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lokine | eNotes Newbie

Posted October 31, 2013 at 9:10 PM via web

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted October 31, 2013 at 11:25 PM (Answer #1)

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For starters, check out:

http://www.enotes.com/homework-help/calculus-ii-461874#answer-643292

It will give you an idea of the picture involved, as well as the area enclosed by the inner loop.

Using that information, we have that:

the inner loop encloses area `9 pi - (27 sqrt(3))/2`

and the inner loop corresponds to `theta` from `7/6 pi` to `11/6 pi`

The outer loop consists of `theta` from 0 to `7/6 pi` and from `11/6 pi` to `2 pi`

It is easier, though, to think of the segment from `11/6 pi` to `2pi` as` `

`-1/6 pi` to 0

Thus, the outer loop corresponds to `theta` from `-1/6 pi` to `7/6 pi`

The formula we want is:

`A=int_(-1/6 pi)^(7/6 pi) 1/2 r^2 d theta`

Thus:

`A=int_(-1/6 pi)^(7/6 pi) 1/2 (3+6 sin(theta))^2 d theta`

(The same steps will follow as in the answer in the link)

`A=9/2 int_(-1/6 pi)^(7/6 pi) (1+2 sin(theta)^2 d theta=`

`9/2 int_(-1/6 pi)^(7/6 pi) 1+4 sin(theta) + 4 sin^2(theta) d theta`

`int 1 + 4 sin(theta)+4 sin^2(theta) d theta = `

`theta-4 cos(theta)+4 int sin^2 theta d theta=`

`theta-4 cos theta+4 int (1-cos(2 theta))/2 d theta=`

`theta-4 cos theta+int 2 - 2 cos(2 theta) d theta=`

` ` `theta-4 cos theta+2 theta-sin(2 theta)=`

`3 theta-4 cos theta - sin(2 theta)`

Evaluating at `7/6 pi` and at `-1/6 pi` we have:

at `7/6 pi` :

`7/2 pi +(3 sqrt(3))/2`

at `-1/6 pi` :

`-pi/2-(3 sqrt(3))/2`

Thus, the area enclosed by the outer loop is:

`9/2 (4 pi+3 sqrt(3))=36 pi+(27 sqrt(3))/2`

Taking the area enclosed by the outer loop minus the area enclosed by the inner loop is:

`27 pi+27sqrt(3)`

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