Find the area enclosed by the graph of the parametric equations \(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)
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
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 \(\pi r^2 = 3^2\pi = \mathbf{9 \pi} \)