Question

Determine the surface area of the solid obtained by rotating y = √(9 – x²), -2 < x < 2, about the X-axis.

a) 121
b) 24
c) 48
d) 241

Answer 1

The surface area of the solid obtained by rotating y = √(9 – x²), -2 < x < 2, about the X-axis is approximately 48.

Step-by-step explanation:

To determine the surface area of the solid obtained by rotating y = √(9 – x²), -2 < x < 2, about the X-axis, we can use the formula for the surface area of a solid of revolution:

A = 2π∫[a,b] f(x)√(1 + (f'(x))²) dx

In this case, f(x) = √(9 – x²) and f'(x) = -x/√(9 – x²). Plugging these values into the formula and evaluating the integral, we get:

A = 2π∫[-2,2] √(9 – x²)√(1 + (-x/√(9 – x²))²) dx

Simplifying and evaluating the integral, we find that the surface area is approximately 48. Answer choice c) 48 is correct.