Polar Coordinates problem area of region

[muddyjch]

Homework Statement

Find the area of the region inside: r = 9 sinθ but outside: r = 1

Homework Equations

The Attempt at a Solution

r = 9 sinθ is a circle with center at (0, 4/2) and radius 4/2 while r= 1 is a circle with center at (0, 0) and radius 1. The two curves intersect where sin(θ)= 1/9. For 0 ≤ θ ≤ π that is satified by θ=sin^-1(1/9) and θ=π-sin^-1(1/9).
This is where i get lost setting up the differential for the area?

 

[Raskolnikov]

r = 9 sinθ is not what you say. I don't know where you went astray in your thinking, but if polar coordinates are giving you trouble, then think of it this way:

r = 9 sinθ is the same thing as r^2 = 9r sinθ, which is equivalent to x^2 + y^2 = 9y, or...
x^2 + (y-9/2)^2 = (9/2)^2. That's a circle centered at (0,9/2) with a radius of 9/2 units.

Can you finish the rest by yourself? I hope this helps.

EDIT: There, I fixed it :D

 

[Raskolnikov]

For 0 ≤ θ ≤ π that is satified by θ=sin^-1(1/9) and θ=π-sin^-1(1/9).

Also, how did you arrive at this conclusion? Using that logic, we could let n = sin^-1(1/9) - k, where k is any real number. Then, using your logic, θ would also equal k in all these cases, i.e., sin^-1(1/9) has infinitely many solutions on a quite finite interval.

 

[muddyjch]

Also, how did you arrive at this conclusion? Using that logic, we could let n = sin^-1(1/9) - k, where k is any real number. Then, using your logic, θ would also equal k in all these cases, i.e., sin^-1(1/9) has infinitely many solutions on a quite finite interval.

It is actually a pi sign but a really bad one sorry for the confusion

 

[HallsofIvy]

Using Latex, \theta= sin^{-1}(1/9) and \theta= \pi- sin^{-1}(1/9).

Of course, for any \alpha, sin(\alpha)= sin(\pi- \alpha).

You should have learned when you first learned integration in polar coordinates that "dxdy" becomes "r dr d\theta".