44.5k views
1 vote
What is the surface area of the solid generated by revolving the curve √y=4x around the x-axis?

a) 32π
b) 16π
c) 64π
d) 128π

User Melonie
by
7.9k points

1 Answer

3 votes

Final answer:

To calculate the surface area of the solid generated by revolving the curve √y=4x around the x-axis, the formula for the surface area of a solid of revolution is used. However, the correct answer requires the limits of integration, which are not provided in the question.

Step-by-step explanation:

To find the surface area of the solid generated by revolving the curve √y=4x around the x-axis, we use the formula for the surface area of a solid of revolution. For a function y=f(x) revolved around the x-axis, the surface area S is given by:

S = 2π ∫_{a}^{b} f(x) √(1+[f'(x)]^2) dx

First, we write the given equation in terms of y: y = 16x^2. Then we find the derivative of y with respect to x, which is dy/dx = 32x. Plugging the function y = 16x^2 and its derivative into the formula and integrating:

S = 2π ∫ f(x) √{1+(32x)^2} dx

After performing the integration from the appropriate limits (which, based on the equation are from x=0 to x=√(y)/4), we calculate the surface area. However, to provide a correct answer, we need to know the limits of integration (or additional context to determine them), which are not specified in the question.

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