I'm trying to compute this function $y=\frac{1}{2}e^x+\frac{1}{12}e^{-3x}$ to get the volume of the solid of revolution when the area under the curve is revolved once around the x axis. And than get the curved surface area of the solid of revolution from the previous part.

This is the equation needed to get the volume $\int \pi y^2dx$. And this is how I computed the function in Mathematica

Pi*Integrate[((1/2)*Exp[x] + (1/12)*Exp[-3x])^2, {x, 0, 1}]

What did I do wrong? Because the solution comes out without $x$ and I can't differentiate it and get the covered surface area with this equation $2\pi \int_{a}^{b}y \sqrt{1+(\frac{dy}{dx})^2}dx$

  • $\begingroup$ What you did wrong was to compute a definite integral with numeric bounds. If such an integral can evaluated it will always produce a number. Try using a symbol (not x) as the upper bound. $\endgroup$ – m_goldberg Apr 30 '14 at 16:32
  • $\begingroup$ Sorry I'm new to Mathematica and I don't have a solid math knowledge yet. If I did put a symbol, how do I get the volume between x=0 and x=1? $\endgroup$ – user1123975 Apr 30 '14 at 16:40
  • $\begingroup$ @user1123975 Can you check you equations first? Do you mean 1/2*Exp[x] or 1/3*Exp[x]? Do you mean 1/12*Exp[-3]*x or 1/12*Exp[-3x]? $\endgroup$ – halirutan Apr 30 '14 at 16:45
  • $\begingroup$ You already have the volume between zero and one from what you have already done. Use the symbolic form to compute the surface area function. Alternatively, say the symbolic result uses symbol u, then symbolic-result /. u -> 1 should work. $\endgroup$ – m_goldberg Apr 30 '14 at 16:48
  • $\begingroup$ And isn't there supposed to be a "+" in the surface area equation? $\endgroup$ – bobthechemist Apr 30 '14 at 16:49

One way to get at the answers you seek is to first define your function:

y[x_] := 1/2 Exp[x] + 1/12 Exp[-3 x]

and then perform the desired operations on that function. For instance, the volume:

Integrate[Pi y[x]^2, x]

Mathematica graphics

or if you want it evaluated over the region from 0 to 1:

Integrate[Pi y[x]^2,{x,0,1}] //N
(* 2.62579 *)

And the surface area can be obtained from:

2 Pi NIntegrate[y[x] Sqrt[1 + y'[x]^2], {x, 0, 1}]
(* 7.44677 *)

I use NIntegrate in this last example since your function cannot be readily integrated symbolically.


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