- #1
will_lansing
- 19
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[SOLVED] Area of Polar Coordinates
Find the area of the region bounded by r=6-4sin[tex]\Theta[/tex]
A=(1/2)[tex]\int[/tex] r[tex]^{2}[/tex] d[tex]\Theta[/tex]
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)[tex]\int[/tex] [36-48sin[tex]\Theta[/tex]+16sin^2[tex]\Theta[/tex] d[tex]\Theta[/tex]
A=(1/2)[36[tex]\Theta[/tex]-48cos[tex]\Theta[/tex]+8[tex]\Theta[/tex]-4sin2[tex]\Theta[/tex]]
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[tex]\Theta[/tex]
Homework Equations
A=(1/2)[tex]\int[/tex] r[tex]^{2}[/tex] d[tex]\Theta[/tex]
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)[tex]\int[/tex] [36-48sin[tex]\Theta[/tex]+16sin^2[tex]\Theta[/tex] d[tex]\Theta[/tex]
A=(1/2)[36[tex]\Theta[/tex]-48cos[tex]\Theta[/tex]+8[tex]\Theta[/tex]-4sin2[tex]\Theta[/tex]]
and i got the answer to be 63.5 where did i go wrong?