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# help

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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
0

The area is 7 pi.

Nov 6, 2019
#2
+109202
+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

Nov 6, 2019
#3
+24364
+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  $$\pi r^2 = 3^2\pi = \mathbf{9 \pi}$$

Nov 6, 2019
edited by heureka  Nov 6, 2019