# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y=1/x and 2x+2y=5 rotated about the y=1/2?

Jan 1, 2017

$= \pi \left(2 \ln 2 + \frac{21}{8}\right) = 12.60$ cubic units, nearly.

#### Explanation:

graph{ ( xy-1)(x+y-5/2) = 0 [-5.37, 5.37, -2.684, 2.684]}

The rectangular hyperbola H; $x y = 1$ and the L: straight line

$x + y = \frac{5}{2}$ meet at $\left(\frac{1}{2.} 2\right) \mathmr{and} \left(2 , \frac{1}{2}\right)$.

The area is depicted by the graph.

The volume is

$\pi \int \left({\left({y}_{L} - \frac{1}{2}\right)}^{2} - {\left({y}_{H} - \frac{1}{2}\right)}^{2}\right) \mathrm{dx}$, with x from $\frac{1}{2}$ to 2

$= \pi \int \left({\left(\frac{5}{2} - x - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{x} - \frac{1}{2}\right)}^{2}\right) \mathrm{dx}$,.with x from $\frac{1}{2}$ to 2

$= \pi \int \left(\left(4 - 4 x + {x}^{2}\right) - \left(\frac{1}{4} - \frac{1}{4 x} + \frac{1}{x} ^ 2\right)\right) \mathrm{dx}$,

with x from $\frac{1}{2}$ to 2

$= \pi \int \left(\frac{1}{4 x} - \frac{1}{x} ^ 2 + \frac{15}{4} - 4 x + x 2\right) \mathrm{dx}$, with x from $\frac{1}{2}$ to 2

$= \pi \left[\left(\frac{1}{4}\right) \left(\ln 2 - \ln \left(\frac{1}{2}\right)\right) + \left(\frac{1}{2} - 2\right) + \frac{15}{4} \left(2 - \frac{1}{2}\right) - 2 \left({2}^{2} - \frac{1}{2} ^ 2\right) - 15 + \frac{1}{3} \left({2}^{3} - \frac{1}{2} ^ 3\right)\right]$

$= \pi \left[\frac{1}{2} \ln 2 - \frac{3}{2} + \frac{45}{8} - \frac{15}{2} - 15 + 21\right]$

$= \pi \left(2 \ln 2 + \frac{21}{8}\right)$, cubic units