Which of these integrals can be used to calculate the surface area of the solid formed by revolving the region bounded by $f ( x ) = \sqrt [ 3 ] { x },$ the -axis, x = 1, and x = 5 about the x-axis?

A) $\int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \left( \frac { 1 } { 3 } x ^ { - 2 / 3 } \right) ^ { 2 } } d x$
B) $2 \pi \int _ { 1 } ^ { 5 } \sqrt { 1 + \left( \frac { 1 } { 3 } x ^ { - 2 / 3 } \right) ^ { 2 } } d x$
C) $2 \pi \int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \left( \frac { 1 } { 3 } x ^ { - 2 / 3 } \right) ^ { 2 } } d x$
D) $2 \pi \int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \frac { 1 } { 3 } x ^ { - 2 / 3 } } d x$
E) $2 \pi \int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \left( x ^ { 1 / 3 } \right) ^ { 2 } } d x$

Question

Which of these integrals can be used to calculate the surface area of the solid formed by revolving the region bounded by $f ( x ) = \sqrt [ 3 ] { x },$ the -axis, x = 1, and x = 5 about the x-axis?

A) $\int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \left( \frac { 1 } { 3 } x ^ { - 2 / 3 } \right) ^ { 2 } } d x$
B) $2 \pi \int _ { 1 } ^ { 5 } \sqrt { 1 + \left( \frac { 1 } { 3 } x ^ { - 2 / 3 } \right) ^ { 2 } } d x$
C) $2 \pi \int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \left( \frac { 1 } { 3 } x ^ { - 2 / 3 } \right) ^ { 2 } } d x$
D) $2 \pi \int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \frac { 1 } { 3 } x ^ { - 2 / 3 } } d x$
E) $2 \pi \int _ { 1 } ^ { 5 } x ^ { 1 / 3 } \sqrt { 1 + \left( x ^ { 1 / 3 } \right) ^ { 2 } } d x$

Quizplus · Accepted Answer

The formula for the surface area of a solid of revolution about the x-axis is $2 \pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} dx$, where $f(x)$ is the function being revolved and $f'(x)$ is its derivative. For $f(x) = x^{1/3}$, $f'(x) = \frac{1}{3}x^{-2/3}$. Substituting these into the formula gives the integral in choice C.

Which of These Integrals Can Be Used to Calculate the Surface