Area under a parametric curve

Substituting $x=f(t)$ into $S=\int_a^b y\,dx$ gives $dx=\dfrac{dx}{dt}\,dt$. Find $\dfrac{dx}{dt}$ first, multiply by $y$, and integrate in $t$. The limits are the values $\alpha$, $\beta$ of $t$ that correspond to $x=a$ and $x=b$.

For example, for $x=t^2$, $y=t$ $(0\le t\le 1)$, $\dfrac{dx}{dt}=2t$ and $S=\displaystyle\int_0^1 t\cdot 2t\,dt=\dfrac23$. The area between the cycloid $x=t-\sin t$, $y=1-\cos t$ $(0\le t\le 2\pi)$ and the $x$-axis is $\displaystyle\int_0^{2\pi}(1-\cos t)^2\,dt=3\pi$.

Examples

1. 01Evaluate this definite integral (give the coefficient of $\pi$):$\displaystyle \begin{matrix} \int_{0}^{2\pi} (1-2\cos t+\cos^{2}t)\,dt \\ \int_{0}^{2\pi}\cos t\,dt=0,\; \int_{0}^{2\pi}\cos^{2}t\,dt=\pi \end{matrix}$AnswerHide

   Answer$\displaystyle 3$

   Steps

   1. Split it term by term: $\int_{0}^{2\pi}1\,dt=2\pi$, the $\cos t$ term is 0, and the $\cos^{2}t$ term is $\pi$
   2. $\displaystyle 2\pi-2\cdot 0+\pi$$\displaystyle = 3\pi$
   3. The question asks for the coefficient of $\pi$, so the answer is 3
2. 02What is dx/dt?$\displaystyle x=t^{4}, \quad y=t$AnswerHide

   Answer$\displaystyle 4t^{3}$

   Steps

   1. Just differentiate $x$ with respect to $t$
   2. $\displaystyle x$$\displaystyle = t^{4}$$\displaystyle \frac{dx}{dt}$$\displaystyle = 4t^{3}$
   3. Drop the exponent by one and bring it in front; $\frac{t^{5}}{5}$ would be the integral instead
3. 03What is the integrand y·(dx/dt)?$\displaystyle y=t, \quad \frac{dx}{dt}=2t$AnswerHide

   Answer$\displaystyle 2t^{2}$

   Steps

   1. Just multiply $y$ by $\frac{dx}{dt}$
   2. $\displaystyle t \cdot 2t$$\displaystyle = 2t^{2}$
   3. Multiplying by $t$ raises the exponent by one
4. 04Area enclosed by a parametric curve. What goes in □?$\displaystyle S = \int y \cdot \square \,dt$AnswerHide

   Answer$\displaystyle \frac{dx}{dt}$
5. 05When finding the area for a parametric curve, what do you find first?$\displaystyle$A$\displaystyle \frac{dy}{dt}$B$\displaystyle \frac{dx}{dt}$C$\displaystyle \frac{dy}{dx}$D$\displaystyle \frac{d^{2}y}{dx^{2}}$AnswerHide

   Answer$\displaystyle \frac{dx}{dt}$
6. 06What is the integrand y·(dx/dt)?$\displaystyle y=t, \quad \frac{dx}{dt}=4t^{3}$AnswerHide

   Answer$\displaystyle 4t^{4}$

   Steps

   1. Just multiply $y$ by $\frac{dx}{dt}$
   2. $\displaystyle t \cdot 4t^{3}$$\displaystyle = 4t^{4}$
   3. Multiplying by $t$ raises the exponent by one

FAQ

Q1Can I just use $\int y\,dt$?AnswerHide

A

No. Area is the integral in $x$, $\int y\,dx$, so to integrate in $t$ you replace $dx$ with $\dfrac{dx}{dt}\,dt$. Forgetting the factor $\dfrac{dx}{dt}$ gives a different value.

Q2What if $x$ decreases as $t$ increases?AnswerHide

A

Use the $t$ for $x=a$ as the lower limit and the $t$ for $x=b$ as the upper limit, even if that puts the larger $t$ at the bottom. Once the limits are set, also check that $y\ge 0$.