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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. if the answer does not exist, enter dne. otherwise, round to four decimal places.

User Pinei
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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).

User JBaruch
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