Question

Use the method cylinders to determine the volume of the solid obtained by rotating the region bounded by the given curves about the given axis of y=e^1/2 x/x+2, y=5 - 1/4 x, x=-1 abd x=6 about the line x=-2.

Answer 1

We have to determine the volume of solid obtained the region bounded by

$y= 2 \sqrt{x-1}$ and $y=x-1$ about the line $x=6$.

After drawing figure of the following curves we obtain that,

The cross sectional area is then,

$A(x) = 2\pi (radius)(height)\\$

$= 2\pi (6-x)(2\sqrt{x-1}-x+1)$

$= 2\pi (x^2-7x+6+12\sqrt{x-1}-2x\sqrt{x-1})$

Now the first cylinder will cut into the solid at x=1 and the final cylinder will cut into the solid at x=5 so there are our limits.

Here is the volume,

$V =\int_{a}^{b} A(x )dx$

$= 2\pi \int_{1}^{5} (x^2-7x+6+12\sqrt{x-1}-2x\sqrt{x-1})dx$

$= 2\pi[( \dfrac{1}{3}x^3- \dfrac{7}{2}x^2+6x+8(x-1)^{\dfrac{3}{2}}- \dfrac{4}{3}(x-1)^{\dfrac{3}{2}}- \dfrac{4}{5}(x-1)^{\dfrac{5}{2}}]_{1}^{5}$

$= 2\pi(\dfrac{136}{15})$

$= \dfrac{272\pi}{15}$