# How compute with cylindrical shells the volume of the solid obtained by revolving around y axis the area under the curve y=e^x and over 0<=x<=1?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the volume of the solid of revolution using the method of cylindrical shells with the following formula, such that:

`V = int_a^b 2pi*x*f(x)dx`

`2pi*x` represents the circumference of the circle base of cylinder

`f(x)` represents the height of cylinder

Since the problem provides the equation of the curve `y = f(x) = e^x` and the limits of integration `a = 0, b = 1` , yields:

`V = int_0^1 2pi*x*e^x dx`

Taking out the constant `2pi` yields:

`V = 2pi int_0^1 x*e^x dx`

You need to use integration by parts, such that:

`int_a^b udv = uv|_a^b - int_a^b vdu`

`u = x => du = dx`

`dv = e^xdx => v = e^x`

`int_0^1 x*e^x dx = x*e^x|_0^1 - int_0^1 e^x dx`

`int_0^1 x*e^x dx = x*e^x|_0^1 - e^x|_0^1`

Factoring out `e^x`   yields:

`int_0^1 x*e^x dx = e^x(x - 1)|_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 x*e^x dx = e^1(1 - 1) - e^0(0 - 1)`

`int_0^1 x*e^x dx = 0 + 1`

`int_0^1 x*e^x dx = 1`

`V = 2p*1 => V = 2pi`

Hence, evaluating the volum of solid obtained by revolving around y axis the area under the curve `y=e^x` , over `x in [0,1]` , using the method of cylindrical shells, yields `V = 2pi.`

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