Final answer:
To find the surface area of the curve f(x) = √ x revolved around the x-axis, the appropriate integral is S = 2π ∫_0^2 √ x √{1+(1/2x^{-1/2})^2}dx, which accounts for both the shape of the function and its rotation about the x-axis.
Step-by-step explanation:
The student is asking which integral would be used to calculate the surface area of the curve f(x) = √ x for 0 ≤ x ≤ 2 revolved about the x-axis. To find this surface area, we use the surface area formula for a curve revolved around the x-axis given by the integral S = 2π ∫_a^b f(x)√{1+(f'(x))^2}dx, where a and b are the bounds of x and f'(x) is the derivative of f(x).
For the curve f(x) = √ x, the derivative f'(x) = ½x^{-½}. Substituting f(x) and f'(x) into the surface area formula gives us the integral S = 2π ∫_0^2 √ x √{1+(½x^{-½})^2}dx, which upon simplification becomes S = 2π ∫_0^2 √ x √{1+¼ x^{-1}}dx. This integral will find the desired surface area.