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SUROJ
 2 years ago
Anybody know how to find the integral of following figure?
SUROJ
 2 years ago
Anybody know how to find the integral of following figure?

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SUROJ
 2 years ago
Best ResponseYou've already chosen the best response.0I mean area of the triangle using polar coordinates

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0using a double integral?

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1rcostheta + 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
 2 years ago
Best ResponseYou've already chosen the best response.0@TuringTest yes @Algebraic! how did u got that? can you explain some more?

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1x=rcostheta y=rsintheta

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1area = integral (r^2)/2 d theta

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0the 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
 2 years ago
Best ResponseYou've already chosen the best response.0but we need to find area of triangle

SUROJ
 2 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! isn't that integral from 0 to pi/4

SUROJ
 2 years ago
Best ResponseYou've already chosen the best response.0Oh yea it's pi/2 thnks @Algebraic! ......but why dA is r^2/s d(theta)

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.1because the area of a circular sector is r^2 /2 times theta

SUROJ
 2 years ago
Best ResponseYou've already chosen the best response.0Finally I got it thnks guys

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1352076292545:dw the small triangle is approximated by a sector of a circle.
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