# Finding the volume of a solid generated by revolving around the line y=6

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!
• @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! – Noble Mushtak Jun 30 '16 at 21:19