Question

area shared between the 2 circles r = 6 cos(theta), r = 6 sin(theta)

Answer 1

okokok

Answer 2

wat grade r u in

Answer 3

well

Answer 4

college calc 3 but doing calc 2 review but I took calc 2 years ago and forgot everything

Answer 5

oh got u

Answer 6

wats ur queestion

Answer 7

sooooooooooooooo

Answer 8

I need the area shared between the 2 circles, r = 6 cos(theta) and r = 6 sin(theta) , and I am having a hard time determing the interval that is the shared area , I believe you set the 2 equations equal to each other like this, 6cos(theta) = 6sin(theta) ,

Answer 9

yes that is write than y do u hv this question here

Answer 10

well................................................

Answer 11

is it pi/4?

Answer 12

yes yes yes yes

Answer 13

im a 7th grader and i know this very well and im a strat a studenet

Answer 14

maybe you should invest a little time in grammar then

Answer 15

@dany10 , do you have the answer?...i am also having trouble setting up polar integral but i solved area using rectangular coord

Answer 16

wat

Answer 17

oh yes i know how to do it i had told him

Answer 18

@vk278 , haha if you are in 7th grade and can do calc2 you my friend are a genius
...im guessing english is not your first language

Answer 19

oh no that is not my first lauguage lol

Answer 20

well

Answer 21

well I came up with this A =(1/2) intergral from -pi/4 to pi/4 (6cos(theta)^2) - intergral from -pi/4 to pi/4 (6sin(theta)^2 )

Answer 22

that is apsullutlly right air five high air five i mean

Answer 23

@vk278 I think your just trolling lol and don't know what i am taking about

Answer 24

i do ido lol

Answer 25

so any other questions i could answer for u

Answer 26

welllllllllllll................................

Answer 27

ok your limits are off, just figured it out

1.
\[\frac{1}{2} \int\limits _0 ^{\pi/4} (6\sin \theta)^{2} d \theta\]
2.
\[\frac{1}{2} \int\limits\limits _{\pi/4} ^{\pi/2} (6\cos \theta)^{2} d \theta\]
add them together to get total area