2.9k views
4 votes
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
by
7.8k points

1 Answer

2 votes

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
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories