Question

using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles x^2 + y^2 =144 and x^2 - 12x + y^2 =0

Answer 1

first I think it might we better if we write that other circle in what's it called...
let's just called it center-radius form

\[x^2-12x+y^2=0 \\ x^2-12x+6^2+y^2=6^2 \text{ note: added } 6^2 \text{ on both sides } \\ (x-6)^2+y^2=36 \]


hmm so the equations look like this when we graph them:


anyways I thought it would be nice to have a visual
but your thingy says to use polar coordinates
\[r^2=x^2+y^2 \\ x= r \cos(\theta) \\ y= r \sin(\theta)\]
so your equations can be translated into polar coordinates by saying this:

\[r^2=144 \\ r^2=12 r \cos(\theta) \]


we can solve both of these equations for r
\[r^2=144 \text{ means } r=12 \text{ or } r=-12 \\ \text{ though these really are both the same equation } \\ \text{ we will use } r=12 \text{ since I plan to use positive \theta} \\ \text{ and we are in the first quadrant }\]

\[r^2=12 r \cos(\theta) \\ r^2-12 r \cos(\theta)=0 \\ r(r-12 \cos(\theta))=0 \\ r=0 \text{ or } r-12 \cos(\theta)=0 \\ \\ r=0 \text{ or } r=12 \cos(\theta) \\ \]

So we are looking at theta between 0 and pi/2
and the equations r=12 and r=12 cos(theta)