Expert: Paul Klarreich - 1/26/2007

Question
Hi, I'm not sure if this type of problem is the type you wanted to answer, but I've tried many ways to do the integration for natural log but I didn't have much success...Here goes...

Find the volume to three decimals when the area between y=e^(x^2)and y=x from x=0 to x=2 is revolved around the y-axis.

Thank you for considering it!

Answer
Questioner:   Joyce
Category:  Calculus
 
Subject:  Math Analysis/Calculus
Question:  Hi, I'm not sure if this type of problem is the type you wanted to answer, but I've tried many ways to do the integration for natural log but I didn't have much success...Here goes...

Find the volume to three decimals when the area between y=e^(x^2)and y=x from x=0 to x=2 is revolved around the y-axis.

Thank you for considering it!
.........................................
Hi, Joyce,

This looks like a 'volumes by cylinders' example.  In any of these, you take a 'sample' piece and then integrate it.  For your sample, which is a thin cylindrical shell,

The height will be the difference between the function values:  e^(x^2) - x

The radius will be the x-coordinate:   x

The thickness is  dx.

The volume is equal to the surface area times the thickness.  For a cylindrical shell, the surface area is the circumference times the height:

dV = 2 pi r h dx = 2 pi x (e^(x^2) - x)dx
  = 2 pi (x e^(x^2) - x^2) dx

OK, now we can integrate this. [I will save some typing and put the 2 pi in at the end:]

{2
| (x e^(x^2) - x^2) dx =
}0

{2
| x e^(x^2) dx   minus
}0

{2
|  x^2 dx =
}0


To handle the first integral,  let  u = x^2, then  du = 2x dx,  x dx = du/2

{  1
| --- e^u du =
}  2

1
--- e^(x^2)
2

The second integral is just  x^3/3.  From here on it's just substitution, so I'll leave that to you. [Don't forget the 2 pi.]