Polar Curves, Area and Length

Coordinates with direction and distance

Polar coordinates satisfy $x=r\cos\theta$ and $y=r\sin\theta$ . A negative radius places the point in the opposite direction: $(r,\theta)$ and $(-r,\theta+\pi)$ describe the same location. Angles differing by $2\pi$ also represent the same direction.

For $r=r(\theta)$ , differentiate the Cartesian parametrization to find tangent slopes:

\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta},

where the denominator is nonzero. Sketching sample angles and identifying zeros of $r$ helps determine which loop an interval traces.

Visual guide

VISUAL GUIDERadius sweeps out small sectors

r = 1 + cos θ

For r = 1 + cos θ, the curve is a cardioid. The shaded sector follows 0 ≤ θ ≤ π/3. Polar area adds sectors of size ½r²dθ, not rectangles of height r.

Worked example: area of a circle

A narrow polar sector has area approximately $r^2\Delta\theta/2$ . Thus a region swept once from $\alpha$ to $\beta$ has area $A=\tfrac12\int_\alpha^\beta r^2d\theta$ .

The curve $r=2\cos\theta$ satisfies $x^2+y^2=2x$ , a circle of radius $1$ centered at $(1,0)$ . It is traced once with nonnegative radius for $-\pi/2\le\theta\le\pi/2$ . Its area is

A=2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=\pi.

Using $0\le\theta\le2\pi$ would trace this circle twice and give twice its area.

Worked example: polar arc length

Since $x'^2+y'^2=r'^2+r^2$ , polar arc length is

L=\int_\alpha^\beta\sqrt{r^2+(r')^2}\,d\theta.

For the spiral $r=\theta$ , $0\le\theta\le1$ , this becomes $\int_0^1\sqrt{1+\theta^2}d\theta=\tfrac12(\sqrt2+\ln(1+\sqrt2))$ . The derivative term accounts for outward radial motion as the angle changes.

Area between two nonnegative radii uses $\tfrac12\int(R^2-r^2)d\theta$ on intervals where the ordering is known. Intersections may occur at the origin under different angles, so solving $r_1(\theta)=r_2(\theta)$ alone can miss geometric intersections.

Practice

Convert $(r,\theta)=(2,\pi/3)$ to Cartesian coordinates.
Find the area inside $r=3$ over $0\le\theta\le\pi/2$ .
Find the length of the same circular arc.

Show worked solutions

$(x,y)=(1,\sqrt3)$ .
$A=\tfrac12\int_0^{\pi/2}9d\theta=9\pi/4$ .
$r'=0$ , so $L=\int_0^{\pi/2}3d\theta=3\pi/2$ .