Answer to Question #348181 in Calculus for Migz

Question

Find the volume by revolving the hyperbola xy = 6 from x=2 to x=4

about the x-axis.

Answer 1

We can rrewrite the equation of the hyperbola:

$y=\frac{6}{x}.$

We know, that the volume of the solid generated by a region under f(x) bounded by the x-axis and vertical lines x=a and x=b, which is revolved about the x-axis is 

$V=\pi\int_a^b[f(x)]^2dx.$

So we have:

$V=\pi\int_2^4(\frac{6}{x})^2dx=\pi\int_2^4\frac{36}{x^2}dx=36\pi\int_2^4\frac{dx}{x^2}=$

$=36\pi(\frac{-1}{x})|_2^4=36\pi(-\frac{1}{4}+\frac{1}{2})=36\pi\cdot\frac{1}{4}=$

$=9\pi \approx 28.27$ cubic units.

Answer: $=9\pi (\approx 28.27)$ cubic units.