Answer to Question #176686 in Calculus for Joshua

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,

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

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

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

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

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