Area with Parametric Equations

Main idea

Area under a curve is still heights times widths, $\int y\,dx$ , no matter how the curve is described. On a parametric curve the height is $g(t)$ and the width $dx$ is $f'(t)\,dt$ , so the area is $\int g(t)\,f'(t)\,dt$ .

A = \int_{\alpha}^{\beta} {\htmlData{sym=0}{g(t)}}\,{\htmlData{sym=1}{f'(t)\,dt}}

What the symbols mean

$g(t)$
Height
y written in terms of t.
$f'(t)\,dt$
Width of a slice
dx written in terms of t.

Explanation

The area under a parametric curve is simply the amount of space between the path a moving point traces and the x-axis below it. Nothing about it is new: it's still a pile of thin rectangles, the same idea as the area under any graph. It's good for shapes that are easiest to describe by motion, like the arch a rolling wheel's rim traces or an ellipse. The integral $\int g(t)\,f'(t)\,dt$ looks unfamiliar, but it's just two pieces multiplied: $g(t)$ (the height of a rectangle, which is the point's $y$ ) and $f'(t)\,dt$ (its width, which is how far right the point moves in a sliver of time). Add up height times width and you have the area; it's the same $\int y\,dx$ you've always used, written in $t$ .

It's just a substitution. Think back to how area under a curve worked: slice the region into thin rectangles of height $y$ and width $dx$ , and add them up, $A = \int_a^b y\,dx$ . Nothing about that cares whether the curve came from $y = f(x)$ or from a moving point. The only issue is that our curve is written in $t$ , not $x$ . So do what you'd do with any $u$ -substitution: put $x = f(t)$ into the integral. Then $y$ becomes $g(t)$ , $dx$ becomes $f'(t)\,dt$ , and the $x$ -limits become the matching $t$ -limits.

Direction matters. The curve should be traced once, left to right, as $t$ goes from $\alpha$ to $\beta$ . Think about why. If the point is moving left, $x$ is decreasing, so $f'(t)$ is negative and every slice counts as negative area. That's the same thing that happens when you integrate $\int_b^a$ instead of $\int_a^b$ . So if the curve is traced right to left, either swap the limits or take the absolute value at the end. And if it's traced more than once, you'd count the area more than once.

Use symmetry for closed curves. For the ellipse $x = a\cos t$ , $y = b\sin t$ , compute the first-quadrant piece and multiply by 4. It's less work and it keeps you away from regions where $y$ is negative.

See it

Here's the cycloid again, traced by a point on a rolling wheel of radius 1, with the area under the arch adding up as it goes. Once the wheel has made a full turn, the total is ∫(1 − cos t)² dt from 0 to 2π = 3π, about 9.42. That's exactly three times the area of the wheel itself. Galileo is said to have guessed that ratio by weighing metal cutouts.

Worked examples

Example 1

Find the area under one arch of the cycloid $x = r(t - \sin t)$ , $y = r(1 - \cos t)$ , $0 \le t \le 2\pi$ .

1
The height is $y = r(1 - \cos t)$ , and the width is $dx = r(1 - \cos t)\,dt$ . As $t$ goes from $0$ to $2\pi$ the point moves left to right across one arch, so the sign will come out positive.

Example 2

Find the area inside the ellipse $x = a\cos t$ , $y = b\sin t$ .

1
Do the first quadrant and multiply by 4. To go left to right there, from $x = 0$ to $x = a$ , $t$ has to run from $\pi/2$ down to $0$ , since the ellipse is traced counterclockwise.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the area under $x = t^2$ , $y = t^3$ for $0 \le t \le 1$ .

Problem 2+5 XP

Find the area under one arch of the cycloid with $r = 2$ .

Problem 3+5 XP

Find the area enclosed by the astroid $x = \cos^3\theta$ , $y = \sin^3\theta$ .

Problem 4+5 XP

Find the area inside $x = 2\cos t$ , $y = 3\sin t$ .

Problem 5+5 XP

A landscaper adds a ripple to an elliptical flower bed, giving the edge $x = 3\cos t$ , $y = 2\sin t + 0.5\sin 2t$ for $0 \le t \le 2\pi$ (meters). Find the bed's area using $A = \int_0^{2\pi}x\,\frac{dy}{dt}\,dt$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

The curve $x = t^2 - 1$ , $y = t^3 - t$ crosses itself at the origin and forms a loop for $-1 \le t \le 1$ . Find the area of the loop.

Where it's used

Planimeters, the mechanical area meters used on maps and blueprints, trace a boundary and compute exactly this kind of integral; Green's theorem in Calculus III explains why.

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