Question

write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the x-axis. y = 49 − x² , −5 ≤ x ≤ 5

Answer 1

To find the area of the surface generated by revolving the curve y = 49 - x² about the x-axis on the interval -5 ≤ x ≤ 5, we can use the formula for finding the area of a surface of revolution. The formula is given by A = ∫(2πy√(1+dy/dx)) dx. Evaluating the integral over the given interval will give us the area of the surface.

Step-by-step explanation:

To find the area of the surface generated by revolving the curve y = 49 - x² about the x-axis on the interval -5 ≤ x ≤ 5, we can use the formula for finding the area of a surface of revolution. The formula is given by:

A = ∫(2πy√(1+dy/dx)) dx

First, let's find dy/dx. Differentiating y = 49 - x² with respect to x, we get dy/dx = -2x. Substituting this into the formula, we have:

A = ∫(2π(49-x²)√(1+(-2x))) dx

Evaluating the integral over the given interval (-5 ≤ x ≤ 5) will give us the area of the surface.