Write an integral giving the area of the surface obtained by revolving the curve about the x-axis. (Do not evaluate the integral.) y = $\frac{4}{x}$ on [3, 6]
A) (  )
B) 8  $\int_{3}^{6} x^{3} \sqrt{x^{4}+16} d x$
C) 8  $\int_{3}^{6} \frac{\sqrt{x^{4}+16}}{x^{3}} d x$
D) (  $\int_{3}^{6} \frac{4}{x}\left(1+\left(-\frac{4}{x^{2}}\right)^{2}\right) d x$ )

Question

Write an integral giving the area of the surface obtained by revolving the curve about the x-axis. (Do not evaluate the integral.) y = $\frac{4}{x}$ on [3, 6]
A) (  ) 
B) 8  $\int_{3}^{6} x^{3} \sqrt{x^{4}+16} d x$ 
C) 8  $\int_{3}^{6} \frac{\sqrt{x^{4}+16}}{x^{3}} d x$ 
D) (  $\int_{3}^{6} \frac{4}{x}\left(1+\left(-\frac{4}{x^{2}}\right)^{2}\right) d x$ )

Quizplus · Accepted Answer

The formula for finding the surface area of a curve revolved around the x-axis is given by:
$$A = 2\pi\int_{a}^{b} f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$
In this case, $f(x) = \frac{4}{x}$ and $\frac{dy}{dx} = -\frac{4}{x^2}$. Substituting these values, we get:
$$A = 2\pi\int_{3}^{6} \frac{4}{x} \sqrt{1+ \left(-\frac{4}{x^2}\right)^2}dx = 8\pi\int_{3}^{6} \frac{\sqrt{x^4+16}}{x^3}dx$$
The closest option is C.

Write an Integral Giving the Area of the Surface Obtained 4x\frac{4}{x}x4​

Write an Integral Giving the Area of the Surface Obtained $\frac{4}{x}$