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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 6, 2019
 #1
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0

The area is 7 pi.

 Nov 6, 2019
 #2
avatar+105238 
+1

x = 6cos t sin t

y  = 6cos^2 t   

 

The curve is traced  out  once from  0  to  pi

 

Let   g  =  6cost sint

Let  f  =  6cos^2 t

Let f '   = -12cost sint

 

So  we have  that

 

 

                              pi

absolute value  of  ∫   g * f '   dt   =     

                              0

 

 

       pi

- 72 ∫   (cos^2 t) (sin^2 t)  dt    = 

       0  

 

 

       pi

- 72 ∫   (cos t * sin t)^2  dt    = 

       0  

 

 

       pi

- 72 ∫   (sin^2t  *cos^2t)  dt    = 

       0  

 

    

Note  .....  (sin^2 t * cos^2 t )   =    [ 1 - cos (4t) ]  /  8

 

               pi

(- 72 / 8  ) ∫   1  - cos(4t)  dt    = 

              0

 

                                      pi

- 9    [    t    +  sin (4t)/4 ]            =

                                     0

 

-9  [ ( pi +  0 ) -  ( 0 + 0) ]   =

 

 

-9 pi  

 

absolute value   of this  = 

 

9pi  units^2

 

 

This is a circle with a radius of 3 centered at (0, 3) 

 

See  the graph here....https://www.desmos.com/calculator/6me6a7tpzr

 

 

cool cool cool

 Nov 6, 2019
 #3
avatar+23516 
+2

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

 

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{6 \cos t \sin t} \\ x &=& 3\cdot 2 \cos t \sin t \quad & | \quad \small{ 2\cos(t)\sin(t)=\sin(2t)} \\ \mathbf{x} &=& \mathbf{3\sin(2t)} \\ \hline \end{array} \)

 

\(\begin{array}{|rclrcl|} \hline \mathbf{y} &=& \mathbf{6 \cos^2(t) } \\\\ && & \cos(2t) &=& \cos^2(t)-\sin^2(t) \\ && & &=& \cos^2(t)- \left(1-\cos^2(t) \right) \\ && & &=& \cos^2(t)- 1 +\cos^2(t) \\ && & &=& 2\cos^2(t)- 1 \\ &&& 2\cos^2(t) &=& 1+\cos(2t) \\ &&& \mathbf{ \cos^2(t)} &=& \mathbf{\dfrac{1+\cos(2t)}{2}} \\ \\ y &=& 6 \left( \dfrac{1+\cos(2t)}{2} \right) \\ y &=& 3 \Big( 1+\cos(2t) \Big) \\ \mathbf{y} &=& \mathbf{3+3 \cos (2t) } \\ \hline \end{array} \)

 

\(\begin{align*} x &= 3\sin(2t), \\ y &= 3+3 \cos (2t) \end{align*}\)  this is a circle with center \((0,3)\) and radius \(= 3 \)

 

The area is  \(\pi r^2 = 3^2\pi = \mathbf{9 \pi} \)

 

laugh

 Nov 6, 2019
edited by heureka  Nov 6, 2019

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