anonymous
  • anonymous
Find the area of the region that lies inside both curves r = A*sin(theta) and r = B*sin(theta), A > 0, B > 0?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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perl
  • perl
what does your notes say
dumbcow
  • dumbcow
these are polar equations?
anonymous
  • anonymous
yes

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lgbasallote
  • lgbasallote
...i am thinking...integration???
anonymous
  • anonymous
yes
lgbasallote
  • lgbasallote
i hate integration -____- we haven't taken that lesson yet. Sorry. Can't help :))
perl
  • perl
when does A sin t = B sin t
dumbcow
  • dumbcow
well with rectangular coordinates, you just integrate f(x) -g(x) from where they intersect does it apply with polar coordinates as well? not as familiar working with polar :|
perl
  • perl
nope
perl
  • perl
cool picture
perl
  • perl
thats very creative :)
dumbcow
  • dumbcow
haha thanks
perl
  • perl
try an example,
perl
  • perl
r = sin theta, r = 2 sin theta
perl
  • perl
here is Paul's notes http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx
dumbcow
  • dumbcow
wait they would never intersect right?
perl
  • perl
they should
dumbcow
  • dumbcow
ha except at theta = npi
perl
  • perl
integral 1/2 (r1^2 - r2^2 )
perl
  • perl
where r = f(theta)
perl
  • perl
ok so i graphed r = sin theta, and r= |dw:1328796260702:dw|2 sin theta
perl
  • perl
and sin theta = 2 sin theta when theta = 0, 2pi
perl
  • perl
so in your problem , we can assume without loss of generality, A > B. in which case we have |dw:1328796488104:dw|
perl
  • perl
|dw:1328796701325:dw|
perl
  • perl
|dw:1328796760182:dw|
perl
  • perl
you can even make it less restrictive by saying |dw:1328799289713:dw|
perl
  • perl
did you test your solution?
perl
  • perl
your solution does not work for r = 2 sin theta, r = 1 sin theta

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