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### Section 9.3 : Area with Parametric Equations

2. Determine the area of the region below the parametric curve given by the following set of parametric equations. You may assume that the curve traces out exactly once from right to left for the given range of $$t$$. You should only use the given parametric equations to determine the answer.

$x = 3 - {\cos ^3}\left( t \right)\hspace{0.25in}\,\,y = 4 + \sin \left( t \right)\hspace{0.25in}0 \le t \le \pi$ Show Solution

There really isn’t too much to this problem. Just recall that the formula from the notes assumes that $$x = f\left( t \right)$$ and $$y = g\left( t \right)$$. So, the area under the curve is,

\begin{align*}A & = \int_{0}^{\pi }{{\left( {4 + \sin \left( t \right)} \right)\left( {3{{\cos }^2}\left( t \right)\sin \left( t \right)} \right)\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + 3{{\cos }^2}\left( t \right){{\sin }^2}\left( t \right)\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + 3{{\left[ {\frac{1}{2}\sin \left( {2t} \right)} \right]}^2}\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + \frac{3}{4}{{\sin }^2}\left( {2t} \right)\,dt}}\\ & = \int_{0}^{\pi }{{12{{\cos }^2}\left( t \right)\sin \left( t \right) + \frac{3}{8}\left( {1 - cos\left( {4t} \right)} \right)\,dt}}\\ & = \left. {\left( { - 4{{\cos }^3}\left( t \right) + \frac{3}{8}t - \frac{3}{{32}}\sin \left( {4t} \right)} \right)} \right|_{0}^{\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{8 + \frac{3}{8}\pi }}\end{align*}

You did recall how to do all the trig manipulations and trig integrals to do this integral correct? If not you should go back to the Integrals Involving Trig Functions section to do some review/problems.