# Calculus II

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Before answering the question about "inner loop", let's just get a picture of this thing.

Sine is periodic with period `2pi` , so the function

`3+6 sin(theta)` has period `2 pi`

and if we figure out everything that happens from 0 to `2 pi` we'll know everything about the graph

At `theta=0` , r=3

(note: in the following pictures, the red arrow points in the direction of `theta` )

As `theta` goes from 0 to ` pi/2` , sine is positive and increasing. So the radius is getting larger, up to:

`3+6*sin(pi/2)=9`

Thus:

Then, as `theta` goes from `pi/2` to `pi` , sine decreases back down to 0, so r decreases from 9 back down to 3, and you get a mirror image:

Then, as you increase `theta` a little bit larger than `pi` , sine becomes negative. Eventually, you get to the point where `r=0`

To find where this happens, solve:

`0=3+6 sin theta`

`-3 = 6 sin theta`

`-1/2 = sin theta`

`theta = 7/6 pi, theta = 11/6 pi`

so this first happens at `theta=7/6 pi`

Increase `theta` a little more, and r becomes negative. Thus, the graph is pointing in the opposite direction as `theta`

This continues, until once again r=0, which we already know happens at `11/6 pi`

r is again positive from `11/6 pi` to `2 pi`

so the graph points in the same direction as `theta`

also, ` `sine is increasing there, so, as we would expect, r goes from 0 back up to 3

Thus, the inner loop corresponds to `theta` from `7/6 pi` up to `11/6 pi` , it starts at (0,0),

and the direction of increasing `theta` is:

Ahh! I misread the question!

We want to use the following:

`A=int_(theta_1)^(theta_2) 1/2 r^2 d theta`

As already shown, we want to consider `theta` from `7/6 pi` to `11/6 pi`

So:

`int_(7/6 pi)^(11/6 pi) 1/2 (3+6 sin(theta))^2 d theta`

`=9/2 int_(7/6 pi)^(11/6 pi) (1+2 sin(theta))^2 d theta `

to make it easier to type, I'm going to write this as an indefinite integral:

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

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

(here, use the double angle formula for cosine)

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

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

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

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

Now, evaluating at `7/6 pi` and `11/6 pi` , we have:

at `11/6 pi` :

`3(11/6 pi) - 4 (sqrt(3)/2) - (-sqrt(3)/2)=11/2 pi -3sqrt(3)/2`

at `7/6 pi` :

`3(7/6 pi) - 4 (-sqrt(3)/2)-(sqrt(3)/2)=7/2 pi +3 sqrt(3)/2`

So, the area enclosed is:

`9/2(2 pi-3 sqrt(3))=9pi - (27 sqrt(3))/2`