Area with Parametric Curves
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.2: “Calculus of Parametric Curves” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-2-calculus-of-parametric-curves |
| Textbook used in class | Stewart, Calculus, Section 10.2: “Calculus with Parametric Curves” |
Opening Scenario
The area under $y = f(x)$ from $x = a$ to $x = b$ is $\int_a^b f(x)\,dx$. If the curve is given parametrically as $x = x(t)$, $y = y(t)$, then $dx = x'(t)\,dt$ and the substitution converts the area integral to a $t$-integral: $$A = \int_\alpha^\beta y(t)\,x'(t)\,dt,$$ where $\alpha$ and $\beta$ are the $t$-values corresponding to $x = a$ and $x = b$. The direction of travel (whether $x$ increases or decreases as $t$ increases) determines whether to negate.
Quick Reference
Area under a parametric curve (region between the curve $y = y(t) > 0$ and the $x$-axis):
If $x$ increases as $t$ goes from $\alpha$ to $\beta$ (i.e., $x'(t) > 0$): $$A = \int_\alpha^\beta y(t)\,x'(t)\,dt.$$
If $x$ decreases as $t$ goes from $\alpha$ to $\beta$ (i.e., $x'(t) < 0$, curve traced right to left): $$A = -\int_\alpha^\beta y(t)\,x'(t)\,dt = \int_\beta^\alpha y(t)\,x'(t)\,dt.$$
Rule of thumb: Take $A = \int y(t)|x'(t)|\,dt$ (use the absolute value), or set up with the smaller $x$ as the lower limit.
Key Concepts
1. The Substitution
The area formula $A = \int_a^b y\,dx$ converts via $x = x(t)$, $dx = x'(t)\,dt$: $$A = \int_\alpha^\beta y(t)\,x'(t)\,dt.$$
The limits $\alpha$ and $\beta$ are the $t$-values where $x = a$ and $x = b$ respectively.
2. Direction Matters
If the curve sweeps from left to right (as in a typical $y = f(x)$ graph), $x'(t) > 0$ and the formula gives a positive area. If the curve sweeps right to left, $x'(t) < 0$, so $y(t) x'(t)\,dt$ is negative and the integral gives a negative area. Negate to get the positive area, or swap the limits.
Practical approach: Identify which direction (left-to-right or right-to-left) the curve is traced for the portion of interest. If left-to-right, use $A = \int_\alpha^\beta y x'(t)\,dt$. If right-to-left, either negate or swap limits.
3. Closed Curves and Enclosed Area
For a closed parametric curve (where the start and end points coincide), the enclosed area can be computed using: $$A = \left|\int_\alpha^\beta y\,x'(t)\,dt\right|$$ (up to sign depending on orientation). The sign is positive if the curve is traced counterclockwise and negative if clockwise (by Green’s theorem in vector calculus, though for MATH163 just use careful setup).
Worked Example
Find the area of the region enclosed by the cycloid $x = r(\theta - \sin\theta)$, $y = r(1-\cos\theta)$, $0 \leq \theta \leq 2\pi$ (one arch) and the $x$-axis.
Here the parameter is $\theta$ (not $t$, but the formulas work identically).
$x' = dx/d\theta = r(1-\cos\theta)$, $y = r(1-\cos\theta)$.
The curve starts and ends on the $x$-axis ($y = 0$ at $\theta = 0$ and $\theta = 2\pi$). As $\theta$ increases from $0$ to $2\pi$, $x$ increases from $0$ to $2\pi r$ (left to right).
$$A = \int_0^{2\pi} y\,\frac{dx}{d\theta}\,d\theta = \int_0^{2\pi} r(1-\cos\theta)\cdot r(1-\cos\theta)\,d\theta = r^2\int_0^{2\pi}(1-\cos\theta)^2\,d\theta.$$
$(1-\cos\theta)^2 = 1 - 2\cos\theta + \cos^2\theta = 1 - 2\cos\theta + \dfrac{1+\cos 2\theta}{2}$.
$$\int_0^{2\pi}(1-\cos\theta)^2\,d\theta = \left[\theta - 2\sin\theta + \frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{2\pi} = \left(\frac{3\theta}{2} - 2\sin\theta + \frac{\sin 2\theta}{4}\right)\Bigg|_0^{2\pi} = 3\pi.$$
$A = 3\pi r^2$.
