Question

What is the volume of the solid produced by revolving `f(x)=e^x-xlnx, x in [1,5]`around the x-axis?

Answer 1

The volume is `29486.554` `"u"^3`, or the sort of exact result is:

`V = pi/2(e^(10) - e^2) - 1696.084pi + (125pi)/3[ln^2(5) - 2/3ln5 + 2/9] - (2pi)/27` `"u"^3`

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To do these kinds of solid of revolution problems around the `x` axis, just remember that you are projecting a circle whose radius varies as `r = f(x)`.

The volume is then given by:

`V = pi int_(a)^(b) f^2(x)dx`

The `f(x)` serves as the varying radius, and the integral over `[a,b]` gives the length of the object, sliced up into discs of thickness `dx`.

In this case, we have:

`V = pi int_(1)^(5) (e^x - xlnx)^2dx`

`= pi int_(1)^(5) e^(2x) - 2xe^xlnx + x^2ln^2xdx`

`= pi overbrace(int_(1)^(5) e^(2x)dx)^("Integral 1") - 2pi overbrace(int_(1)^(5) xe^xlnxdx)^"Integral 2" + pi overbrace(int_(1)^(5) x^2ln^2xdx)^"Integral 3"`

• The first integral is `1/2e^(2x)` evaluated from `x = 1` to `5`, which is `1/2(e^(10) - e^2)`.
• The second integral is not possible with elementary functions, and the numerical solution is `848.042`.
• The third integral is hard but doable here.

`int x^2ln^2xdx = ?`

Let:

• `u = ln^2x`
• `dv = x^2dx`
• `du = (2lnx)/xdx`
• `v = x^3/3`

Thus,

`int x^2ln^2xdx`

`= (x^3ln^2x)/3 - int x^3/3 (2lnx)/xdx`

`= 1/3 x^3ln^2x - 2/3int x^2 lnxdx`

Repeat with:

• `u = lnx`
• `dv = x^2dx`
• `du = 1/xdx`
• `v = x^3/3`

And we get:

`=> 1/3 x^3ln^2x - 2/3[1/3x^3lnx - int x^3/3 1/xdx]`

`= 1/3 x^3ln^2x - 2/3[1/3x^3lnx - 1/3 int x^2dx]`

`= 1/3 x^3ln^2x - [2/9x^3lnx - 2/9 int x^2dx]`

`= 1/3 x^3ln^2x - 2/9x^3lnx + 2/9 int x^2dx`

`= {:[1/3 x^3ln^2x - 2/9x^3lnx + 2/9 x^3/3]|:}_(1)^(5)`

`= [1/3 (5)^3ln^2(5) - 2/9(5)^3ln(5) + 2/9(5)^3/3] - [cancel(1/3 (1)^3ln^2(1))^(0) - cancel(2/9(1)^3ln(1))^(0) + 2/9 (1)^3/3]`

`= [1/3 125ln^2(5) - 2/9(125ln5 - 125/3)] - 2/9(1/3)`

`= 125/3[ln^2(5) - 2/3ln5 + 2/9] - 2/27`

So, adding up all the results,

`V = pi { 1/2(e^(10) - e^2)} - 2pi {848.042} + pi {125/3[ln^2(5) - 2/3ln5 + 2/9] - 2/27}`

`= color(blue)(pi/2(e^(10) - e^2) - 1696.084pi + (125pi)/3[ln^2(5) - 2/3ln5 + 2/9] - (2pi)/27)`

or `29486.554`.