anonymous
  • anonymous
area shared between the 2 circles r = 6 cos(theta), r = 6 sin(theta)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
okokok
anonymous
  • anonymous
wat grade r u in
anonymous
  • anonymous
well

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anonymous
  • anonymous
college calc 3 but doing calc 2 review but I took calc 2 years ago and forgot everything
anonymous
  • anonymous
oh got u
anonymous
  • anonymous
wats ur queestion
anonymous
  • anonymous
sooooooooooooooo
anonymous
  • anonymous
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) ,
anonymous
  • anonymous
yes that is write than y do u hv this question here
anonymous
  • anonymous
well................................................
anonymous
  • anonymous
is it pi/4?
anonymous
  • anonymous
yes yes yes yes
anonymous
  • anonymous
im a 7th grader and i know this very well and im a strat a studenet
anonymous
  • anonymous
maybe you should invest a little time in grammar then
dumbcow
  • dumbcow
@dany10 , do you have the answer?...i am also having trouble setting up polar integral but i solved area using rectangular coord
anonymous
  • anonymous
wat
anonymous
  • anonymous
oh yes i know how to do it i had told him
dumbcow
  • dumbcow
@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
anonymous
  • anonymous
oh no that is not my first lauguage lol
anonymous
  • anonymous
well
anonymous
  • anonymous
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 )
anonymous
  • anonymous
that is apsullutlly right air five high air five i mean
anonymous
  • anonymous
@vk278 I think your just trolling lol and don't know what i am taking about
anonymous
  • anonymous
i do ido lol
anonymous
  • anonymous
so any other questions i could answer for u
anonymous
  • anonymous
welllllllllllll................................
dumbcow
  • dumbcow
ok your limits are off, just figured it out |dw:1378245523052:dw| 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
anonymous
  • anonymous
yo
anonymous
  • anonymous
hello u there
goformit100
  • goformit100
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