## SUROJ Group Title Anybody know how to find the integral of following figure? one year ago one year ago

1. SUROJ

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2. SUROJ

I mean area of the triangle using polar co-ordinates

3. TuringTest

using a double integral?

4. Algebraic!

rcostheta + rsintheta =1 r=1/(costheta +sintheta) $\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ r^2 }{2 } d \theta = \frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ 1 }{(\cos \theta +\sin \theta )^2} d \theta$

5. SUROJ

@TuringTest yes @Algebraic! how did u got that? can you explain some more?

6. Algebraic!

x=rcostheta y=rsintheta

7. Algebraic!

area = integral (r^2)/2 d theta

8. SUROJ

why is that?

9. 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 $A=\int\limits_{a}^{b}\frac{ 1 }{ 2 }r^2 d \theta$.

10. SUROJ

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11. SUROJ

but we need to find area of triangle

12. SUROJ

@Algebraic! isn't that integral from 0 to pi/4

13. Algebraic!

pi/2

14. Algebraic!

are you in calc. 2?

15. SUROJ

Oh yea it's pi/2 thnks @Algebraic! ......but why dA is r^2/s d(theta)

16. Algebraic!

because the area of a circular sector is r^2 /2 times theta

17. SUROJ

Finally I got it thnks guys

18. sirm3d

|dw:1352076292545:dw| the small triangle is approximated by a sector of a circle.