Which of these integrals can be used to calculate the surface area of the solid formed by revolving the region bounded by $f ( x ) = \left( 1 - x ^ { 2 } \right) ^ { 3 },$ the x-axis, x = -2, and x = 0 about the x-axis?

A) $\int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 3 } \sqrt { 1 + 36 x ^ { 2 } \left( 1 - x ^ { 2 } \right) ^ { 4 } } d x$
B) $2 \pi \int _ { - 1 } ^ { 4 } 6 x \left( 1 - x ^ { 2 } \right) ^ { 2 } \sqrt { 1 + \left( 1 - x ^ { 2 } \right) ^ { 6 } } d x$
C) $2 \pi \int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 3 } \sqrt { 1 + 6 x \left( 1 - x ^ { 2 } \right) ^ { 2 } } d x$
D) $\int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 6 } \sqrt { 1 + 6 x \left( 1 - x ^ { 2 } \right) ^ { 2 } } d x$
E) $2 \pi \int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 3 } \sqrt { 1 + 36 x ^ { 2 } \left( 1 - x ^ { 2 } \right) ^ { 4 } } d x$

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 ) = \left( 1 - x ^ { 2 } \right) ^ { 3 },

the x-axis, x = -2, and x = 0 about the x-axis?

A) $\int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 3 } \sqrt { 1 + 36 x ^ { 2 } \left( 1 - x ^ { 2 } \right) ^ { 4 } } d x$
B) $2 \pi \int _ { - 1 } ^ { 4 } 6 x \left( 1 - x ^ { 2 } \right) ^ { 2 } \sqrt { 1 + \left( 1 - x ^ { 2 } \right) ^ { 6 } } d x$
C) $2 \pi \int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 3 } \sqrt { 1 + 6 x \left( 1 - x ^ { 2 } \right) ^ { 2 } } d x$
D) $\int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 6 } \sqrt { 1 + 6 x \left( 1 - x ^ { 2 } \right) ^ { 2 } } d x$
E) $2 \pi \int _ { - 1 } ^ { 4 } \left( 1 - x ^ { 2 } \right) ^ { 3 } \sqrt { 1 + 36 x ^ { 2 } \left( 1 - x ^ { 2 } \right) ^ { 4 } } d x$

Quizplus · Accepted Answer

The correct integral for calculating the surface area of a solid formed by revolving a function $f(x)$ about the x-axis is given by $2 \pi \int y \sqrt{1 + (f'(x))^2} dx$, where $y = f(x)$. The derivative of $f(x) = (1 - x^2)^3$ is $f'(x) = -6x(1 - x^2)^2$, so the integral becomes $2 \pi \int (1 - x^2)^3 \sqrt{1 + 36x^2(1 - x^2)^4} dx$. The bounds should be from x = -2 to x = 0, but none of the options correctly reflect these bounds, indicating a potential error in the question or choices. However, option E most closely matches the formula for surface area despite the incorrect bounds.

Which of These Integrals Can Be Used to Calculate the Surface