will_lansing
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[SOLVED] Area of Polar Coordinates
Find the area of the region bounded by r=6-4sin\Theta
A=(1/2)\int r^{2} d\Theta
I'm not sure what the bounds are but I thought they were 0 to 2pi. Am I wrong if so how then do you go about finding the bounds?
A=(1/2)\int [36-48sin\Theta+16sin^2\Theta d\Theta
A=(1/2)[36\Theta-48cos\Theta+8\Theta-4sin2\Theta]
and i got the answer to be 63.5 where did i go wrong?
Homework Statement
Find the area of the region bounded by r=6-4sin\Theta
Homework Equations
A=(1/2)\int r^{2} d\Theta
The Attempt at a Solution
I'm not sure what the bounds are but I thought they were 0 to 2pi. Am I wrong if so how then do you go about finding the bounds?
A=(1/2)\int [36-48sin\Theta+16sin^2\Theta d\Theta
A=(1/2)[36\Theta-48cos\Theta+8\Theta-4sin2\Theta]
and i got the answer to be 63.5 where did i go wrong?