# Find the volume of the solid generated by rotating about the x-axis the region bounded by the curves

I'm new to this site, and this will be my last question post. My boyfriend has a few problems that he's unable to complete because of a family emergency and I have decided to try and help him. His professor hasn't responded to his emails asking for an extension on the deadline, which is tonight. I haven't done calculus in a long time, so these are pretty difficult for me to do on my own.

Find the volume of the solid generated by rotating about the x-axis the region bounded by the curves:

y=e^(10x)+e^(−10x), x=0, x=6, and the x-axis.

He had already started working on this problem and got the answer: pi(12+(sinh(120)/10)). However, the online system won't accept this answer. If you could help me figure out where he went wrong or what I can change in his answer to make it acceptable by the system, I would really appreciate that!

First, as your friend likely noticed, we can rewrite the main curve as $y=2\cosh 10x$. Also, we are going from $x=0$ to $x=6$, so we integrate from $0$ to $6$. This solid of revolution can be solved using method of rings, where our radius is simply $y=2\cosh 10x$, so we get: $$\int_0^6 \pi(2\cosh 10x)^2dx$$ Now, if we calculate this out, we get the same answer as your friend, so I'm pretty sure he's right. However, notice how they did not use hyperbolic trig in the original question. Therefore, I think what they would like is an alternate form of the equivalent answer without hyperbolic trig by using the definition $\sinh x=\frac{e^x-e^{-x}}{2}$: $$\left(12+\frac{e^{120}-e^{-120}}{20}\right)\pi$$ Please try entering this into the computer and see if it works or not.
Also, if you know anything about what format they want, tell us so we can help you. For the last problem, did you need an exact answer or did an approximation work? If an approximation will work, try $2.0486025*10^{51}$.
• @Rupali I was referring to the $y=6$ problem. – Noble Mushtak Jun 30 '16 at 22:20