231k views
1 vote
Let the region R be the area enclosed by the function f(x)=ln(x) and g(x)= 1/2x-2. Find the volume of the solid generated when the region R is revolved about the line y=-5.

User Asmundur
by
8.3k points

1 Answer

4 votes

Answer:


V=61.66

Explanation:

This problem can be solved by using the expression for the Volume of a solid with the washer method


V=\pi \int \limit_a^b[R(x)^2-r(x)^2]dx

where R and r are the functions f and g respectively (f for the upper bound of the region and r for the lower bound).

Before we have to compute the limits of the integral. We can do that by taking f=g, that is


f(x)=g(x)\\ln(x)=(1)/(2)x-2

there are two point of intersection (that have been calculated with a software program as Wolfram alpha, because there is no way to solve analiticaly)

x1=0.14

x2=8.21

and because the revolution is around y=-5 we have


R=ln(x)-(-5)\\r=(1)/(2)x-2-(-5)\\

and by replacing in the integral we have


V=\pi \int \limit_(x1)^(x2)[(lnx+5)^2-((1)/(2)x+3)^2]dx\\


V=\pi [28x+(1)/(x)+xln^2x-12xlnx-6lnx]

and by evaluating in the limits we have


V=61.66

Hope this helps

regards

User Alissa
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories