Areas enclosed by parametric curves: cardioids, astroids, and a check

Start with the shape and the direction

For $x=f(t)$, $y=g(t)$, a practical procedure is to understand the curve, write an area integral using its branches, and substitute $dx=f'(t)\,dt$. The parametric-sketching guide explains how to determine the shape.

Symmetry often reduces the calculation. Signed integrals require attention to which branch is above another and whether $x$ increases or decreases along the parameter interval. When one branch doubles back, split it at the turning point before interpreting the integral geometrically.

Consecutive pieces may combine neatly after substitution, as in the first example. This is a useful consequence of their orientations, not a universal rule that arbitrary area pieces must become a single integral.

Example 1: a cardioid

Find the enclosed area of

$$x=\cos t(1+\cos t),\qquad y=\sin t(1+\cos t),\qquad0\le t\le2\pi.$$

Replacing $t$ by $2\pi-t$ leaves $x$ unchanged and negates $y$, so the curve is symmetric about the $x$-axis. On the upper half, $0\le t\le\pi$,

$$x'=-\sin t(1+2\cos t),\qquad y'=\cos t+\cos2t=2\cos(3t/2)\cos(t/2).$$

The path starts at $(2,0)$, reaches its greatest height at $t=\pi/3$, turns in the $x$-direction at $t=2\pi/3$, and ends at the origin:

$$P(\pi/3)=\left(\frac34,\frac{3\sqrt3}{4}\right),\qquad P(2\pi/3)=\left(-\frac14,\frac{\sqrt3}{4}\right).$$

Call the height on $0\le t\le2\pi/3$ $y_+$ and that on $2\pi/3\le t\le\pi$ $y_-$. For negative $x$ the upper half lies between those branches; for positive $x$ it lies between $y_+$ and zero. Its area $H$ is therefore

$$H=\int_{-1/4}^{2}y_+\,dx-\int_{-1/4}^{0}y_-\,dx.$$

The first branch runs toward decreasing $x$, so its parameter bounds reverse. Substitution gives

$$H=-\int_0^{2\pi/3}y(t)x'(t)\,dt -\int_{2\pi/3}^{\pi}y(t)x'(t)\,dt =-\int_0^\pi y(t)x'(t)\,dt.$$

Consequently,

$$\begin{aligned} H&=\int_0^\pi\sin^2t(1+\cos t)(1+2\cos t)\,dt\\ &=\int_0^\pi\{\sin^2t+3\sin^2t\cos t+2\sin^2t\cos^2t\}\,dt\\ &=\left[\frac{3t}{4}-\frac{\sin2t}{4}+\sin^3t-\frac{\sin4t}{16}\right]_0^\pi =\frac{3\pi}{4}. \end{aligned}$$

Reflecting across the axis doubles the area:

$$\boxed{A=\frac{3\pi}{2}.}$$

Example 2: an astroid

Find the enclosed area of

$$x=\cos^3t,\qquad y=\sin^3t,\qquad0\le t\le2\pi.$$

Reflection in either coordinate axis preserves the curve, so it suffices to find its first-quadrant area and multiply by four. On $0\le t\le\pi/2$,

$$x'=-3\cos^2t\sin t,\qquad y'=3\sin^2t\cos t.$$

The point moves from $(1,0)$ to $(0,1)$, with $x$ decreasing and $y$ increasing.

The first-quadrant area is

$$Q=\int_0^1y\,dx =3\int_0^{\pi/2}\sin^4t\cos^2t\,dt =3(I_4-I_6),$$

where $I_n=\int_0^{\pi/2}\sin^nt\,dt$. For $n\ge2$, integration by parts gives

$$\begin{aligned} I_n&=[-\cos t\sin^{n-1}t]_0^{\pi/2} +(n-1)\int_0^{\pi/2}\cos^2t\sin^{n-2}t\,dt\\ &=(n-1)(I_{n-2}-I_n). \end{aligned}$$

Hence $I_n=(n-1)I_{n-2}/n$. Starting with $I_0=\pi/2$, we obtain $I_4=3\pi/16$ and $I_6=5\pi/32$. Therefore $Q=3\pi/32$ and

$$\boxed{A=4Q=\frac{3\pi}{8}.}$$

The integration-by-parts recurrence is worth understanding as well as remembering. See also integration by parts.

A second calculation using Green's area formula

For a simple closed, piecewise smooth boundary traversed once counterclockwise, the enclosed area is

$$\boxed{A=\frac12\int_\alpha^\beta\{f(t)g'(t)-f'(t)g(t)\}\,dt.}$$

Clockwise traversal changes the sign. Isolated cusps do not prevent applying the formula piecewise and taking limits. For a self-intersecting path, this expression gives signed area with multiplicity; taking its absolute value need not give the ordinary area of the union of its loops. Multiple traversals also count area multiple times.

Check the cardioid

Writing $r=1+\cos t$, $x=r\cos t$, and $y=r\sin t$, direct differentiation gives $xy'-x'y=r^2$. Thus

$$A=\frac12\int_0^{2\pi}(1+\cos t)^2\,dt=\frac{3\pi}{2}.$$

Check the astroid

Here the integrand simplifies to

$$xy'-x'y =3\cos^4t\sin^2t+3\cos^2t\sin^4t =3\sin^2t\cos^2t.$$

Therefore

$$A=\frac32\int_0^{2\pi}\sin^2t\cos^2t\,dt =\frac3{16}\int_0^{2\pi}(1-\cos4t)\,dt =\frac{3\pi}{8}.$$

This independent setup is often more useful for checking than repeating the same long calculation.

Why the formula works

For the examples above, the polar angle increases continuously, and the region is swept once by radial segments. A small angular change $d\theta$ sweeps area $r^2d\theta/2$. Since $x=r\cos\theta$ and $y=r\sin\theta$,

$$x\,dy-y\,dx=r^2d\theta.$$

Integrating proves the formula for this radial setting. The cusp of the cardioid has radius zero and is handled by a limit.

The determinant of two nearby position vectors gives the same local calculation:

$$\frac12\{f(t)g(t+\Delta t)-g(t)f(t+\Delta t)\} =\frac12(fg'-gf')\Delta t+o(\Delta t).$$

For the general simple-boundary statement, Green's theorem supplies a direct proof. Choose $P=-y/2$ and $Q=x/2$. Then

$$\oint_{\partial R}(P\,dx+Q\,dy) =\iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dx\,dy =\iint_R1\,dx\,dy=A.$$

Substitution of $dx=f'\,dt$ and $dy=g'\,dt$ yields the claimed sign $fg'-f'g$. The radial argument explains the two examples without requiring the full theorem; it should not be mistaken for a proof for every possible boundary.

If an examination expects only elementary methods, use the branch integrals as your main solution, or supply whatever justification is required for an additional theorem. The determinant interpretation is related to the cross product.