0
$\begingroup$

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 $2\pi \int_{a}^{b}y \sqrt{1+(\frac{dy}{dx})^2}dx$

Improve this question
$\endgroup$
7
  • $\begingroup$ 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
    Commented Apr 30, 2014 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
    Commented Apr 30, 2014 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
    Commented Apr 30, 2014 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
    Commented Apr 30, 2014 at 16:48
  • $\begingroup$ And isn't there supposed to be a "+" in the surface area equation? $\endgroup$
    – bobthechemist
    Commented Apr 30, 2014 at 16:49

1 Answer 1

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.