r6sintheta Outside the circle r3 Find the area of region ins

Solution

just replace r = 1 with r = 3 and get the whole scenrio same

How do we find the area of the region inside r=6sin(), but outside r=1?

So, here\'s my work thus far:

First off, we know: r2=x2+y2 and sin()=y/r

Therefore,

r=x2+y2 and 6sin()=6yr

The original equation turns into: x2+y2=6y/r ; or equivalently x2+y2=6y

Completing the square turns this into x2+(y3)2=9, a circle centered at (0,3) with a radius of 3

The area of that circle is 9, so the answer is going to be 9 - something. I\'m having a hard time figuring out the something.

The other circle is x2+y2=1, a circle centered at the origin with a radius of 1.

Well, I was thinking, let\'s find the points of intersection by setting the two equations equal to eachother. So set x2+y21=x2+y26y (because they both = 0).

So cancel out the x2 and y2, so we get 1=6y or y=1/6. Plug that back in and find the two x coordinates. Plugging it back into x2+y2=1, so you get +/- the 35/36. So the two intersection points are (35/36,1/6) and (35/36,1/6). So I\'m thinking, alright, just integrate the following from 35/36 to 35/36: x2+y21(x2+y26y)dx. But I get the integral of (6y1)dx.