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Find the area of the surface generated when the curve y = 4 sqrt x is revolved about the x-axis for 60 < x < 96?

User Antar
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Final answer:

To find the surface area of a solid formed by revolving y = 4 √ x about the x-axis from x = 60 to x = 96, set up and evaluate the integral A = 2π ∫ y √dx using the formula for the surface area of a solid of revolution.

Step-by-step explanation:

The question is asking to calculate the surface area of a solid of revolution formed by revolving the graph of the curve y = 4 √ x around the x-axis between x = 60 and x = 96. The formula for the surface area of a solid of revolution about the x-axis is given by the integral A = 2π ∫ y √dx, where y is the function of x, and ds is the differential arc length which can be expressed as √ds = √(1 + (dy/dx)^2)dx. Here, dy/dx is the derivative of y with respect to x, which for the function y = 4 √ x is dy/dx = 2/√ x. Applying these, the integral becomes A = 2π ∫ (4 √ x) √(1 + (2/√ x)^2)dx from x = 60 to x = 96. Evaluating this definite integral gives the surface area.

User Chachi
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