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mathstina Group Title

Evaluate the integral ∬sin(x^2+y^2 )dA where R is the region that lies above the x-axis within the circle x^2 +y^2 =16 by changing into polar coordinates.

  • 2 years ago
  • 2 years ago

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  1. mathstina Group Title
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    as for radius is from -4 to 4. for angle is from 0 to 180 ?

    • 2 years ago
  2. experimentX Group Title
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    yep!!

    • 2 years ago
  3. mathstina Group Title
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    differentiate by r dr dθ ?

    • 2 years ago
  4. experimentX Group Title
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    use radiant (0 to pi)!! and dA = r dr d(theta)

    • 2 years ago
  5. mathstina Group Title
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    ok. let me work out....

    • 2 years ago
  6. experimentX Group Title
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    \[ \int_0^4 \int_0^\pi \sin ( r^2 \cos^2 \theta + r^2 \sin^2 \theta) \; r \; dr\; d\theta \]

    • 2 years ago
  7. mathstina Group Title
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    could u pls explain to me why e radius is from 0 to 4 nt -4 to 4? and hw to get e function?

    • 2 years ago
  8. mathstina Group Title
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    ok i understood. hw abt e integration part?

    • 2 years ago
  9. experimentX Group Title
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    \[ \int_0^4 \int_0^\pi \sin ( r^2) \; r \; d\theta\; dr\ \\ \int_0^4 \sin ( r^2) \; r \; \left [ \theta \right ]_0^\pi dr\\ \] just use subs r^2 = u

    • 2 years ago
  10. mathstina Group Title
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    ok...thanks. let me work out

    • 2 years ago
  11. mathstina Group Title
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    is the ans \[- \pi \cos8

    • 2 years ago
  12. mathstina Group Title
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    \[-\pi \cos 8\]

    • 2 years ago
  13. mathstina Group Title
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    am i right??

    • 2 years ago
  14. experimentX Group Title
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    http://www.wolframalpha.com/input/?i=Integrate[Integrate[Sin[r^2]r%2C+{theta%2C+0%2C+pi}]%2C+{r%2C0%2C+4}] check it out again

    • 2 years ago
  15. experimentX Group Title
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    Integrate[Integrate[Sin[r^2]r, {theta, 0, pi}], {r,0, 4}]

    • 2 years ago
  16. mathstina Group Title
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    hw?

    • 2 years ago
  17. mathstina Group Title
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    i need guidance pls...

    • 2 years ago
  18. experimentX Group Title
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    |dw:1353425265831:dw|

    • 2 years ago
  19. experimentX Group Title
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    do your integration and arrive at \[ \pi [ \cos 0 - \cos 16] \over 2\\ = {\pi(1 - \cos 16) \over 2 }\]just use half angle formula

    • 2 years ago
  20. mathstina Group Title
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    do i need to use mehod by substitution for r^2

    • 2 years ago
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