To find the volume of the solid generated by revolving about the y-axis the region under the curve y = 5e^-2x in the first quadrant, we can use the method of cylindrical shells.
The formula for the volume generated by a cylindrical shell is V = 2πrh Δx, where r is the radius of the shell, h is the height of the shell, and Δx is the thickness of the shell.
In this case, we are rotating about the y-axis, so our cylindrical shells will have a height of y and a thickness of Δy. The radius of each shell will be the distance from the y-axis to the curve, which is x.
We can solve for x in terms of y by rearranging the equation y = 5e^-2x:
y/5 = e^-2x
ln(y/5) = -2x
x = -ln(y/5)/2
So the volume of each cylindrical shell is:
V = 2πx * y * Δy
Substituting for x:
V = 2π(-ln(y/5)/2) * y * Δy
V = -πln(y/5) * y * Δy
To find the total volume, we need to integrate this expression from y = 0 to y = infinity:
V = ∫[0,∞] -πln(y/5) * y dy
Using integration by parts:
u = ln(y/5), dv = y dy
du = 1/y dy, v = 1/2 y^2
∫[0,∞] -πln(y/5) * y dy = [-π(1/2y^2 ln(y/5) - 1/4y^2)] [0,∞]
= [π/4]
Therefore, the volume of the solid generated by revolving about the y-axis in the region under the curve y = 5e^-2x in the first quadrant is π/4, which is a finite value.
Hence, the answer is 0.7854 (rounded to four decimal places).