Boxed answer: Area of one cycloid arch $= 3\pi r^2$ (three times the area of the rolling circle of radius $r$).
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting direction: using $\int y\,x'(t)\,dt$ when the curve goes right to left | Getting a negative area for a clearly positive region | Check the sign of $x'(t)$ over the interval; negate or swap limits if the curve goes right to left |
| Using $t$-limits in the wrong order | Placing the larger $t$-value as the lower limit | $\alpha$ is the $t$-value at the starting $x$ (left), $\beta$ at the ending $x$ (right) |
| Mixing up $x'(t)\,dt$ with $y'(t)\,dt$ | Writing $A = \int y(t)\,y'(t)\,dt$ | The area formula comes from $\int y\,dx$; the substitution introduces $dx = x'(t)\,dt$ |
Common Misconceptions
the area formula is $\int y(t)\,dt$ rather than $\int y(t)\,x'(t)\,dt$.
This is the multiplicative-not-additive error applied to the substitution. The Cartesian area formula is $\int y\,dx$; when $x = x(t)$, the differential $dx$ becomes $x'(t)\,dt$, introducing the factor $x'(t)$. Dropping that factor mistakes the area integral for a simple integral of $y$ over time and gives an answer with the wrong units and value.
the sign of $x'(t)$ does not affect the area.
This is the rate-as-fixed-number error. When the curve is traced from right to left, $x'(t) < 0$, and the integral $\int y(t) x'(t)\,dt$ produces a negative value. Ignoring the sign yields a wrong sign on the area. The fix is to negate the integral or swap the limits so that area is computed as a positive quantity.
Leveled Practice
Level 1 -- Ellipse Area
Problem 1. Find the area of the ellipse $x = a\cos t$, $y = b\sin t$, $0 \leq t \leq 2\pi$.
Show answer
$dx/dt = -a\sin t$, $y = b\sin t$.
The ellipse is closed; trace it counterclockwise. Use $A = -\int_0^{2\pi} y\,x'(t)\,dt$ (the negative because the full ellipse traversal gives a negative integral for counterclockwise curves by the standard orientation convention -- alternatively, note that integrating $y\,dx$ counterclockwise gives the negative of the area, so $A = -\int$):
$A = -\int_0^{2\pi} b\sin t\cdot(-a\sin t)\,dt = ab\int_0^{2\pi}\sin^2 t\,dt = ab\cdot\pi = \pi ab$.
Boxed answer: Area $= \pi ab$.
Level 2 -- Curve Segment
Problem 2. Find the area under the curve $x = t^2$, $y = t^3$, from $t = 0$ to $t = 1$ (above the $x$-axis).
Show answer
$x'(t) = 2t$, $y = t^3$.
$A = \displaystyle\int_0^1 t^3\cdot 2t\,dt = 2\int_0^1 t^4\,dt = 2\cdot\dfrac{1}{5} = \dfrac{2}{5}$.
Mastery Checklist
Mental Model
Area under a curve is $\int y\,dx$. A parametric curve gives $dx$ as $x'(t)\,dt$, so the area integral becomes $\int y(t)\,x'(t)\,dt$. The area is the same integral as always, just with a different variable of integration. The only subtlety is direction: if the curve moves backward ($x'(t) < 0$), the integral picks up a sign flip, just as $\int_b^a f = -\int_a^b f$.
Connections
Looking back
- Area under a curve (Section 5.2): The formula $A = \int_a^b y\,dx$ is extended here by substitution.
- Parametric direction (Section 10.1): The direction of the curve determines whether to negate.