Volume and Surface Integration in Polar Coordinates

Core Definition

If a three-dimensional solid $E$ is bounded below by a planar region $D$ in the $xy$ -plane and above by a continuous surface $z = f(x, y) \ge 0$ , its volume in polar coordinates is given by:

V = \iint_D f(x, y) \, dA = \iint_D f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta

More generally, when the solid lies between a top surface

z_{\text{top}}(r, \theta)

and a bottom surface

z_{\text{bottom}}(r, \theta)

over the domain

D

, the volume is:

V = \iint_D \big[ z_{\text{top}}(r, \theta) - z_{\text{bottom}}(r, \theta) \big] \, r \, dr \, d\theta

Concept Explanation

1. Cylindrical & Rotational Symmetry in 3D Solids

Many real-world engineering components and physical structures possess rotational symmetry about the $z$ -axis. When the height of a solid depends primarily or exclusively on the radial distance from the vertical axis ( $r = \sqrt{x^2 + y^2}$ ), converting the base domain $D$ to polar coordinates dramatically simplifies the double integral.

Common rotationally symmetric surface equations in polar coordinates include:

Circular Paraboloid: $z = c - (x^2 + y^2) \implies z = c - r^2$
Circular Cone: $z = c \sqrt{x^2 + y^2} \implies z = c r$
Hemisphere: $z = \sqrt{R^2 - (x^2 + y^2)} \implies z = \sqrt{R^2 - r^2}$
Circular Cylinder: $x^2 + y^2 = R^2 \implies r = R$

2. Identifying the Base Domain $D$ in the $xy$ -Plane

To determine the limits of integration for the double integral:

Find Intersection with Base Plane ( $z = 0$ ): If the solid is bounded below by the $xy$ -plane, set $z_{\text{top}} = 0$ and solve for $r$ . For example, the paraboloid $z = 16 - r^2$ meets $z = 0$ at $r^2 = 16 \implies r = 4$ , defining a full disk domain $0 \le r \le 4$ and $0 \le \theta \le 2\pi$ .
Find Curve of Intersection Between Surfaces: If the solid is bounded between two surfaces (such as a cone $z = r$ and a paraboloid $z = 2 - r^2$ ), equate their heights: $r = 2 - r^2 \implies r^2 + r - 2 = 0 \implies (r + 2)(r - 1) = 0$ Since $r \ge 0$ , the surfaces intersect along the circle $r = 1$ . The projection onto the $xy$ -plane is the disk $0 \le r \le 1$ .

3. The Role of the Jacobian $r$ in Volume Calculations

When setting up volume integrals, the integrand consists of the surface height difference $[z_{\text{top}} - z_{\text{bottom}}]$ multiplied by the differential area element $dA = r \, dr \, d\theta$ .

Notice that the factor of $r$ from the Jacobian is essential: for surfaces involving radicals like $\sqrt{R^2 - r^2}$ or exponentials $e^{-r^2}$ , the extra factor of $r$ supplies the exact derivative needed for immediate $u$ -substitution ( $du = -2r \, dr$ ), making analytical volume evaluation straightforward.

Engineering Applications

Civil & Geotechnical Engineering: Calculating the storage volume and fluid retention capacity of paraboloid retention ponds, dome water reservoirs, and conical aggregate stockpiles.
Mechanical Engineering: Computing the solid material volume and component mass $M = \iint \rho \, z(r) \, dA$ of hemispherical pressure vessel heads, pistons, and parabolic dish reflectors.
Aerospace Engineering: Evaluating aerodynamic drag and internal propellant volume for parabolic or conical rocket nose cones.
Manufacturing & Casting: Sizing mold cavities and computing volumetric thermal shrinkage for rotational die castings.

Example Problems

Example 1: Volume Enclosed Under a Paraboloid

Find the volume of the solid enclosed below the paraboloid $z = 16 - x^2 - y^2$ and above the $xy$ -plane ( $z = 0$ ).

Solution

Step 1: Determine the Base Domain in the $xy$ -Plane
The solid is capped below by $z = 0$ . Equate the surface equation to zero:

16 - x^2 - y^2 = 0 \implies x^2 + y^2 = 16

In polar coordinates,

x^2 + y^2 = r^2

, so:

r^2 = 16 \implies r = 4

The domain

D

is the circular disk of radius

4

centered at the origin:

0 \le r \le 4, \quad 0 \le \theta \le 2\pi

Step 2: Express Surface Height in Polar Coordinates
State the Paraboloid Polar Form:

z = 16 - (x^2 + y^2) = 16 - r^2

Step 3: Formulate the Volume Integral
State the Double Integral Volume Formula in Polar Coordinates:

V = \iint_D z \, dA = \int_{0}^{2\pi} \int_{0}^{4} (16 - r^2) \, r \, dr \, d\theta

Distribute the Jacobian

r

V = \int_{0}^{2\pi} \int_{0}^{4} (16r - r^3) \, dr \, d\theta

Step 4: Evaluate via Separability
Apply Fubini's Separability Theorem:

V = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{4} (16r - r^3) \, dr \right)

Evaluate the radial integral:

\int_{0}^{4} (16r - r^3) \, dr = \left[ 8r^2 - \frac{r^4}{4} \right]_{0}^{4} = \left( 8(16) - \frac{256}{4} \right) - 0 = 128 - 64 = 64

Evaluate the angular integral:

\int_{0}^{2\pi} d\theta = 2\pi

Multiply:

V = 2\pi \times 64 = 128\pi

Example 2: Volume of a Hemispherical Dome

Compute the volume of the solid lying below the hemispherical dome $z = \sqrt{9 - x^2 - y^2}$ and above the $xy$ -plane.

Solution

Step 1: Determine the Base Domain
The dome terminates at the $xy$ -plane where $z = 0$ :

\sqrt{9 - x^2 - y^2} = 0 \implies x^2 + y^2 = 9 \implies r = 3

The integration region

D

is the circular disk:

0 \le r \le 3, \quad 0 \le \theta \le 2\pi

Step 2: Formulate the Integral with the Jacobian
Substitute $z = \sqrt{9 - r^2}$ and $dA = r \, dr \, d\theta$ :

V = \int_{0}^{2\pi} \int_{0}^{3} \sqrt{9 - r^2} \, r \, dr \, d\theta

Step 3: Evaluate via $u$ -Substitution
For the radial integral, let $u = 9 - r^2$ , then $du = -2r \, dr \implies r \, dr = -\frac{1}{2} du$ . When $r = 0$ , $u = 9$ ; when $r = 3$ , $u = 0$ .

\int_{0}^{3} \sqrt{9 - r^2} \, r \, dr = \int_{9}^{0} -\frac{1}{2} u^{1/2} \, du = \frac{1}{2} \int_{0}^{9} u^{1/2} \, du

Evaluate the antiderivative:

= \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{0}^{9} = \frac{1}{3} \left( 9^{3/2} - 0 \right) = \frac{1}{3}(27) = 9

Step 4: Complete Volume Calculation
Multiply by the angular integral:

V = \left( \int_{0}^{2\pi} d\theta \right) \times 9 = 2\pi \times 9 = 18\pi

(Note: The geometric formula for a hemisphere of radius $R=3$ confirms this: $V = \frac{2}{3}\pi R^3 = \frac{2}{3}\pi (27) = 18\pi$ .)

Connections & References

Parent Concepts: Double Integrals in Polar Coordinates, Multiple Integrals
Related Notes: Polar Rectangles and Annular Sectors, Separability in Polar Integrals
Course Sessions: Session 15 - Double Integrals in Polar Coordinates