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Math Help - Area of a 2D surface

  1. #1
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    Area of a 2D surface

    I was tutoring this morning and I was stuck in this problem shamefully...

    Please help me answer it.

    The region bounded by y=x^8 , y=1, and revolved along y=4

    My solution was taking the integral of pi*[y^(1/8)]^2 dy

    But the answer is not correct, any help please?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    I was tutoring this morning and I was stuck in this problem shamefully...

    Please help me answer it.

    The region bounded by y=x^8 , y=1, and revolved along y=4

    My solution was taking the integral of pi*[y^(1/8)]^2 dy

    But the answer is not correct, any help please?
    Have you considered shifted down the curve 4 units?
    The advantage of that is that you now have,
    x^8-4 revolved about y=0 and use the standard formula.
    Find the intersection between y=1 and x=8 are x=+/- 1
    Thus,
    INTEGRAL^1_{-1} pi*(x^8-4)^2 dx
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  3. #3
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    Hello, tttcomrader!

    The region bounded by y = x^8 , y = 1, and revolved about y = 4

    You revolved the region about the x-axis . . . . We want "washers".

    . . . . . . . .
    1
    . . V .= .π
    [(x^8 - 4) - 3] dx
    . . . . . . .
    -1

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