11.3 Applications of polar double integrals

Area and Volume Calculations

Calculating Areas Using Polar Double Integrals

The area of a region $D$ bounded by polar curves is given by:

$A = \iint_D r \, dr \, d\theta$

That extra factor of $r$ in the integrand is the Jacobian from the coordinate transformation. It accounts for the fact that polar "grid cells" get larger as you move away from the origin.

To set up the integral:

1. Sketch the region and identify the bounding curves in polar form.
2. Determine the range of $\theta$ by finding where the curves intersect (solve $r_1(\theta) = r_2(\theta)$).
3. For each $\theta$, identify the inner radius $r_{\text{in}}(\theta)$ and outer radius $r_{\text{out}}(\theta)$.
4. Write the integral as $A = \int_{\alpha}^{\beta} \int_{r_{\text{in}}(\theta)}^{r_{\text{out}}(\theta)} r \, dr \, d\theta$.
5. Evaluate the inner integral first, then the outer.

This approach works naturally for cardioids ($r = a(1 + \cos\theta)$), limaçons, and rose curves ($r = a\cos(n\theta)$), where rectangular coordinates would require messy algebra.

Calculating Volumes Using Polar Double Integrals

For a solid whose height above the $xy$-plane is given by $z = f(r, \theta)$, the volume is:

$V = \iint_D f(r, \theta) \, r \, dr \, d\theta$

The region $D$ here is the projection (shadow) of the solid onto the $xy$-plane. Setting up the limits follows the same logic as the area case: figure out the $\theta$-range and the $r$-bounds for each $\theta$.

Example: To find the volume of a paraboloid $z = 1 - r^2$ above the $xy$-plane, note that $z \geq 0$ when $r \leq 1$. The projection is the unit disk, so:

$V = \int_0^{2\pi} \int_0^1 (1 - r^2) \, r \, dr \, d\theta = \int_0^{2\pi} \left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 d\theta = \int_0^{2\pi} \frac{1}{4} \, d\theta = \frac{\pi}{2}$

Hemispheres and cones with circular bases follow the same pattern. Whenever the base region is a disk or annulus, polar coordinates will save you significant work.

Calculating Mass Using Polar Double Integrals

For a flat lamina (thin plate) occupying region $D$ with density function $\delta(r, \theta)$, the total mass is:

$M = \iint_D \delta(r, \theta) \, r \, dr \, d\theta$

The density $\delta$ has units like kg/m², so integrating it over the area gives mass. If $\delta$ is constant, this reduces to $\delta \cdot A$.

The first moments about the coordinate axes are:

• $M_x = \iint_D y \, \delta(r, \theta) \, r \, dr \, d\theta$
• $M_y = \iint_D x \, \delta(r, \theta) \, r \, dr \, d\theta$

Since $x = r\cos\theta$ and $y = r\sin\theta$, these become:

• $M_x = \iint_D r\sin\theta \, \delta(r, \theta) \, r \, dr \, d\theta$
• $M_y = \iint_D r\cos\theta \, \delta(r, \theta) \, r \, dr \, d\theta$

These moments feed directly into the center of mass calculation below.

Moment and Center of Mass

Center of Mass in Polar Coordinates

The center of mass $(\bar{x}, \bar{y})$ is the "balance point" of the lamina. Once you have $M$, $M_x$, and $M_y$:

$\bar{x} = \frac{M_y}{M}, \qquad \bar{y} = \frac{M_x}{M}$

Notice the subscript swap: $\bar{x}$ uses $M_y$ and vice versa. This is a common source of errors on exams. The reason is that $M_y$ measures the moment about the $y$-axis, which depends on the $x$-coordinate.

Steps to find the center of mass:

1. Compute $M = \iint_D \delta(r,\theta) \, r \, dr \, d\theta$.
2. Compute $M_y = \iint_D r\cos\theta \, \delta(r,\theta) \, r \, dr \, d\theta$.
3. Compute $M_x = \iint_D r\sin\theta \, \delta(r,\theta) \, r \, dr \, d\theta$.
4. Divide: $\bar{x} = M_y / M$ and $\bar{y} = M_x / M$.

Before computing, check for symmetry. If the region and density are symmetric about the $x$-axis, then $\bar{y} = 0$ and you can skip the $M_x$ integral entirely.

Moment of Inertia in Polar Coordinates

The moment of inertia measures how mass is distributed relative to an axis of rotation. For a lamina with density $\delta(r, \theta)$, the moment of inertia about the origin (equivalently, about the $z$-axis) is:

$I_O = \iint_D r^2 \, \delta(r, \theta) \, r \, dr \, d\theta = \iint_D r^3 \, \delta(r, \theta) \, dr \, d\theta$

Note that the integrand contains $r^2$ (the squared distance from the $z$-axis) multiplied by the usual $\delta \cdot r$ from the polar area element, giving $r^3 \, \delta$.

You can also compute moments about individual axes:

• $I_x = \iint_D y^2 \, \delta \, r \, dr \, d\theta = \iint_D r^2 \sin^2\theta \, \delta \, r \, dr \, d\theta$
• $I_y = \iint_D x^2 \, \delta \, r \, dr \, d\theta = \iint_D r^2 \cos^2\theta \, \delta \, r \, dr \, d\theta$

A useful check: $I_O = I_x + I_y$. This is the perpendicular axis theorem for laminas, and it's a quick way to verify your calculations.

The parallel axis theorem lets you shift to a different axis: if you know $I_{\text{cm}}$ (moment about an axis through the center of mass), then the moment about a parallel axis a distance $d$ away is $I = I_{\text{cm}} + Md^2$.

Probability and Symmetry

Probability Distributions in Polar Coordinates

For a joint probability density function $f(r, \theta)$, the probability that a random point lands in region $D$ is:

$P(D) = \iint_D f(r, \theta) \, r \, dr \, d\theta$

The classic application is the Gaussian integral. The standard bivariate normal distribution $f(x,y) = \frac{1}{2\pi} e^{-(x^2+y^2)/2}$ converts beautifully to polar form as $f(r,\theta) = \frac{1}{2\pi} e^{-r^2/2}$, and the $r$ factor in the integrand makes the radial part integrable in closed form.

Other applications include modeling radial error distributions (e.g., distance from a target center) where the density depends only on $r$, not $\theta$.

For any valid probability density, the total probability over the entire domain must equal 1:

$\iint_{\text{all space}} f(r, \theta) \, r \, dr \, d\theta = 1$

Exploiting Symmetry in Polar Double Integrals

Symmetry can cut your computation in half (or more). Before evaluating any polar double integral, check for these patterns:

• Symmetry about the polar axis ($\theta = 0$): If both the region and the integrand are unchanged when $\theta$ is replaced by $-\theta$, integrate over $0 \leq \theta \leq \pi$ and double the result.
• Symmetry about $\theta = \pi/2$: If the region and integrand are symmetric about the $y$-axis, integrate over one side and double.
• Full rotational symmetry: If the integrand depends only on $r$ (not $\theta$) and the region is a full disk or annulus, the $\theta$-integral just contributes a factor of $2\pi$.

Rose curves $r = a\cos(n\theta)$ have $n$-fold or $2n$-fold symmetry depending on whether $n$ is odd or even. You can integrate over a single petal and multiply by the number of petals.

Cardioids $r = a(1 + \cos\theta)$ are symmetric about the polar axis, so you only need $\theta \in [0, \pi]$ and then double.

Symmetry arguments also tell you when certain integrals vanish. For instance, if a region is symmetric about the $x$-axis and the density is too, then $M_x = 0$ automatically because $\sin\theta$ is odd over symmetric $\theta$-intervals.