Question

Find the area of the region that lies inside both curves r = A*sin(theta) and r = B*sin(theta), A > 0, B > 0?

Answer 1

\[r_1=A\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{A}{2}\right)^2=\left(\frac{A}{2}\right)^2\]\[r_2=B\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{B}{2}\right)^2=\left(\frac{B}{2}\right)^2\]\[S=\frac{\pi}{4}\cdot\left(\min{\left\{A,B\right\}}\right)^2\]

Answer 2

this does not work

Answer 3

try A = 2 , B = 1

Answer 4

should be 3pi/2 , not sure how you got your answer

Answer 5

ok looks like you did a polar conversion
r = A sin theta
r^2 = r* A sin theta
x^2 + y^2 = A r* sin theta
x^2 + y^2 = Ay
x^2 + y^2 - Ay = 0 complete square
x^2 + (y -A/2)^2 = (A/2)^2
similiarly
x^2 + (y-B/2)^2 = (B/2)^2
directly integrating polar is much easier, btw
so now you have two cartesian equations , and you did ... ?

Answer 6

im not sure how to to integrate it using cartesian equations, hmmm

Answer 7

ok i guess it can be done, but it is tricky

Answer 8

wlog A > B > 0 so

Answer 9

These are circles. Why do you integrate?