anonymous
  • anonymous
Anybody know how to find the integral of following figure?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1352073105071:dw|
anonymous
  • anonymous
I mean area of the triangle using polar co-ordinates
TuringTest
  • TuringTest
using a double integral?

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anonymous
  • anonymous
rcostheta + rsintheta =1 r=1/(costheta +sintheta) \[\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ r^2 }{2 } d \theta = \frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ 1 }{(\cos \theta +\sin \theta )^2} d \theta \]
anonymous
  • anonymous
@TuringTest yes @Algebraic! how did u got that? can you explain some more?
anonymous
  • anonymous
x=rcostheta y=rsintheta
anonymous
  • anonymous
area = integral (r^2)/2 d theta
anonymous
  • anonymous
why is that?
sirm3d
  • sirm3d
the differential of of the area of a sector is given by \[dA=\frac{ 1 }{ 2 }r^2 d \theta \] so the area of the region is a definite integral \[A=\int\limits_{a}^{b}\frac{ 1 }{ 2 }r^2 d \theta \].
anonymous
  • anonymous
|dw:1352074998837:dw|
anonymous
  • anonymous
but we need to find area of triangle
anonymous
  • anonymous
@Algebraic! isn't that integral from 0 to pi/4
anonymous
  • anonymous
pi/2
anonymous
  • anonymous
are you in calc. 2?
anonymous
  • anonymous
Oh yea it's pi/2 thnks @Algebraic! ......but why dA is r^2/s d(theta)
anonymous
  • anonymous
because the area of a circular sector is r^2 /2 times theta
anonymous
  • anonymous
Finally I got it thnks guys
sirm3d
  • sirm3d
|dw:1352076292545:dw| the small triangle is approximated by a sector of a circle.

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