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Trigonometric identity double definite integral

  1. Jul 8, 2014 #1
    Double integral of (52-x^2-y^2)^.5

    2<_ x <_ 4
    2<_ y <_ 6

    I get up to this simplicity that results in a zero!!!

    1-cos^2(@) - sin^2(@) = 0

    This identity seems to be useless.

    HELP PLEASE.
     
  2. jcsd
  3. Jul 8, 2014 #2

    ehild

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    The identity is true for angles. But you can not take x=cosθ and y=sinθ here.

    ehild
     
  4. Jul 8, 2014 #3
    Where am I supposed to start?
     
  5. Jul 8, 2014 #4
    And why can't I assume the trig substitutions here? Is it because the substitution results in that zero?
     
  6. Jul 8, 2014 #5

    pasmith

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    Your integral is [tex]
    \int_2^4 \left( \int_2^6 \sqrt{52 - x^2 - y^2}\,dy\right)\,dx.
    [/tex]

    To do the inner integral: substitute [itex]y = \sqrt{52 - x^2}\sin\theta[/itex]. Remember that for this integral [itex]x[/itex] is a constant.

    However I suspect that the [itex]x[/itex]-integral you obtain will be nasty.

    You can instead set [itex]x = r \cos \theta[/itex], [itex]y = r\sin\theta[/itex], but the limits of the [itex]r[/itex]-integral depend on [itex]\theta[/itex] in a complicated way.
     
  7. Jul 8, 2014 #6
    Thanks. I'll try it. It looks nasty
     
  8. Jul 9, 2014 #7
    I've tried

    X=(52)^.5 [cos(@)]

    Y=(52)^.5 [sin(@)]

    I can't get the substitution to match
     
  9. Jul 9, 2014 #8

    BiGyElLoWhAt

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    I like what pasmith said.

    Via pythagoreans theorem -(x^2 + y^2) = -r^2
     
  10. Jul 9, 2014 #9

    BiGyElLoWhAt

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    Also don't forget to change your limits.
     
  11. Jul 9, 2014 #10

    HallsofIvy

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    x and y are independent variables. (x, y) can be any point on the rectangle [itex]2\le x\le 4[/itex], [itex]2\le y\le 6[/itex]. They do NOT lie on a circle of radius 52 about the origin.
     
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