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\textbf{Problem} ($\#$102, pg. 43) Find the area enclosed by the ellipse 
\[ \left\{ \begin{array}{ll}
x = a \cos(\theta) \\
y = b \sin(\theta) \\
0 \leq t \leq 2\pi.
\end{array} \right.\]
\textit{Solution}: Using Theorem 1.2, we first compute
\[ x'(\theta) = -a\sin(\theta).\]
Now the theorem tells us that the area is given by (note: the $a$ and $b$ in the theorem are the bounds on the parameter $t$ -- they are not the same as the $a$ and $b$ in this problem!!)
\[ \begin{array}{ll}
\mathrm{Area} &= \displaystyle\int_0^{2\pi} y(\theta) x'(\theta) \mathrm{d}\theta \\ 
&= \displaystyle\int_0^{2\pi} (-a \sin(\theta)) (b\sin(\theta)) \mathrm{d}\theta \\
&= -ab \displaystyle\int_0^{2\pi} \underbrace{\sin^2(\theta)}_{=\frac{1-\cos(2\theta)}{2}} \mathrm{d}\theta \\
&=   -ab \displaystyle\int_0^{2\pi} \dfrac{1}{2} \mathrm{d}\theta  - ab \displaystyle\int_0^{2\pi} \dfrac{\cos(2\theta)}{2} \mathrm{d}\theta \\
&=-\dfrac{ab}{2} \theta\Bigg|_{\theta=0}^{\theta=2\pi} - \dfrac{ab}{4} \sin(2\theta) \Bigg|_{\theta=0}^{\theta=2\pi} \\
&=-\dfrac{ab}{2}(2\pi - 0) - \dfrac{ab}{4} (\underbrace{\sin(4\pi)}_{=0} - \underbrace{\sin(0)}_{=0}) \\
&\stackrel{?}{=} -ab \pi
\end{array} \]

Note: again we see that the ``area" became negative. The actual answer is $ab\pi$. The sign error is a consequence of the direction of our parametrization of the ellipse! Some texts write Theorem 1.2 with a $|x'(t)|$ inside the integral to accomodate this, but our text does not. In short -- if you end up with a negative answer for area, just make it positive to get the correct answer.
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