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Tony's Intertwingled Memex • Reference Manual

Volume and Surface Integration in Polar Coordinates

Formulas & Definitions
Core Definition

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

V=∬Df(x,y) dA=∬Df(rcos⁡θ,rsin⁡θ) r dr dθV = \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 ztop(r,θ)z_{\text{top}}(r, \theta) and a bottom surface zbottom(r,θ)z_{\text{bottom}}(r, \theta) over the domain DD, the volume is:
V=∬D[ztop(r,θ)−zbottom(r,θ)] r dr dθV = \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 zz-axis. When the height of a solid depends primarily or exclusively on the radial distance from the vertical axis (r=x2+y2r = \sqrt{x^2 + y^2}), converting the base domain DD to polar coordinates dramatically simplifies the double integral.

Common rotationally symmetric surface equations in polar coordinates include:

2. Identifying the Base Domain DD in the xyxy-Plane

To determine the limits of integration for the double integral:

  1. Find Intersection with Base Plane (z=0z = 0): If the solid is bounded below by the xyxy-plane, set ztop=0z_{\text{top}} = 0 and solve for rr. For example, the paraboloid z=16−r2z = 16 - r^2 meets z=0z = 0 at r2=16  ⟹  r=4r^2 = 16 \implies r = 4, defining a full disk domain 0≤r≤40 \le r \le 4 and 0≤θ≤2π0 \le \theta \le 2\pi.
  2. Find Curve of Intersection Between Surfaces: If the solid is bounded between two surfaces (such as a cone z=rz = r and a paraboloid z=2−r2z = 2 - r^2), equate their heights:
    r=2−r2  ⟹  r2+r−2=0  ⟹  (r+2)(r−1)=0r = 2 - r^2 \implies r^2 + r - 2 = 0 \implies (r + 2)(r - 1) = 0
    Since r≥0r \ge 0, the surfaces intersect along the circle r=1r = 1. The projection onto the xyxy-plane is the disk 0≤r≤10 \le r \le 1.

3. The Role of the Jacobian rr in Volume Calculations

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

Notice that the factor of rr from the Jacobian is essential: for surfaces involving radicals like R2−r2\sqrt{R^2 - r^2} or exponentials e−r2e^{-r^2}, the extra factor of rr supplies the exact derivative needed for immediate uu-substitution (du=−2r drdu = -2r \, dr), making analytical volume evaluation straightforward.

Engineering Applications

Example Problems

Example 1: Volume Enclosed Under a Paraboloid

Find the volume of the solid enclosed below the paraboloid z=16−x2−y2z = 16 - x^2 - y^2 and above the xyxy-plane (z=0z = 0).

Solution

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

16−x2−y2=0  ⟹  x2+y2=1616 - x^2 - y^2 = 0 \implies x^2 + y^2 = 16
In polar coordinates, x2+y2=r2x^2 + y^2 = r^2, so:
r2=16  ⟹  r=4r^2 = 16 \implies r = 4
The domain DD is the circular disk of radius 44 centered at the origin:
0≤r≤4,0≤θ≤2π0 \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−(x2+y2)=16−r2z = 16 - (x^2 + y^2) = 16 - r^2

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

V=∬Dz dA=∫02π∫04(16−r2) r dr dθV = \iint_D z \, dA = \int_{0}^{2\pi} \int_{0}^{4} (16 - r^2) \, r \, dr \, d\theta
Distribute the Jacobian rr:
V=∫02π∫04(16r−r3) dr dθV = \int_{0}^{2\pi} \int_{0}^{4} (16r - r^3) \, dr \, d\theta

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

V=(∫02πdθ)(∫04(16r−r3) dr)V = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{4} (16r - r^3) \, dr \right)
Evaluate the radial integral:
∫04(16r−r3) dr=[8r2−r44]04=(8(16)−2564)−0=128−64=64\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:
∫02πdθ=2π\int_{0}^{2\pi} d\theta = 2\pi
Multiply:
V=2π×64=128πV = 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=9−x2−y2z = \sqrt{9 - x^2 - y^2} and above the xyxy-plane.

Solution

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

9−x2−y2=0  ⟹  x2+y2=9  ⟹  r=3\sqrt{9 - x^2 - y^2} = 0 \implies x^2 + y^2 = 9 \implies r = 3
The integration region DD is the circular disk:
0≤r≤3,0≤θ≤2π0 \le r \le 3, \quad 0 \le \theta \le 2\pi

Step 2: Formulate the Integral with the Jacobian
Substitute z=9−r2z = \sqrt{9 - r^2} and dA=r dr dθdA = r \, dr \, d\theta:

V=∫02π∫039−r2 r dr dθV = \int_{0}^{2\pi} \int_{0}^{3} \sqrt{9 - r^2} \, r \, dr \, d\theta

Step 3: Evaluate via uu-Substitution
For the radial integral, let u=9−r2u = 9 - r^2, then du=−2r dr  ⟹  r dr=−12dudu = -2r \, dr \implies r \, dr = -\frac{1}{2} du. When r=0r = 0, u=9u = 9; when r=3r = 3, u=0u = 0.

∫039−r2 r dr=∫90−12u1/2 du=12∫09u1/2 du\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:
=12[23u3/2]09=13(93/2−0)=13(27)=9= \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=(∫02πdθ)×9=2π×9=18πV = \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=3R=3 confirms this: V=23πR3=23π(27)=18πV = \frac{2}{3}\pi R^3 = \frac{2}{3}\pi (27) = 18\pi.)

Connections & References