Question

Find the area of the surface generated when the given curve is revolved about the given axis. y = square root of 8x - x^2, for 1 <= x <= 7; about the x-axis 

          Find the area of the surface generated when the given curve is revolved about the given axis.
y = square root of 8x - x^2, for 1 <= x <= 7; about the x-axis

Added by Jessica M.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the area of the surface generated when the given curve is revolved about the given axis. y = square root of 8x - x^2, for 1 <= x <= 7; about the x-axis
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Transcript

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00:01 Hi, today we are solving the question in which given function is y is equals to 5x whole raised to the power 1 by 3 for x lying between 0 and 25 about the y -axis.
00:18 So the given limits are 0 and 25.
00:23 Now differentiating y with respect to x.
00:27 So dy by dx is equals to 5 raised to the power 1 by 3 and 3x raised by 3x raised to the power 2 by 3.
00:37 That implies 1 plus dy by dx whole square is equals to 1 plus 5 raised to the power 2 by 3 by 9x raised to the power 4 by 3.
00:55 Here under root 1 plus dy by dx whole square is equals to under root 9x raised to the power 4 by 3 plus 5 raised to the power 2 by 3 divided by 9x raised to the power 4 by 3.
01:14 Now let us find the limit.
01:17 So when y is equals to 0.
01:29 When here given as when x is equals to 0...
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