calc

Find the exact area of the surface obtained by rotating the curve about the x-axis.

9x = y2 + 36, 4 ≤ x ≤ 8

I keep on getting 4pi (3^(3/2) +1)

so I tried to put the answer in that way and also 24.78pi
But it is wrong

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  1. 9x = y^2 + 36

    I assume you just mean the part above the x-axis, since the curve is symmetric about that axis. So, using discs, the volume is

    v = ∫[4,8] πr^2 dx
    where r^2 = y^2 = 9x-36
    v = π∫[4,8] 9x-36 dx
    = π(9/2 x^2 - 36x) [4,8]
    = 72π

    Using shells, we have

    v = ∫[0,6] 2πrh dy
    where r = y and h = (8-x) = 8-(y^2+36)/9)
    v = 2π∫[0,6] y(8-(y^2+36)/9) dy
    2π(2y^2 - y^4/36) [0,6]
    = 72π

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  2. I am just suppose to find the surface area. We are studying arc length

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  3. Oops. Oh well. The observation about symmetry still holds. The surface area is just the sum of circumferences of many circles of radius y. So,

    dS = 2πy ds
    S = 2π∫[4,8] y√(1+y'^2) dx

    9x = y^2+36, so y = 3√(x-4)
    18 = 2yy'
    y' = 9/y = 9/√(9x-36)
    so, y'^2 = 81/(9x-36) = 9/(x-4)

    S = 2π∫[4,8] (3√(x-4))√((x+5)/(x-4)) dx
    = 6π∫[4,8] √(x+5) dx
    = 4π(x+5)^(3/2) [4,8]
    = 79.49 π

    Better double-check my math.

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  4. I redid the problem using polar coordinates, where I set the (0,0,0) at the base of the paraboloid section, so that

    z = 4 - (x^2+y^2)/9

    That means that we have only to evaluate

    S = 1/9 ∫[0,2π]∫[0,6] r√(81+4r^2) dr dθ
    and we get a nice tidy 49π

    Not sure where I went wrong using x = g(y) and rectangular coordinates. You can see a discussion of the coordinate change at

    http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/surface/surface.html

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