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dhokes
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The arc of the parabola with equation y²=4ax, where a is a positive constant, from the origin O to A (a, 2a) is rotated through 2π radians about the x-axis. How do I show that the area of the surface generated is:

8πa² (2√2-1) / 3

:confused: ??
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dvs
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y^2 = 4ax => dy/dx = 2a/y = a/sqrt(ax)

So that:
sqrt(1 + (dy/dx)^2) = sqrt(1 + a/x)

Then:
surface area = 2pi ∫ y sqrt(1 + (dy/dx)^2) dx
= 2pi ∫ 2 sqrt(ax) sqrt(1 + a/x) dx
= 4pi ∫ sqrt(ax + a^2) dx
= 4pi a ∫(x + a)^(1/2) dx
= 4pi a [(2/3) (x + a)^(3/2)]{0, a}
= (8/3)pi a [(2a)^(3/2) - a^(3/2)]
= (8/3)pi a^2 (2sqrt(2) - 1)
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dhokes
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thanks dvs
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Angel Interceptor
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(Original post by dhokes)
The arc of the parabola with equation y²=4ax, where a is a positive constant, from the origin O to A (a, 2a) is rotated through 2π radians about the x-axis. How do I show that the area of the surface generated is:

8πa² (2√2-1) / 3

:confused: ??
Please post future mathematical problems in the Maths section of Academic Help.
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Comp_Genius
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(Original post by Angel Interceptor)
Please post future mathematical problems in the Maths section of Academic Help.
lol, it is already:confused: :rolleyes:
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