# Volume of the revolution solid

Let $K \subset \mathbb R^2$ be area bounded by curves $x=2$, $y=3$, $xy=2$ and $xy=4$.

What is the volume of the solid we get after rotating $K$ around $y$-axis?

I tried to express the curves in terms of $y$ and solved the integral $\displaystyle \pi\left(\int_2^3 \frac{16}{y^2}dy - \int_1^3\frac{4}{y^2}dy \right)$ but it came out zero. I got the integral by subtracting the volume of the inner shape from the volume of the outer shape, if that makes any sense.

## 2 Answers

Do this in two parts. First find the section of the volume between $y=3$ and $y=2$, then the section of the volume between $y=2$ and $y=1$. For the first volume, use the integral $$\pi\int_{2}^{3} \frac{16}{x^2}dx-\pi\int_{2}^{3} \frac{4}{x^2}dx$$ This is the difference of the volumes formed by rotating the regions between $xy=4$ and the y-axis and between $xy=2$ and the y-axis between $y=2$ and $y=3$. When you evaluate these integrals, you get $$\pi\int_{2}^{3} \frac{12}{x^2}dx$$ $$\pi(-\frac{12}{3}+\frac{12}{2})$$ $$\pi(6-4)$$ $$2\pi$$ Now for the second region. For this I use the integrals $$4\pi-\pi\int_{1}^{2} \frac{4}{x^2}dx$$ Which is the difference of the volumes of the cylinder made by rotating a rectangle about the x-axis and the volume of the area under $xy=2$ rotated about the x-axis. This gives us $$4\pi-\pi(-\frac{4}{2}+\frac{4}{1})$$ $$4\pi-2\pi$$ $$2\pi$$ The total volume is the sum of the volumes, which is $$4\pi$$ Is this the correct answer?

• Nicely done. I have tried indeed to do something similar, but failed in finding the volume of the second region, as you refer to it in your post. Thank you very much. – windircurse May 2 '17 at 21:47
• No problem! I was glad to do it. In your book (or wherever you got this problem from) is there a solution listed? – Frpzzd May 2 '17 at 21:51
• Nope, unfortunately. – windircurse May 2 '17 at 21:53
• If my solution is correct, please consider accepting it as the correct answer. – Frpzzd May 2 '17 at 23:10

$y = 3$ intersects $xy = 2$ at $x = \frac 23$ and $xy = 4$ at $x = \frac 43$

By shells I get.

$2\pi\int_\frac {2}{3}^{\frac 43} (3-\frac 2x)x \ dx + 2\pi\int_{\frac 43}^2 (\frac 4x -\frac 2x)x \ dx\\ 2\pi\int_\frac {2}{3}^{\frac 43} 3x-2 \ dx + 2\pi\int_{\frac 43}^2 2\ dx\\ 2\pi(\frac{3}{2}x^2 -2x|_\frac {2}{3}^{\frac 43} + 2x| _{\frac 43}^2)\\ 2\pi(\frac 83 - \frac 23 - \frac 83 + \frac 43 + 4 - \frac 83) \\ 4\pi$

• Can you please explain how you got those integrals? – Frpzzd May 2 '17 at 22:42
• @FranklinP.Dyer what part is not clear? I think I have explained where the limits came from. as far is the what is inside. The volume of each shell equals $2\pi$ times the height of each shell times the radius of each shell. – Doug M May 2 '17 at 22:47
• @FranklinP.Dyer thanks for making me take a second look, I had the wrong limit (x=3 not x =2) – Doug M May 2 '17 at 22:50
• Oh, I see. No problem! Glad we are now in agreement. – Frpzzd May 2 '17 at 23:00