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CheapFungus
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1.Where did I go wrong in finding the area enclosed inside r = 3 cos θ?
I used the formula 1/2 ∫ ((f(θ)) squared dθ from alpha to beta
I looked for the area of the semicircle from 0 to pi and then multiplied the whole thing by 2, since the equation is a circle, so it looks like this:
=>2*1/2 ∫ from 0 to pi (3 cos θ)squared dθ
=> ∫ from 0 to pi (9 cos squared θ) dθ
=> ∫ from 0 to pi (9 (1/2) (1+cos 2θ) dθ [Using half angle identities]
=> 9/2 ∫ from 0 to pi (1+cos 2θ) dθ
Integrating I get :
=> 9/2 (θ + 1/2 sin 2θ ] from 0 to pi)
=> 9/2 [(pi + 1/2 sin 2pi) - (0 + 1/2 sin 2o)]
=> 9/2 [pi + 0 - 0-0]
So I get the final answer of 9/2pi. Thing is, the answer in the book is 9/4pi and I can't figure out they go there. Help, anyone?
Homework Equations
I used the formula 1/2 ∫ ((f(θ)) squared dθ from alpha to beta
The Attempt at a Solution
I looked for the area of the semicircle from 0 to pi and then multiplied the whole thing by 2, since the equation is a circle, so it looks like this:
=>2*1/2 ∫ from 0 to pi (3 cos θ)squared dθ
=> ∫ from 0 to pi (9 cos squared θ) dθ
=> ∫ from 0 to pi (9 (1/2) (1+cos 2θ) dθ [Using half angle identities]
=> 9/2 ∫ from 0 to pi (1+cos 2θ) dθ
Integrating I get :
=> 9/2 (θ + 1/2 sin 2θ ] from 0 to pi)
=> 9/2 [(pi + 1/2 sin 2pi) - (0 + 1/2 sin 2o)]
=> 9/2 [pi + 0 - 0-0]
So I get the final answer of 9/2pi. Thing is, the answer in the book is 9/4pi and I can't figure out they go there. Help, anyone?
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