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Find the area enclosed by the graph of the parametric equations\(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)
 

RektTheNoob  Nov 28, 2018

Best Answer 

 #2
avatar+20680 
+6

Find the area enclosed by the graph of the parametric equations
\(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)

 

\(\text{Second attempt:} \\ \begin{array}{|rcll|} \hline x &=& 6 \cos( t) \sin( t) \\ y &=& 6 \cos^2( t) \\ \hline x^2+y^2 &=& \Big( 6\cos( t) \sin( t) \Big)^2 + \Big( 6 \cos^2( t) \Big)^2 \\ x^2+y^2 &=& 6^2 \cos^2( t) \sin^2( t) + 6^2 \cos^2( t)\cos^2( t) \\ x^2+y^2 &=& 6^2 \cos^2( t)\Big( \underbrace{\sin^2( t) + \cos^2( t)}_{=1} \Big) \\ x^2+y^2 &=& 6^2 \cos^2( t) \\ x^2+y^2 &=& 6\cdot 6 \cos^2( t) \quad & | \quad y = 6 \cos^2( t) \\ \mathbf{x^2+y^2} & \mathbf{=} & \mathbf{6y} \\\\ y^2-6y+x^2 &=& 0 \\ (y-3)^2-9+x^2 &=& 0 \\ (y-3)^2+x^2 &=& 9 \\ \mathbf{(y-3)^2+(x+0)^2} & \mathbf{=} & \mathbf{3^2} \\ \hline \end{array}\\ \begin{array}{|l|} \hline \text{This is a circle with center $(0,3)$ and radius = $3$ } \\ \text{The area enclosed is $ \pi r^2 = \pi \cdot 3^2 = \mathbf{9 \pi} $ } \\ \hline \end{array}\)

 

laugh

heureka  Nov 28, 2018
 #1
avatar+20680 
+7

Find the area enclosed by the graph of the parametric equations

\(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)

 

\(\text{Formula 1:} \\ \begin{array}{|rcll|} \hline 2 \cos( t) \sin( t) &=& \sin(2t) \quad & | \quad \cdot 3 \\ \mathbf{6 \cos( t) \sin( t) } & \mathbf{=} & \mathbf{3\sin(2t)} \\ \hline \end{array} \)

 

\(\text{Formula 2:} \\ \begin{array}{|rcll|} \hline 2 \cos^2( t) &=& 1+\cos(2t) \quad & | \quad \cdot 3 \\ \mathbf{6 \cos^2( t) } & \mathbf{=} & \mathbf{3+3\cos(2t)} \\ \hline \end{array}\)

 

\(\text{Substitute:} \\ \begin{array}{|rcll|} \hline x &=& 6 \cos( t) \sin( t) \quad & | \quad \mathbf{6 \cos( t) \sin( t) =3\sin(2t)} \\ &=& 3\sin(2t) \\ &=& 0+ 3\sin(2t) \\\\ y &=& 6 \cos^2( t) \quad & | \quad \mathbf{6 \cos^2( t)=3+3\cos(2t)} \\ &=& 3+3\cos(2t) \\\\ \hline x &=& 0+3\sin(2t) \\ y &=& 3+3\cos(2t) \\ \hline \end{array}\\ \begin{array}{|l|} \hline \text{This is a circle with center $(0,3)$ and radius = $3$ } \\ \text{The area enclosed is $ \pi r^2 = \pi \cdot 3^2 = \mathbf{9 \pi} $ } \\ \hline \end{array}\)

 

 

laugh

heureka  Nov 28, 2018
edited by heureka  Nov 28, 2018
 #2
avatar+20680 
+6
Best Answer

Find the area enclosed by the graph of the parametric equations
\(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)

 

\(\text{Second attempt:} \\ \begin{array}{|rcll|} \hline x &=& 6 \cos( t) \sin( t) \\ y &=& 6 \cos^2( t) \\ \hline x^2+y^2 &=& \Big( 6\cos( t) \sin( t) \Big)^2 + \Big( 6 \cos^2( t) \Big)^2 \\ x^2+y^2 &=& 6^2 \cos^2( t) \sin^2( t) + 6^2 \cos^2( t)\cos^2( t) \\ x^2+y^2 &=& 6^2 \cos^2( t)\Big( \underbrace{\sin^2( t) + \cos^2( t)}_{=1} \Big) \\ x^2+y^2 &=& 6^2 \cos^2( t) \\ x^2+y^2 &=& 6\cdot 6 \cos^2( t) \quad & | \quad y = 6 \cos^2( t) \\ \mathbf{x^2+y^2} & \mathbf{=} & \mathbf{6y} \\\\ y^2-6y+x^2 &=& 0 \\ (y-3)^2-9+x^2 &=& 0 \\ (y-3)^2+x^2 &=& 9 \\ \mathbf{(y-3)^2+(x+0)^2} & \mathbf{=} & \mathbf{3^2} \\ \hline \end{array}\\ \begin{array}{|l|} \hline \text{This is a circle with center $(0,3)$ and radius = $3$ } \\ \text{The area enclosed is $ \pi r^2 = \pi \cdot 3^2 = \mathbf{9 \pi} $ } \\ \hline \end{array}\)

 

laugh

heureka  Nov 28, 2018
 #3
avatar+92814 
0

Thanks, heureka

 

Your second way is more elegant.....it translates easily into a circular function...

 

cool cool cool

CPhill  Nov 28, 2018

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