Area with Parametric Equations — Question 1

Question 1 – Solution

Solution

1. The interval $0\le t\le\pi/2$ traces the ellipse’s first-quadrant arc from $(4,0)$ to $(0,3)$. By symmetry, the total area is four times the area in that quadrant.

2. Differentiate the horizontal coordinate: $$\frac{dx}{dt}=-4\sin t.$$ Because $x$ decreases on this interval, the first-quadrant area is $$A_1=-\int_0^{\pi/2}y(t)x'(t)\,dt.$$

3. Substitute $y=3\sin t$ and $x'=-4\sin t$: $$A_1=-\int_0^{\pi/2}(3\sin t)(-4\sin t)\,dt =12\int_0^{\pi/2}\sin^2t\,dt.$$

4. Using $\sin^2t=(1-\cos2t)/2$, $$\int_0^{\pi/2}\sin^2t\,dt=\frac\pi4.$$ Hence $A_1=3\pi$.

5. Multiply by four quadrants: $$A=4A_1=\boxed{12\pi}.$$