Areas Bounded by Parametric Curves, calculus lesson

A curve drawn by a moving point does not come with a formula $y = F(x)$ , yet it can still fence off a region, like the loop a skater carves on the ice. You can find that area without ever solving for $y$ in terms of $x$ . The trick is to let the parameter do the bookkeeping, and to watch which way the point is moving.

Key terms

Signed area integral

Definition Let $x = x(t)$ and $y = y(t)$ for $\alpha \le t \le \beta$ , where $x'$ is continuous on $[\alpha, \beta]$ and $y$ is continuous on $[\alpha, \beta]$ . Suppose the points lie on the graph of a function $F$ that is continuous on an interval containing every value $x(t)$ , so that $y(t) = F(x(t))$ . Then the substitution rule gives

\int_{\alpha}^{\beta} y(t)\, x'(t)\, dt = \int_{x(\alpha)}^{x(\beta)} F(x)\, dx.

When $x(t)$ increases from $a = x(\alpha)$ to $b = x(\beta)$ and $y(t) \ge 0$ , this number is the area under the curve and above the interval $[a, b]$ of the $x$ -axis.

In plain words The area under a graph is the integral of $y$ with respect to $x$ . Along a parametric curve $dx = x'(t)\, dt$ , so you integrate $y(t)\, x'(t)$ in $t$ . The result is a signed area, so parts below the axis or traced leftward count as negative.

Example One arch of the cycloid $x = r(\theta - \sin\theta)$ , $y = r(1 - \cos\theta)$ is traced for $0 \le \theta \le 2\pi$ , moving left to right with $y \ge 0$ . Since $x'(\theta) = r(1 - \cos\theta)$ ,

A = \int_0^{2\pi} r^2 (1 - \cos\theta)^2\, d\theta = r^2 \int_0^{2\pi} \left(1 - 2\cos\theta + \cos^2\theta\right) d\theta = r^2 (2\pi + \pi) = 3\pi r^2.

The arch encloses three times the area of the rolling circle.

Orientation

Definition The orientation of a parametric curve is the direction in which the point $(x(t), y(t))$ moves as $t$ increases. With the hypotheses above, if $x(t)$ decreases from $x(\alpha)$ to $x(\beta) < x(\alpha)$ while $y(t) \ge 0$ , then

\int_{\alpha}^{\beta} y(t)\, x'(t)\, dt = -\int_{x(\beta)}^{x(\alpha)} F(x)\, dx,

which is the negative of the area under the curve.

In plain words If the point runs from right to left, $x'(t)$ is negative and the integral comes out negative. The size is still the area. Only the sign tells you about the direction.

Example The upper half of the unit circle is $x = \cos t$ , $y = \sin t$ for $0 \le t \le \pi$ , traced from $(1, 0)$ to $(-1, 0)$ . Then

\int_0^{\pi} \sin t \,(-\sin t)\, dt = -\int_0^{\pi} \sin^2 t\, dt = -\frac{\pi}{2}.

The half disk has area $\frac{\pi}{2}$ , and the minus sign records the leftward motion.

Closed curves

Definition A parametric curve on $[\alpha, \beta]$ is closed if $(x(\alpha), y(\alpha)) = (x(\beta), y(\beta))$ , and a simple closed curve if in addition it does not cross itself, that is, $(x(s), y(s)) \neq (x(t), y(t))$ whenever $\alpha \le s < t < \beta$ . Suppose a simple closed curve is piecewise smooth, meaning $x$ and $y$ are continuous on $[\alpha, \beta]$ and have continuous derivatives on each of finitely many subintervals. If it is traced exactly once counterclockwise, the area of the region it encloses is

A = -\int_{\alpha}^{\beta} y(t)\, x'(t)\, dt = \int_{\alpha}^{\beta} x(t)\, y'(t)\, dt.

The two integrals agree because integration by parts turns one into the other, and the boundary term $x(t)y(t)$ takes the same value at both ends of a closed curve. If the curve is traced exactly once clockwise, the signs flip and the area is

A = \int_{\alpha}^{\beta} y(t)\, x'(t)\, dt.

In plain words Going counterclockwise, the top of the region is traced leftward and the bottom rightward. So the integral of $y\, x'$ adds the area under the bottom and subtracts the area under the top, which gives minus the area inside. The minus sign in front turns it back into the area.

Example The ellipse $x = a\cos t$ , $y = b\sin t$ with $a, b > 0$ and $0 \le t \le 2\pi$ is traced once counterclockwise. Its area is

A = -\int_0^{2\pi} b\sin t \,(-a\sin t)\, dt = ab \int_0^{2\pi} \sin^2 t\, dt = \pi a b.

With $a = b = r$ this is the familiar $\pi r^2$ .

Intersections

Definition A curve crosses itself at a point $P$ when there are parameter values $s \neq t$ in the parameter interval with

x(s) = x(t) \quad \text{and} \quad y(s) = y(t).

Two curves $(x_1(s), y_1(s))$ and $(x_2(t), y_2(t))$ meet where $x_1(s) = x_2(t)$ and $y_1(s) = y_2(t)$ , with independent parameters $s$ and $t$ . These parameter values become the limits of integration for a loop or a region between the curves.

