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anonymous

  • 4 years ago

what is the volume of the region bounded by y=e^-x^2, y=0, x=0, and x=2 when it is rotated around the y axis?

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  1. TuringTest
    • 4 years ago
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    \[y=e^{-x^2}\]?

  2. anonymous
    • 4 years ago
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    yes

  3. TuringTest
    • 4 years ago
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    I don't really see how to integrate this is this how the problem was given originally? or did you get this function from somewhere?

  4. anonymous
    • 4 years ago
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    ya thats the equations we were given :/

  5. dumbcow
    • 4 years ago
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    |dw:1327620023966:dw| the volume is sum of the cross-sectional areas from 0 to 1. each area is a circle with radius of x. From y=0 to e^-4, the radius is constant at 2 y = e^-x^2 lny = -x^2 -ln(y) = x^2 Area = pi*r^2 = pi*x^2 = -pi*ln(y) \[\large V = \pi \int\limits_{0}^{e^{-4}} 4 dy + \pi \int\limits_{e^{-4}}^{1}-\ln (y) dy\]

  6. anonymous
    • 4 years ago
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    where did you get the first integral from?

  7. dumbcow
    • 4 years ago
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    \[= 4\pi e^{-4} -\pi \ln(y)(y-1) {e^{-4} \to 1}\] ... from the part where radius is a constant 2 pi*2^2 = 4pi

  8. anonymous
    • 4 years ago
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    ahhh ok i see makes sense. thank you very much i was having problems with this question lol

  9. TuringTest
    • 4 years ago
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    me too, nice one dumbcow :D

  10. dumbcow
    • 4 years ago
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    your welcome

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