Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.
-$$y = x ^ { 3 } , 0 \leq x \leq 2 ; x \text {-axis }$$
A) $2 \pi \int _ { 0 } ^ { 2 } x ^ { 3 } \sqrt { 1 + 9 x ^ { 4 } } d x$
B) $\pi \int _ { 0 } ^ { 2 } x ^ { 2 } \sqrt { 1 + 9 x ^ { 4 } } d x$
C) $\pi \int _ { 0 } ^ { 2 } x ^ { 3 } \sqrt { 1 + 9 x ^ { 4 } } d x$
D) $2 \pi \int _ { 0 } ^ { 2 } x ^ { 2 } \sqrt { 1 + 9 x ^ { 4 } } d x$

Question

Quizplus · Accepted Answer

The formula for the surface area generated by revolving a curve $y = f(x)$ about the x-axis from $x = a$ to $x = b$ is $2 \pi \int_{a}^{b} y \sqrt{1 + (dy/dx)^2} dx$. For $y = x^3$, $dy/dx = 3x^2$, so the integral becomes $2 \pi \int_{0}^{2} x^3 \sqrt{1 + (3x^2)^2} dx$, which matches choice A.

Set Up an Integral for the Area of the Surface y=x3,0≤x≤2;x-axis y = x ^ { 3 } , 0 \leq x \leq 2 ; x \text {-axis }y=x3,0≤x≤2;x-axis

Set Up an Integral for the Area of the Surface $y = x ^ { 3 } , 0 \leq x \leq 2 ; x \text {-axis }$