SUROJ
Anybody know how to find the integral of following figure?
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SUROJ
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|dw:1352073105071:dw|
SUROJ
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I mean area of the triangle using polar co-ordinates
TuringTest
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using a double integral?
Algebraic!
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rcostheta + rsintheta =1
r=1/(costheta +sintheta)
\[\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ r^2 }{2 } d \theta = \frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ 1 }{(\cos \theta +\sin \theta )^2} d \theta \]
SUROJ
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@TuringTest yes @Algebraic! how did u got that? can you explain some more?
Algebraic!
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x=rcostheta
y=rsintheta
Algebraic!
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area = integral (r^2)/2 d theta
SUROJ
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why is that?
sirm3d
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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 \[A=\int\limits_{a}^{b}\frac{ 1 }{ 2 }r^2 d \theta \].
SUROJ
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|dw:1352074998837:dw|
SUROJ
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but we need to find area of triangle
SUROJ
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@Algebraic! isn't that integral from 0 to pi/4
Algebraic!
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pi/2
Algebraic!
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are you in calc. 2?
SUROJ
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Oh yea it's pi/2 thnks @Algebraic! ......but why dA is r^2/s d(theta)
Algebraic!
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because the area of a circular sector is r^2 /2 times theta
SUROJ
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Finally I got it thnks guys
sirm3d
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|dw:1352076292545:dw| the small triangle is approximated by a sector of a circle.