In plain words Before you can integrate over a loop you need to know where the loop starts and ends in terms of $t$ . You find that by asking when two different times give the same point.

Example Take $x = t^2$ , $y = t^3 - 3t$ . Equal $x$ values with $s \neq t$ force $s = -t$ . Then $y(-t) = -y(t)$ , so equal $y$ values need $y(t) = 0$ , which gives $t = \pm\sqrt{3}$ . The curve crosses itself at $(3, 0)$ , and the loop is traced for $-\sqrt{3} \le t \le \sqrt{3}$ . At $t = -1$ the point is $(1, 2)$ , above the axis, and $x$ is decreasing there, so the top is traced leftward and the loop runs counterclockwise. Its area is

A = -\int_{-\sqrt{3}}^{\sqrt{3}} (t^3 - 3t)(2t)\, dt = -\int_{-\sqrt{3}}^{\sqrt{3}} (2t^4 - 6t^2)\, dt = -\left(\frac{36\sqrt{3}}{5} - 12\sqrt{3}\right) = \frac{24\sqrt{3}}{5}.

Splitting intervals

Definition Suppose $x'$ and $y$ are continuous on $[\alpha, \beta]$ and each has only finitely many zeros there. If $x'(t)$ or $y(t)$ changes sign, split the interval into pieces $[t_{k-1}, t_k]$ with $\alpha = t_0 < t_1 < \dots < t_n = \beta$ , where the inner points $t_1, \dots, t_{n-1}$ are the parameter values at which $x'(t) = 0$ or $y(t) = 0$ . On each piece the curve moves in one horizontal direction and stays on one side of the $x$ -axis, so

\left\lvert \int_{t_{k-1}}^{t_k} y(t)\, x'(t)\, dt \right\rvert

is the geometric area between that piece of the curve and the $x$ -axis. Add or subtract these pieces according to the picture.

In plain words Cut the curve wherever it turns around horizontally or crosses the axis. Each piece then behaves like an ordinary graph, and you can see whether its area should be added or removed.

Example For the loop above, split at $t = 0$ , where $x'(t) = 2t = 0$ . On $[-\sqrt{3}, 0]$ the curve is the top half, moving left, with $y \ge 0$ . On $[0, \sqrt{3}]$ it is the bottom half, moving right, with $y \le 0$ . Since

\int_0^{\sqrt{3}} (2t^4 - 6t^2)\, dt = \frac{18\sqrt{3}}{5} - 6\sqrt{3} = -\frac{12\sqrt{3}}{5},

and the integrand is even, each half contributes $12\sqrt{3}/5$ of area. Together they give $\frac{24\sqrt{3}}{5}$ , matching the one-step answer.

Geometric area and repeated tracing

Definition The geometric area of a region is a nonnegative number that does not depend on any parametrization. Suppose a piecewise smooth parametrization on $[\alpha, \beta]$ runs around a simple closed curve exactly $k$ times, always in the same direction. Then

-\int_{\alpha}^{\beta} y(t)\, x'(t)\, dt

equals $k$ times the enclosed area when the direction is counterclockwise, and $-k$ times it when clockwise. To get the geometric area, restrict to an interval that traces the boundary exactly once, then take the absolute value.

In plain words The integral counts every lap. Run around the fence twice and it reports twice the field.

Example The circle $x = \cos t$ , $y = \sin t$ for $0 \le t \le 4\pi$ goes around twice counterclockwise, and

-\int_0^{4\pi} \sin t\,(-\sin t)\, dt = \int_0^{4\pi} \sin^2 t\, dt = 2\pi.

The disk has area $\pi$ . Using $0 \le t \le 2\pi$ gives the correct value.

Check yourself

Use $x = t^2$ , $y = t$ for $0 \le t \le 1$ to find the area under $y = \sqrt{x}$ from $x = 0$ to $x = 1$ .

Show the answer

Here $x'(t) = 2t$ , so the area is

\int_0^1 t \cdot 2t\, dt = \frac{2}{3}.

The curve moves right with $y \ge 0$ , so this integral is the area itself.

Find the area enclosed by the astroid $x = \cos^3 t$ , $y = \sin^3 t$ , $0 \le t \le 2\pi$ , which is traced once counterclockwise.

Show the answer

With $x'(t) = -3\cos^2 t \sin t$ the area is

A = -\int_0^{2\pi} \sin^3 t \,(-3\cos^2 t \sin t)\, dt = 3\int_0^{2\pi} \sin^4 t \cos^2 t\, dt = 3 \cdot \frac{\pi}{8} = \frac{3\pi}{8}.

Someone takes $x = 2\cos t$ , $y = 2\sin t$ for $0 \le t \le 6\pi$ , computes minus the integral of $y\, x'$ over that interval and gets $12\pi$ . What is the area of the disk, and what went wrong?

Show the answer

The disk has area $4\pi$ . The interval $[0, 6\pi]$ traces the circle three times, so the integral counted the area three times.

Areas Bounded by Parametric Curves

Key terms

Signed area integral

Orientation

Closed curves

Intersections

Splitting intervals

Geometric area and repeated tracing

Check yourself

How this lesson connects

Builds on