Answer to Question #180405 in Calculus for Aditya Sharma

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-

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

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

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

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

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

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