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    Ok, it's me again with a couple of P5 questions that I can't do; this time they relate to arc length. Here goes:

    1. x^(2/3) + y^(2/3) = 1

    a) Show that the perimeter of the curve is 6.
    b) Find, in terms of pi, the surface area generated when the quadrant of the curve for which x ≥ 0, y ≥ 0, when rotated completely about the y-axis (ANS: 6pi/5)

    For a, I differentiated the terms implicitly (possibly wrongly) and got:

    dy/dx = 0.5(3-2x^-1/3)(y^3)

    But I couldn't rearrange [1 + (dy/dx)^2]^1/2 to get me anything that I could integrate, therefore I couldn't do part a) or consequently part b).

    2. A sphere has radius a - find the curved surface area of a zone of the sphere cut off by 2 parallel planes a distance b apart.

    well, I know that a sphere has surface area = (4pi)a^2

    so I suppose you could say dS/da = (8pi)a

    but then I'm not sure about the integral - should it be:

    0b [ 1 + 64(pi)2 a2 ]^1/2

    or something else - I can't find a way of rearranging the above integral into something nice and integratable. Could someone point out where I'm going wrong, please ? I'd be really grateful (rep bonus, albeit a very meagre one... I've only been here for a week or so!)
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    (Original post by Yttrium)
    1. x^(2/3) + y^(2/3) = 1
    a) Show that the perimeter of the curve is 6.
    b) Find, in terms of pi, the surface area generated when the quadrant of the curve for which x ≥ 0, y ≥ 0, when rotated completely about the y-axis (ANS: 6pi/5)
    Q1(a)
    y=[1-x^(2/3)]^(3/2)
    dy/dx=(3/2)[-(2/3)x^(-1/3)][1-x^(2/3)]^(1/2)
    dy/dx=[-x^(-1/3)][1-x^(2/3)]^(1/2)
    (dy/dx)^2=[x^(-2/3)][1-x^(2/3)]
    (dy/dx)^2=x^(-2/3)-1
    1+(dy/dx)^2=x^(-2/3)
    [1+(dy/dx)^2]^0.5=x^(-1/3)
    Since we have an expression for [1+(dy/dx)^2]^0.5 in terms of x we want limits in terms of x. Clearly the extreme values of x are +-1, and as the graph is symmetrical about the x and y axes we can save the arc length is given by:
    S=2∫x^(-1/3) dx, with limits -1, 1.
    S=4∫x^(-1/3) dx, with limits 0, 1.
    S=6[x^(2/3)] with limits 0,1.
    S=6.

    (b) We require:
    SA=2pi ∫y[1+(dy/dx)^2] dx with x limits 0,1.
    SA=2pi ∫[1-x^(2/3)]^(3/2).x^(-1/3) dx with limits 0,1.
    Let u=1-x^(2/3). du/dx=-(2/3)x^(-1/3), du=-(2/3)x^(-1/3) dx
    SA=2pi ∫[u]^(3/2).(-3/2)du with limits 1,0.
    SA=3pi ∫u^(3/2) with limits 0,1.
    SA=3pi[(2/5)u^(5/2)] with limits 0,1.
    SA=3pi[2/5]
    SA=6pi/5.
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    2. A sphere has radius a - find the curved surface area of a zone of the sphere cut off by 2 parallel planes a distance b apart.
    You know that a sphere is formed by rotating a (semi)circle about the axes.
    We need to find the surface area of a length of the corresponding arc of (semi)circle.
    The circle we need to consider has equation x^2+y^2=a^2.
    We could take the limits for the rotation to be k, k+b, where k is any real contant, but as the surface area will be the same for rotation of any part of the curve lets just use the limits 0,b.
    y=[a^2-x^2]^(1/2)
    dy/dx=(1/2)(-2x)(a^2-x^2)^(-1/2)
    dy/dx=-x(a^2-x^2)^(-1/2)
    (dy/dx)^2=x^2(a^2-x^2)^-1.
    1+(dy/dx)^2=x^2/(a^2-x^2)+1
    1+(dy/dx)^2=x^2/(a^2-x^2)+(a^2-x^2)//(a^2-x^2)
    1+(dy/dx)^2=a^2/(a^2-x^2)
    [1+(dy/dx)^2]^0.5=a/(a^2-x^2)^0.5
    SA=2pi∫y[1+(dy/dx)^2]^0.5 dx
    SA=2pi∫[(a^2-x^2)^0.5].[a/(a^2-x^2)^0.5] dx with limits 0,b
    SA=2pi∫a dx with limits 0,b
    SA=2pia[x] with limits 0,b
    SA=2piab.
 
 
 
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