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anonymous

  • 5 years ago

how do i find the volume revolving around y=2 with the line y=2 and y=2sinx between 0 and pi/2

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  1. anonymous
    • 5 years ago
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    i figured it out...but i dont know why it must be calculated that way

  2. anonymous
    • 5 years ago
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    for those who want to know what i got: int(pi(2-2sinx)^2dx,x,0,pi/2)= pi(3pi-8)

  3. anonymous
    • 5 years ago
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    You are going to use the integral 0 o pi /2 of pi (Upper line minus the lower line)^2...remember revolving about a line that is not the x-axis means going from upper function to lower function and then using that as your radius...

  4. anonymous
    • 5 years ago
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    will reflecting the curve y=2sinx over the y=2 line work?

  5. anonymous
    • 5 years ago
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    when i reflected it i got y=-2sinx+2...is that the correct reflection...then tried to find the radius by upper-lower divided by 2

  6. anonymous
    • 5 years ago
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    i think i got the wrong reflection, which is why it didn't work when i did it that way. The reflection of y=2sinx over the line y=2 is really y=-2sinx+4....I see my mistake now. Finding the radius that way actually yields the same radius as just upper line minus lower line.

  7. anonymous
    • 5 years ago
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    If you have reflected correctly then the integral will yield the same result...remember you are simply trying to figure R(x) which is measure from axis of rotation to the outside edge

  8. anonymous
    • 5 years ago
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    sorry i realized...the answer i got doesn't match the book. Can you find the volume of for y=2sinx and y=2 revolving around y=2 between x=0 and pi/2? so i can check my answer?

  9. anonymous
    • 5 years ago
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    maybe the book is wrong

  10. anonymous
    • 5 years ago
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    nevermind found the mistake...when taking integral of -8sinx i forgot to switch the sign..

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