Find the area of the graph enclosed by the parametric equations,
y = 6 cos(t) sin(t)
x= 6 cos^2(t)
Thus far I did some algebra and came up with 6x^2+6y^2= 36y, but am stuck after that.
Thank you!
Did you plot the curve? The answer is pretty obvious if you do.
To do it formally though
$\displaystyle y^2(t)=36cos^2(t)sin^2(t)=6x(t)(1-\frac{x(t)}{6})$
dropping the t for convenience
$\displaystyle y^2=6x(1-\frac{x}{6})=6x-x^2$
$\displaystyle y=\pm\sqrt{6x-x^2}$
so find the area under $\displaystyle y(x)=\sqrt{6x-x^2}$ and double it to include the area under the $\displaystyle -\sqrt{6x-x^2}$
This will all be much clearer if you plot these functions.
As another trickier way of doing this
$\displaystyle y^2(t)=36cos^2(t)sin^2(t)=36cos^2(t)(1-cos^2(t))$
$\displaystyle y^2=6x(1-\frac{x}{6})=6x-x^2=-(x^2-6x+9-9)=-(x-3)^2+9$
$\displaystyle (x-3)^2+y^2=9$
This is immediately recognized as a circle of radius 3 centered at (3,0).
It's area is $\displaystyle \pi(3)^2=9\pi$
Hello, orange!
Find the area of the area enclosed by the parametric equations: .$\displaystyle \begin{Bmatrix}x &=& 6\cos^2\!t \\ y &=&6\cos t\sin t \end{Bmatrix}$
Square both equations: .$\displaystyle \begin{Bmatrix}x^2 &=& 36\cos^4\!t \\ y^2 &=& 36\cos^2\!t\sin^2\!t \end{Bmatrix}$
Add: . . . . .$\displaystyle x^2+y^2 \;=\;36\cos^4\!t + 36\cos^2\!t\sin^2\!t $
. . . . . . . . . $\displaystyle x^2+y^2\;=\;36\cos^2\!t\underbrace{(\cos^2\!t + \sin^2\!t)}_{\text{This is 1}}$
. . . . . . . . .$\displaystyle x^2 + y^2\;=\;36\cos^2\!t$
. . . . . . . . . $\displaystyle x^2+y^2 \;=\;6(6\cos^2\!t)$
. . . . . . . . .$\displaystyle x^2 + y^2 \;=\;6x$
. . . . . $\displaystyle x^2 - 6x + y^2 \;=\;0$
. . $\displaystyle x^2 - 6x + 9 + y^2 \:=\:9$
. . . . $\displaystyle (x-3)^2 + y^2 \;=\;9$
This is a circle with radius 3.