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!

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- Dec 16th 2013, 12:00 PMorangeParametric equations?
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! - Dec 16th 2013, 12:10 PMPlatoRe: Parametric equations?
- Dec 16th 2013, 12:44 PMromsekRe: Parametric equations?
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 $\displaystyle \pi(3)^2=9\pi$ - Dec 16th 2013, 01:06 PMSorobanRe: Parametric equations?
Hello, orange!

Quote:

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.

- Dec 16th 2013, 04:18 PMorangeRe: Parametric equations?
Thank you for all these wonderful explanations!