Finding the Area of the Region Bounded by Two Polar Curves
The region between two polar curves
- Just as with $x,y$ curves, you can find the area between two polar curves.
- Take the area inside the outer curve and subtract the area inside the inner curve.
- Each is a $\tfrac12\int r^2\,d\theta$ integral; the answer is their difference.
- The trick is finding where the curves cross to set the angle limits.
Find the intersection angles first
- Set the two curves equal, $f(\theta)=g(\theta)$, and solve for the intersection angles.
- These angles are the limits of integration $\alpha,\beta$ for the enclosed region.
- (Polar intersections can be sneaky — a point may be reached at different $\theta$; sketch to be sure.)
- The region of interest lies between these angles.
The area between two polar curves is $\tfrac12\int_\alpha^\beta(\ ?\ )\,d\theta$ where the integrand is...
Outer squared minus inner squared.
To set the angle limits, find the ____ angles by solving $f(\theta)=g(\theta)$.
Where the curves cross bounds the region.
Outer squared minus inner squared
- On the interval, identify which curve is outer (larger $r$) and which is inner.
- The area between them is:
-
$$A=\frac{1}{2}\int_{\alpha}^{\beta}\Big(\big[r_{\text{outer}}\big]^2-\big[r_{\text{inner}}\big]^2\Big)\,d\theta$$
- Square each radius, subtract, halve, integrate — the polar washer.
Between two polar curves
The area between two polar curves subtracts the inner sectors from the outer — $\tfrac12\int(r_\text{out}^2-r_\text{in}^2)\,d\theta$.
Over the interval, the outer curve is the one with the...
Outer = farther from the origin = larger $r$.
Square first, then subtract
- Just like Cartesian washers, it's $r_{\text{out}}^2-r_{\text{in}}^2$, not $(r_{\text{out}}-r_{\text{in}})^2$.
- Square each radius before subtracting.
- Check which curve is outer on the interval — it can swap, needing a split.
- The $\tfrac12$ and the squares are both essential.
Area inside $r=2$, outside $r=1$: $\tfrac12\int_0^{2\pi}(4-1)\,d\theta$ (as a multiple of $\pi$).
$\tfrac12\cdot3\cdot2\pi=3\pi$.
You square each radius before subtracting, not $(r_\text{out}-r_\text{in})^2$.
It is $r_\text{out}^2-r_\text{in}^2$.
If the outer and inner curves swap over the interval, you should split the integral at the crossing.
A split keeps outer-minus-inner correct on each piece.
The integrand is $r_{\text{outer}}^2-r_{\text{inner}}^2$ — square before subtracting, never $(r_{\text{out}}-r_{\text{in}})^2$. Get the intersection angles right (sketch the curves), and confirm which curve is outer over the interval; if they trade places, split the integral at the crossing.
Find the area inside $r=2$ but outside $r=1$ (an annulus).
- Outer $r=2$, inner $r=1$, over all angles $0$ to $2\pi$.
- $A=\dfrac12\displaystyle\int_0^{2\pi}\big(2^2-1^2\big)\,d\theta=\dfrac12\int_0^{2\pi}3\,d\theta=\dfrac12\cdot3\cdot2\pi=3\pi$.
- (Matches the ring area $\pi(2^2-1^2)=3\pi$.) ✓
The area between two polar curves is $A=\frac{1}{2}\int_{\alpha}^{\beta}\big(r_{\text{outer}}^2-r_{\text{inner}}^2\big)\,d\theta$. First find the intersection angles for the limits, identify outer vs. inner, and square each radius before subtracting — never $(r_{\text{out}}-r_{\text{in}})^2$.