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

Given curves are-

$f_1(x)=\dfrac{e^{\frac{x}{2}}}{x+2},f_2(x)=5-\dfrac{1}{4}x, x=-1,x=6$

The Volume of the cylinder about the line x=-2 is-

$V=2\pi\int_{-1}^6x[f_1(x)-f_2(x)]dx$

$=2\pi\int_{-1}^6x[\dfrac{e^{\frac{x}{2}}}{x+2}-5+\dfrac{1}{4}x]dx$

$=2\pi [\int_{-1}^6\dfrac{xe^{\frac{x}{2}}}{x+2}dx-\int_{-1}^65xdx+\int_{-1}^6\dfrac{x^2}{4}dx]$

$=2\pi[2e^{\frac{x}{2}}-2(\dfrac{\frac{x}{2}+1}{e})-5\dfrac{x^2}{2}+\dfrac{x^3}{12}|_{-1}^6$

$=2\pi[2e^3-\dfrac{8}{e}-90+18-(2e^{-\frac{x}{2}}-\dfrac{1}{e}-\dfrac{5}{2}-\dfrac{1}{12})]$

$=2\pi[2e^3-\dfrac{7}{e}-\dfrac{833}{12}]$