Which of these integrals can be used to calculate the surface area of the solid formed by revolving the region bounded by $f ( x ) = x ^ { 4 } - 3 x,$ the x-axis, x = -2, and x = 0 about the x-axis?

A) $2 \pi \int _ { - 2 } ^ { 0 } \left( x ^ { 4 } - 3 x \right) \sqrt { 1 + \left( 4 x ^ { 3 } - 3 \right) ^ { 2 } } d x$
B) $2 \pi \int _ { - 2 } ^ { 0 } \left( 4 x ^ { 3 } - 3 \right) \sqrt { 1 + \left( x ^ { 4 } - 3 x \right) ^ { 2 } } d x$
C) $\int _ { - 2 } ^ { 0 } \left( x ^ { 4 } - 3 x \right) \sqrt { 1 + \left( 4 x ^ { 3 } - 3 \right) } d x$
D) $\int _ { 0 } ^ { - 2 } \left( 4 x ^ { 3 } - 3 \right) \sqrt { 1 + \left( x ^ { 4 } - 3 x \right) ^ { 2 } } d x$
E) $2 \pi \int _ { 0 } ^ { - 2 } \left( x ^ { 4 } - 3 x \right) \sqrt { 1 + \left( 4 x ^ { 3 } - 3 \right) ^ { 2 } } d x$

Question

Which of these integrals can be used to calculate the surface area of the solid formed by revolving the region bounded by $f ( x ) = x ^ { 4 } - 3 x,$ the x-axis, x = -2, and x = 0 about the x-axis?

A) $2 \pi \int _ { - 2 } ^ { 0 } \left( x ^ { 4 } - 3 x \right) \sqrt { 1 + \left( 4 x ^ { 3 } - 3 \right) ^ { 2 } } d x$
B) $2 \pi \int _ { - 2 } ^ { 0 } \left( 4 x ^ { 3 } - 3 \right) \sqrt { 1 + \left( x ^ { 4 } - 3 x \right) ^ { 2 } } d x$
C) $\int _ { - 2 } ^ { 0 } \left( x ^ { 4 } - 3 x \right) \sqrt { 1 + \left( 4 x ^ { 3 } - 3 \right) } d x$
D) $\int _ { 0 } ^ { - 2 } \left( 4 x ^ { 3 } - 3 \right) \sqrt { 1 + \left( x ^ { 4 } - 3 x \right) ^ { 2 } } d x$
E) $2 \pi \int _ { 0 } ^ { - 2 } \left( x ^ { 4 } - 3 x \right) \sqrt { 1 + \left( 4 x ^ { 3 } - 3 \right) ^ { 2 } } d x$

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. Here, $f(x) = x^4 - 3x$, and its derivative is $f'(x) = 4x^3 - 3$. Thus, the correct integral is $2 \pi \int_{-2}^{0} (x^4 - 3x) \sqrt{1 + (4x^3 - 3)^2} dx$.

Which of These Integrals Can Be Used to Calculate the Surface