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SUROJ

  • 2 years ago

Anybody know how to find the integral of following figure?

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  1. SUROJ
    • 2 years ago
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    |dw:1352073105071:dw|

  2. SUROJ
    • 2 years ago
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    I mean area of the triangle using polar co-ordinates

  3. TuringTest
    • 2 years ago
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    using a double integral?

  4. Algebraic!
    • 2 years ago
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    rcostheta + rsintheta =1 r=1/(costheta +sintheta) \[\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ r^2 }{2 } d \theta = \frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ 1 }{(\cos \theta +\sin \theta )^2} d \theta \]

  5. SUROJ
    • 2 years ago
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    @TuringTest yes @Algebraic! how did u got that? can you explain some more?

  6. Algebraic!
    • 2 years ago
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    x=rcostheta y=rsintheta

  7. Algebraic!
    • 2 years ago
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    area = integral (r^2)/2 d theta

  8. SUROJ
    • 2 years ago
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    why is that?

  9. sirm3d
    • 2 years ago
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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 \[A=\int\limits_{a}^{b}\frac{ 1 }{ 2 }r^2 d \theta \].

  10. SUROJ
    • 2 years ago
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    |dw:1352074998837:dw|

  11. SUROJ
    • 2 years ago
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    but we need to find area of triangle

  12. SUROJ
    • 2 years ago
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    @Algebraic! isn't that integral from 0 to pi/4

  13. Algebraic!
    • 2 years ago
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    pi/2

  14. Algebraic!
    • 2 years ago
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    are you in calc. 2?

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

  16. Algebraic!
    • 2 years ago
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    because the area of a circular sector is r^2 /2 times theta

  17. SUROJ
    • 2 years ago
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    Finally I got it thnks guys

  18. sirm3d
    • 2 years ago
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    |dw:1352076292545:dw| the small triangle is approximated by a sector of a circle.

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