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
dropping the t for convenience
so find the area under and double it to include the area under the
This will all be much clearer if you plot these functions.
As another trickier way of doing this
This is immediately recognized as a circle of radius 3 centered at (3,0).
It's area is
Hello, orange!
Find the area of the area enclosed by the parametric equations: .
Square both equations: .
Add: . . . . .
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This is a circle with radius 3.