Question #348181

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

about the x-axis.


1
Expert's answer
2022-06-07T15:32:09-0400

We can rrewrite the equation of the hyperbola:

y=6x.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=πab[f(x)]2dx.V=\pi\int_a^b[f(x)]^2dx.





So we have:

V=π24(6x)2dx=π2436x2dx=36π24dxx2=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}=

=36π(1x)24=36π(14+12)=36π14==36\pi(\frac{-1}{x})|_2^4=36\pi(-\frac{1}{4}+\frac{1}{2})=36\pi\cdot\frac{1}{4}=

=9π28.27=9\pi \approx 28.27 cubic units.


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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS