Area in Polar Coordinates Calculator — Theory
1. Polar area idea
In polar coordinates, a curve is written as
\[
r=f(\theta).
\]
A small slice of the region looks like a circular sector. A sector with radius \(r\) and small angle
\(d\theta\) has area
\[
dA=\frac12r^2\,d\theta.
\]
This is the reason for the factor \(\frac12\) in the polar area formula.
2. Area enclosed by one polar curve
If the curve \(r=f(\theta)\) encloses a region as \(\theta\) moves from \(a\) to \(b\), then the area is
\[
A=\frac12\int_a^b r(\theta)^2\,d\theta.
\]
Since \(r^2\) is used, negative values of \(r\) still contribute positive area.
3. Area between two polar curves
If one curve is outside another on the interval, then
\[
A=\frac12\int_a^b
\left(r_{\text{outer}}^2-r_{\text{inner}}^2\right)d\theta.
\]
The outer curve must really be outside the inner curve on the interval. If the curves switch order,
the interval may need to be split at the intersection angles.
4. Why curves may need to be split
For area between curves, the expression
\[
r_1^2-r_2^2
\]
may change sign. This means that the curve that is outside has changed. In that case, the geometric
area is often computed as
\[
A=\frac12\int_a^b
\left|r_1^2-r_2^2\right|\,d\theta.
\]
Another method is to split the integral wherever \(r_1^2=r_2^2\).
5. Example: \(r=4\sin\theta\)
Use
\[
A=\frac12\int_0^{2\pi}(4\sin\theta)^2\,d\theta.
\]
Simplify:
\[
A=\frac12\int_0^{2\pi}16\sin^2\theta\,d\theta
=
8\int_0^{2\pi}\sin^2\theta\,d\theta.
\]
Since
\[
\int_0^{2\pi}\sin^2\theta\,d\theta=\pi,
\]
the result is
\[
A=8\pi.
\]
The same circle is traced once on \(0\le\theta\le\pi\), giving geometric area \(4\pi\). On
\(0\le\theta\le2\pi\), the integral counts the tracing over the full interval.
6. Example: cardioid \(r=2+2\cos\theta\)
The area of the cardioid over \(0\le\theta\le2\pi\) is
\[
A=\frac12\int_0^{2\pi}(2+2\cos\theta)^2\,d\theta.
\]
Expand:
\[
(2+2\cos\theta)^2
=
4+8\cos\theta+4\cos^2\theta.
\]
Then
\[
A=\frac12\left(8\pi+0+4\pi\right)=6\pi.
\]
7. Example: area between two curves
Suppose the outer curve is \(r_1=4\sin\theta\), the inner curve is \(r_2=2\), and the interval is
\(\pi/6\le\theta\le5\pi/6\). The area is
\[
A=\frac12\int_{\pi/6}^{5\pi/6}
\left((4\sin\theta)^2-2^2\right)d\theta.
\]
Simplify:
\[
A=\frac12\int_{\pi/6}^{5\pi/6}
\left(16\sin^2\theta-4\right)d\theta.
\]
The limits \(\pi/6\) and \(5\pi/6\) come from the intersection condition
\[
4\sin\theta=2.
\]
8. Negative radius values
A negative radius is allowed in polar coordinates:
\[
(r,\theta)\equiv(-r,\theta+\pi).
\]
For area, the formula uses \(r^2\), so a negative radius still gives positive sector area.
However, negative radius values can make the visual region harder to interpret, so the graph is important.
10. Numerical integration
Many polar area integrals do not simplify nicely. In that case, numerical integration is used:
- Adaptive Simpson: refines where the integrand changes quickly.
- Composite Simpson: applies Simpson’s rule over equal subintervals.
- Composite trapezoid: approximates the integrand using trapezoids.
Numerical results should be checked with the graph, especially when curves intersect or switch order.
11. Shaded region interpretation
The shaded graph represents the region measured by the integral. For one curve, the shaded region is
swept from the pole to the curve:
\[
0\le r\le f(\theta).
\]
For two curves, the shaded region is between them:
\[
r_{\text{inner}}(\theta)\le r\le r_{\text{outer}}(\theta).
\]
12. Common mistakes
- Forgetting the factor \(\frac12\): polar area is not \(\int r^2\,d\theta\); it is \(\frac12\int r^2\,d\theta\).
- Using \(r\) instead of \(r^2\): area depends on the square of the radius.
- Ignoring intersections: area between curves often requires intersection angles.
- Using the wrong outer curve: check which curve is farther from the pole.
- Overcounting from repeated tracing: the interval may trace the same region more than once.
- Misreading negative radius: negative \(r\) changes the plotted direction, even though \(r^2\) is positive.
