Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve $y = \ln x ^ { 5 }$ about the $x$-axis on the interval $1 \leq x \leq 5$.
A)
$$\int _ { 5 } ^ { 1 } 2 \pi x \ln ( x ) \sqrt { 1 + \left( \frac { 5 } { x } \right) ^ { 2 } } d x$$
B)
$$\int _ { 5 } ^ { 1 } 2 \pi \sqrt { 1 + \left( \frac { 5 } { x } \right) ^ { 2 } } d x$$
C)
$$\int _ { 1 } ^ { 5 } 2 \pi ( 5 \ln ( x ) ) \sqrt { 1 + \left( \frac { 5 } { x } \right) ^ { 2 } } d x$$
D)
$$\int _ { 5 } ^ { 1 } 2 \pi ( 5 x ) \sqrt { 1 + \left( \frac { 5 } { x } \right) ^ { 2 } } d x$$
E)
$$\int _ { 1 } ^ { 5 } 2 \pi ( 5 x ) \sqrt { 1 + \left( \frac { 5 } { x } \right) ^ { 2 } } d x$$

Question

Quizplus · Accepted Answer

The formula for the surface area of a surface of revolution is given by$$S = \int_a^b 2\pi f(x) \sqrt{1+\left(f'(x)\right)^2} dx$$where $f(x)$ is the function that is being rotated and $a$ and $b$ are the bounds of the interval of rotation.In this case, $f(x) = \ln x^5$, so $f'(x) = \frac{5}{x}$. Substituting these into the formula, we get$$S = \int_1^5 2\pi \ln x^5 \sqrt{1+\left(\frac{5}{x}\right)^2} dx$$$$= \int_1^5 2\pi (5\ln x) \sqrt{1+\left(\frac{5}{x}\right)^2} dx$$

Set Up, but Do Not Evaluate, an Integral for the Area