Question

can someone please help me with my question below? thanks!

Answer 1

i'll just put the question up again.

Answer 2

What is the area bounded by

Answer 3

r^2 = 4 cos(2theta)

Answer 4

U can write ur function like this :\[ r^{2} =4(\cos^{2}\theta - \sin^{2} \theta)\]
I dont remmeber now what graph this is, but its something comun.

Answer 5

i don't know what formula to apply though. here are the choices if that helps

a) 8 b) 4 c) 2 d) 1 e) 0.5

Answer 6

Thanks, Sam

Answer 7

yw

Answer 8

Do you know how to find the two theta values from which the integral ranges. I still get kind of confused about that?

Answer 9

I guess just integrate from 0 to 2 theta, and later multiplie by 2, since it is simetric

Answer 10

ok, cool. thank you so much!

Answer 11

i think you integrate from 0 to pi, should give one complete rotation

Answer 12

\[r=f(\theta)\] you want
\[\frac{1}{2}\int_a^bf^2(\theta)d\theta\]
\[\frac{1}{2}\int_0^{\pi}4\cos(2\theta)d\theta\]
\[2\int_0^{\pi}\cos(2\theta)d\theta\]

Answer 13

oh and then you have to double it because you want both parts

Answer 14

ok that is wrong, hold on and let me check what limits we want

Answer 15

cant figure out the damned limits