5.1 Multiple Integrals in Cylindrical and Spherical Coordinates

5.1 Multiple Integrals in Cylindrical and Spherical Coordinates

Multiple integrals generalize single-variable integration to higher-dimensional Euclidean domains, computing volumes, masses, and spatial averages. On the GRE Mathematics Subject Test, questions test reversing integration order via Fubini's Theorem, choosing the optimal coordinate system, and correctly setting up limits for quadric surfaces.

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Double Integrals Over General Regions and Fubini's Theorem

Let $R \subset \mathbb{R}^2$ be a bounded, closed (compact), rectifiable region. The double integral of a bounded function $f: R \to \mathbb{R}$ is defined as the limit of Riemann sums: $\iint_R f(x, y) \, dA = \lim_{\|P\| \to 0} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \, \Delta A_{ij}.$

When $f$ is continuous on $R$, integrability is guaranteed. Regions in the plane are classified into two standard types:

• Type I (Vertically Simple): $R = {(x, y) \in \mathbb{R}^2 : a \le x \le b, ; g_1(x) \le y \le g_2(x)}$.
• Type II (Horizontally Simple): $R = {(x, y) \in \mathbb{R}^2 : c \le y \le d, ; h_1(y) \le x \le h_2(y)}$.

Fubini's Theorem

If $f(x, y)$ is continuous on a rectifiable region $R$, the double integral equals the iterated integral in either order: $\iint_R f(x, y) \, dA = \int_a^b \left( \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \right) dx = \int_c^d \left( \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \right) dy.$

Reversing the Order of Integration

A quintessential GRE technique involves reversing the order of integration when the inner integral possesses no elementary antiderivative, such as $\int e^{x^2} , dx$, $\int \frac{\sin x}{x} , dx$, or $\int \cos(y^2) , dy$. The strategy is strictly geometric:

1. Extract the current boundary inequalities (e.g., $0 \le y \le 1$ and $y \le x \le 1$).
2. Sketch the planar domain $R$.
3. Formulate the alternative slice bounds (e.g., $0 \le x \le 1$ and $0 \le y \le x$).
4. Execute the new iterated integral, where the inner antiderivative is now multiplied by an algebraic power of the variable.

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Polar Coordinates in the Plane

When domains or integrands exhibit rotational symmetry about the origin, planar Cartesian coordinates $(x, y)$ are transformed into polar coordinates $(r, \theta)$: $x = r \cos \theta, \quad y = r \sin \theta, \quad x^2 + y^2 = r^2,$ where $r \ge 0$ and $\theta \in [0, 2\pi)$ (or $[-\pi, \pi)$).

The Area Element

The differential area element transforms with the radial distortion factor $r$: $dA = dx \, dy = r \, dr \, d\theta.$ This factor arises from the Jacobian determinant $\frac{\partial(x, y)}{\partial(r, \theta)} = r$. Forgetting $r$ is the single most common error on multivariable integration questions.

Canonical Polar Curves

• Concentric Circles: $x^2 + y^2 \le a^2 \iff 0 \le r \le a, ; 0 \le \theta \le 2\pi$.
• Offset Circles: $(x - a)^2 + y^2 = a^2 \iff r = 2a \cos \theta$ for $-\pi/2 \le \theta \le \pi/2$.
• Offset Circles along y-axis: $x^2 + (y - a)^2 = a^2 \iff r = 2a \sin \theta$ for $0 \le \theta \le \pi$.
• The Gaussian Integral: Evaluating $I = \int_{-\infty}^\infty e^{-x^2} , dx$ via $I^2 = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)} , dA = \int_0^{2\pi} d\theta \int_0^\infty r e^{-r^2} , dr = 2\pi \left( \frac{1}{2} \right) = \pi \implies I = \sqrt{\pi}$.

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Triple Integrals in Cylindrical Coordinates

Cylindrical coordinates $(r, \theta, z)$ combine polar coordinates in the $xy$-plane with the Cartesian vertical height $z$: $x = r \cos \theta, \quad y = r \sin \theta, \quad z = z.$

Volume Element

$dV = dz \, dA = r \, dz \, dr \, d\theta.$

Quadric Surfaces with Axial Symmetry

• Circular Cylinder: $x^2 + y^2 = R^2 \iff r = R$.
• Circular Paraboloid: $z = c(x^2 + y^2) \iff z = c r^2$.
• Circular Cone: $z = c \sqrt{x^2 + y^2} \iff z = c r$.
• Sphere: $x^2 + y^2 + z^2 = R^2 \iff z = \pm \sqrt{R^2 - r^2}$.

Cylindrical coordinates are optimal when the projection of the 3D solid onto the $xy$-plane is circular, annular, or sectorial, while the vertical bounds $z_1(r, \theta) \le z \le z_2(r, \theta)$ are straightforward algebraic expressions in $r$.

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Triple Integrals in Spherical Coordinates

Spherical coordinates $(\rho, \varphi, \theta)$ represent points in $\mathbb{R}^3$ via radial distance $\rho$, colatitude (polar angle) $\varphi$, and azimuthal angle $\theta$:

• $\rho = \sqrt{x^2 + y^2 + z^2} \ge 0$: distance from the origin.
• $\varphi \in [0, \pi]$: angle measured downward from the positive $z$-axis ($\varphi = 0$ is north pole, $\varphi = \pi/2$ is $xy$-plane, $\varphi = \pi$ is south pole).
• $\theta \in [0, 2\pi)$: azimuthal angle in the $xy$-plane measured from the positive $x$-axis.

