Expert: Steve Holleran - 5/27/2008

Question
Let R be the region bounded by the graphs of y=x^(1/2), y=e^(-x), and the y-axis.  Write an integral that represents the area of R.  Find the area of R.  
Set up the integral that will give the volume of the solid generated when R is revolved about the horizontal line
y=-1.
What is the volume?

Answer
Hi Harman,

Let's do  the area part first.  Using a grapher to find the intersection point, it is x = .4263.

So the area integral is A = INT, 0 to 0.4263, [e^(-x) - rt(x)] dx

and the value should be 0.1615


For the volume , you want to use the washer method.  Since the axis of revolution is y = -1, the outer radius is e^(-x) + 1  and the inner radius is rt(x) + 1.

The integral for the volume is

           V = pi * INT, 0 to 0.4263, [(1+e^(-x))^2 - (1 + rt(x))^2 dx

and the value should be 0.5190

Steve