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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

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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*}

Nov 28, 2018

#2
+23788
+9

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*}

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

Nov 28, 2018

#1
+23788
+10

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*}

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

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

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

Nov 28, 2018
edited by heureka  Nov 28, 2018
#2
+23788
+9

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*}

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

heureka Nov 28, 2018
#3
+106515
0

Thanks, heureka

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

CPhill  Nov 28, 2018