Transformation Equations

$x = \rho \sin \varphi \cos \theta, \quad y = \rho \sin \varphi \sin \theta, \quad z = \rho \cos \varphi.$ Note that the cylindrical radial coordinate relates via $r = \rho \sin \varphi$.

Spherical Volume Element

$dV = \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta.$

Canonical Boundaries in Spherical Coordinates

• Sphere of radius $R$: $\rho = R$.
• Cone with apex at origin: $z = \sqrt{x^2 + y^2} \iff \rho \cos \varphi = \rho \sin \varphi \iff \tan \varphi = 1 \iff \varphi = \pi/4$.
• Ice-cream cone solid (inside sphere $\rho \le R$ and cone $\varphi \le \alpha$): $0 \le \rho \le R, \quad 0 \le \varphi \le \alpha, \quad 0 \le \theta \le 2\pi.$

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Coordinate System Selection Guide

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Step-by-Step Worked Problems

Problem 1: Reversing Integration Order

Evaluate the iterated integral $\int_0^4 \int_{\sqrt{y}}^2 \frac{y}{\sqrt{x^5 + 1}} , dx , dy$.

Solution:

1. The current bounds describe $R = {(x, y) : 0 \le y \le 4, ; \sqrt{y} \le x \le 2}$.
2. The curve $x = \sqrt{y}$ is the parabola $y = x^2$. Since $x$ ranges from $\sqrt{y}$ to $2$ and $y$ ranges from $0$ to $4$, the domain in Type I form is: $0 \le x \le 2, \quad 0 \le y \le x^2.$
3. Interchange the integrals: $\int_0^2 \left( \int_0^{x^2} \frac{y}{\sqrt{x^5 + 1}} \, dy \right) dx = \int_0^2 \frac{1}{\sqrt{x^5 + 1}} \left[ \frac{y^2}{2} \right]_0^{x^2} dx = \int_0^2 \frac{x^4}{2\sqrt{x^5 + 1}} \, dx.$
4. Let $u = x^5 + 1$, so $du = 5x^4 , dx$. The limits transform from $x = 0 \implies u = 1$ to $x = 2 \implies u = 33$: $\int_1^{33} \frac{1}{2\sqrt{u}} \cdot \frac{du}{5} = \frac{1}{10} \left[ 2\sqrt{u} \right]_1^{33} = \frac{\sqrt{33} - 1}{5}.$

Problem 2: Spherical Coordinates Integration

Compute the triple integral $\iiint_B (x^2 + y^2 + z^2) , dV$, where $B$ is the unit ball $x^2 + y^2 + z^2 \le 1$.

Solution:

1. In spherical coordinates, $x^2 + y^2 + z^2 = \rho^2$. The unit ball $B$ is defined by $0 \le \rho \le 1$, $0 \le \varphi \le \pi$, and $0 \le \theta \le 2\pi$.
2. The volume element is $dV = \rho^2 \sin \varphi , d\rho , d\varphi , d\theta$.
3. Substitute and factor into independent single integrals: $\iiint_B \rho^2 (\rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta) = \left( \int_0^{2\pi} d\theta \right) \left( \int_0^\pi \sin \varphi \, d\varphi \right) \left( \int_0^1 \rho^4 \, d\rho \right).$
4. Compute each factor:
   • $\int_0^{2\pi} d\theta = 2\pi$
   • $\int_0^\pi \sin \varphi , d\varphi = [-\cos \varphi]_0^\pi = -(-1) - (-1) = 2$
   • $\int_0^1 \rho^4 , d\rho = \left[ \frac{\rho^5}{5} \right]_0^1 = \frac{1}{5}$
5. The product is $2\pi \times 2 \times \frac{1}{5} = \frac{4\pi}{5}$.

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GRE Exam Traps & Pitfalls

Trap 1: Dropping the Jacobian Metric Factors Always insert $r$ in polar/cylindrical integrals ($dA = r , dr , d\theta, ; dV = r , dz , dr , d\theta$) and $\rho^2 \sin \varphi$ in spherical integrals ($dV = \rho^2 \sin \varphi , d\rho , d\varphi , d\theta$). Forgetting these metric factors is the most frequent computational error under time pressure.

Trap 2: Colatitude $\varphi$ Range Misconception The angle $\varphi$ in spherical coordinates only ranges from $0$ to $\pi$, NEVER $0$ to $2\pi$. The azimuthal angle $\theta$ sweeps out the full $2\pi$ circle around the $z$-axis. Integrating $\varphi$ from $0$ to $2\pi$ doubles the domain and leads to invalid signed cancellations.

Trap 3: Reversing Bounds Without Sketching Never invert integration bounds purely by algebraic transposition. Swapping $\int_0^1 \int_y^1$ into $\int_y^1 \int_0^1$ is mathematically meaningless because the outer integral bounds cannot depend on variables. Always sketch the region.