Area with Parametric Equations — Question 10

Question 10

Problem

For $x=t^3$, $y=t^2$, $0\le t\le1$, compare the area under the curve with the area to its left inside the unit square.

Question 10 – Solution

Solution

1. As $t$ increases from $0$ to $1$, the curve runs from $(0,0)$ to $(1,1)$ with $$x'(t)=3t^2\ge0.$$ The area below the curve is therefore $$A_{\mathrm{below}}=\int_0^1y(t)x'(t)\,dt.$$

2. Substitute $y=t^2$ and $x'=3t^2$: $$A_{\mathrm{below}}=\int_0^1 3t^4\,dt =\left[\frac35t^5\right]_0^1 =\boxed{\frac35}.$$

3. The curve divides the unit square into the area below it and the area to its left (equivalently, above it). Since the square has area $1$, $$A_{\mathrm{left}}=1-A_{\mathrm{below}} =1-\frac35 =\boxed{\frac25}.$$

4. This can also be checked directly with $dy=2t\,dt$: $$A_{\mathrm{left}}=\int_0^1x\,dy =\int_0^1t^3(2t)\,dt =\frac25.$$ Thus the area under the curve is larger by $1/5$.