# 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.

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