Write an integral giving the area of the surface obtained by revolving the curve about the $x$-axis. (Do not evaluate the integral.)
$$y = \frac { 2 } { x } \text { on } [ 4,5 ]$$
A) $2 \pi \int _ { 4 } ^ { 5 } \frac { 2 } { x } \left( 1 + \left( - \frac { 2 } { x ^ { 2 } } \right) ^ { 2 } \right) d x$
B) $2 \pi \int _ { 4 } ^ { 5 } \frac { \sqrt { x ^ { 4 } - 4 } } { x ^ { 3 } } d x$
C) $4 \pi \int _ { 4 } ^ { 5 } x ^ { 3 } \sqrt { x ^ { 4 } + 4 } d x$
D)
$$4 \pi \int _ { 4 } ^ { 5 } \frac { \sqrt { x ^ { 4 } + 4 } } { x ^ { 3 } } d x$$

Question

Quizplus · Accepted Answer

We use the formula for finding the surface area obtained by revolving the curve $y=f(x)$ about the $x$-axis:
$$2\pi\int_{a}^{b}f(x)\sqrt{1+\left(f'(x)\right)^2}dx$$
In this case, $f(x)=\frac{2}{x}$ and $f'(x)=\frac{-2}{x^2}$, so we have
$$2\pi\int_{4}^{5}\frac{2}{x}\sqrt{1+\left(\frac{-2}{x^2}\right)^2}dx=2\pi\int_{4}^{5}\frac{2}{x}\sqrt{1+\frac{4}{x^4}}dx$$
We simplify this by multiplying the numerator and denominator of the square root by $x^4$:
$$2\pi\int_{4}^{5}\frac{2x^4}{x^5}\sqrt{x^4+4}dx=4\pi\int_{4}^{5}\frac{\sqrt{x^4+4}}{x^3}dx$$
This matches the integrand in choice D, so the answer is D.

Write an Integral Giving the Area of the Surface Obtained xxx

Write an Integral Giving the Area of the Surface Obtained $x$