# The following parametric equations define an ellipse. x = 2cos(theta), y = The following...

## Question:

The following parametric equations define an ellipse.

{eq}x = 2 \cos(\theta) {/eq}

, {eq}y = 3 \sin(\theta) {/eq}, {eq}0 \le \theta \le \pi {/eq} .

Compute the area enclosed by this ellipse.

Hint : The ellipse is symmetric over the x-axis . Select limits for integration , so that you are integrating left to right .

## Area under Parametric Curves

The area of a region between the x-axis and a curve given parametrically as {eq}\displaystyle x=x(t), y=y(t), a\leq t\leq b, {/eq}

is given as {eq}\displaystyle \int_a^b y(t) ds=\int_a^b y(t)x'(t)\ dt. {/eq}

The ellipse given parametrically as {eq}\displaystyle x = 2 \cos\theta, y = 3 \sin \theta, 0 \le \theta \le \pi {/eq}

is the top half of an ellipse elongated between -2 and 2, horizontally and from 0 to 3, vertically.

Because {eq}\displaystyle x(\pi)=-2, x(0)=2, {/eq} to integrate from left to right, the limits of integration will be from {eq}\displaystyle \pi {/eq} to 0.

So, the area of the region between the x axis and the given ellipse is

{eq}\displaystyle \begin{align} \text{Area}&=\int_{\pi}{0}y(\theta) x'(\theta)\ d\theta \\ &=\int^{0}_{\pi}3\sin \theta (-2\sin \theta) \ d\theta \\ &=-3\int^{0}_{\pi} 2\sin^2 \theta \ d\theta\\ &=-3\int^{0}_{\pi}\left(1-\cos(2 \theta)\right) \ d\theta, &\left[\text{ using the double angle formula } 2\sin^2 \theta = 1-\cos(2 \theta)\right] \\ &=-3\left(\theta-\frac{1}{2}\sin(2 \theta)\right) \bigg\vert^{0}_{\pi}\\ &=-3\left(-\pi\right) \\ &=\boxed{3\pi}. \end{align} {/eq}

The result obtained is half of the ellipse with semi-axes 2,3, therefore it is confirmed by the formula {eq}\displaystyle \frac{ab \pi}{2}. {/eq}