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I was looking at the book's example. The author left the final integration as an exercise, and I was attempting it.

1/2 integral of [ (2 - 2sin(delta) )^2 - 0 ] d delta from 0 to 2pi

for the sake of work, i will let x = delta

(2-2sin(x))^2 => 4 - 8sinx + 4sin^2(x)

and i know that the half angle formula sin^2(x) = (1-cos(2x)) / 2

rewrote:

4 - 8*sin(x) + 2 - 2*cos(2*x)

1/2 * integrate over x

4x + 8*cos(x) + 2x - sin(2*x)

and from 0 to 2pi

1/2 [12pi + 8 - 0] - [0 - 0 - 0 -1], but the book's answer was 6pi.

i am clueless, where is my computation error?

thank you.

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# Homework Help: Help on integrating polar coordinates

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