I'm new to this site. My boyfriend is unable to complete these problems due to a family emergency and I was going to try to help him because they are due in a few hours and his professor hasn't responded to his emails.

The problem is asking to find the volume of the solid generated by revolving the given region about the line y=6. The region is bounded by the curves $$ y=\frac 5x, \quad y=0, \quad x=1, \quad x=5 $$

I haven't done calculus in a long time and am having a hard time trying to solve this problem. I graphed out the lines given, but I'm unsure how to mathematically revolve this shape and which integral to use.


Since this is between $y=\frac 5 x$ and $y=0$ and we're rotating it around $y=6$, we need to use method of cylinders. Therefore, we're going to take the inverse of this function and say $x=\frac 5 y$.

Now, at $x=1$, we have $y=5$ and at $x=5$, we have $y=1$. Therefore, from $y=1$ to $y=5$, the height of the cylinder is $\frac 5 y-1$ (remember to subtract $1$ because we are starting from $x=1$). Because we are rotating around $y=6$, the radius is $6-y$. Thus, we get: $$\int_1^5 2\pi(6-y)\left(\frac 5 y-1\right)dy$$ However, it said that our boundary was $y=0$, so we also need to consider the region between $y=0$ and $y=1$. Here, the height is $5-1=4$ and the radius is still $6-y$, so, adding this to the above, we get: $$\int_0^1 2\pi(6-y)(4)dy+\int_1^5 2\pi(6-y)\left(\frac 5 y-1\right)dy$$ I'm going to leave the computation of this integral up to you. However, if you need the answer, I will be glad to solve the integral out if I have time. I also suggest Wolfram Alpha if you get stuck because you can type in something like "integral from 0 to 1 of 2*pi*(6-y)(4)" and they will give you the answer. Good luck!

  • $\begingroup$ THANK YOU SO MUCH! I used Wolfram Alpha for the integration! This really really helped! I appreciate it! :-) $\endgroup$ – Rupali Jun 30 '16 at 21:17
  • $\begingroup$ @Rupali Awesome! Also, the reason you got the wrong answer at first is because you likely typed in "2pi(6-y)(4)" which means something different to computers than "2pi*(6-y)(4)". This is because if you don't include the asterisk, they see $\pi([...])$ and think you are talking about the prime-counting function. However, I'm really glad you got the right answer in the end! $\endgroup$ – Noble Mushtak Jun 30 '16 at 21:19
  • $\begingroup$ That makes sense! Thank you, again! $\endgroup$ – Rupali Jun 30 '16 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.