exact volume of the Steinmetz solid formed by the intersection of two solid circular cylinders of equal radius $a$ whose central axes intersect at right angles:

x^2 + y^2 \le a^2 \quad \text{and} \quad x^2 + z^2 \le a^2

Step 1: Exploit Octant Symmetry The solid is invariant under reflections across all three coordinate planes ($x \to -x$, $y \to -y$, $z \to -z$). Therefore, the total volume is 8 times the volume in the first octant ($x \ge 0, y \ge 0, z \ge 0$):

V = 8 V_{octant}

Step 2: Determine First-Octant Limits From the cylinder equations:

Cylinder 1: $y^2 \le a^2 - x^2 \implies 0 \le y \le \sqrt{a^2 - x^2}$
Cylinder 2: $z^2 \le a^2 - x^2 \implies 0 \le z \le \sqrt{a^2 - x^2}$
For both square roots to be real: $0 \le x \le a$

Notice that for any fixed $x \in [0, a]$, the cross-section perpendicular to the $x$-axis is a square of side length $\sqrt{a^2 - x^2}$!

Step 3: Set up the Triple Integral

V_{octant} = \int_0^a \int_0^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2}} 1 \, dz \, dy \, dx

Evaluate the inner $z$-integral:

\int_0^{\sqrt{a^2 - x^2}} dz = \sqrt{a^2 - x^2}

Evaluate the middle $y$-integral:

\int_0^{\sqrt{a^2 - x^2}} \sqrt{a^2 - x^2} \, dy = (\sqrt{a^2 - x^2}) \cdot (\sqrt{a^2 - x^2}) = a^2 - x^2

Step 4: Evaluate the Outer $x$-Integral

V_{octant} = \int_0^a (a^2 - x^2) \, dx = \left[ a^2 x - \frac{x^3}{3} \right]_0^a = a^3 - \frac{a^3}{3} = \frac{2}{3}a^3

Step 5: Compute Total Volume

V = 8 V_{octant} = 8 \left( \frac{2}{3}a^3 \right) = \mathbf{\frac{16}{3}a^3}

This elegant Archimedean result proves that the volume of the intersection of two orthogonal cylinders is $\frac{16}{3}a^3 \approx 5.333 a^3$ cubic units. $\blacksquare$