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Shell Method To be Used?

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid generated by revolving the region bounded by the curve y= e^(x^2) and the lines y=0, x=0, and x=1, about the y-axis.


    2. Relevant equations
    Shell method?


    3. The attempt at a solution
    I don't know which method to pick, disk/washer or shell method?
    I picked the disk method and got this:
    ∏* ∫e ^ (x^4) dx
    but I stopped, because that definitely seems wrong.
     
    Last edited: Jun 13, 2013
  2. jcsd
  3. Jun 13, 2013 #2

    Mark44

    Staff: Mentor

    What's the exact statement of the problem? I can't tell which line the region is to be rotated around.

    On these kinds of problems, you can usually use either method (considering disk/washer to be essentially the same method). However, it's often the case that one method will result in an easier integral.

    It's important to draw a couple of pictures - one that shows the region to be rotated, and one that shows a split-section of the solid of revolution. That picture should show your area element, either a vertical strip or a horizontal strip, and the solid that is formed by rotating that strip.
     
  4. Jun 13, 2013 #3
    I drew a picture, but it's hard to figure out which method to use really. The reason I chose the disk method first was because it seems like it would be a solid with no hole in the middle. But I'm not even sure if this logic is right. I believe the bounded area is a vertical strip, at least, it is according to my diagram.
     
  5. Jun 13, 2013 #4

    Mark44

    Staff: Mentor

    Is the region to be revolved around the y-axis? Your problem statement wasn't clear to me.
     
  6. Jun 13, 2013 #5
    Yes sir, it is to be revolved around the y-axis. (I edited it in the original post)
     
  7. Jun 13, 2013 #6

    Mark44

    Staff: Mentor

    OK. Disks are NOT the way to go. Although the thing doesn't have a hole in it, the top is dished, so when y > 1, your disks would need to become washers.

    The simplest way to do this is to use cylindrical shells. The area elements are vertical strips that extend from the x-axis up to the graph of y = ex2. Each strip gets revolved around the y-axis to form a shell.

    The volume of one of these shells is ##2\pi x * \text{height}*\Delta x##.
     
  8. Jun 13, 2013 #7
    So the form of the cylindrical shell method is:

    2∏ * ∫rh dx ?

    The height is the y distance and the radius is the x distance? But how would those be determined?
     
  9. Jun 13, 2013 #8

    Mark44

    Staff: Mentor

    Yes. The radius is x, the distance from the axis of rotation (y-axis or x = 0) to a point (x, ex2) on the curve. The height is the y value at the point on the curve.

    Since you're using vertical strips (rotated around the y-axis) whose width is Δx, the integral has dx in it. This means that everything needs to be in terms of x. That's why we can leave the radius as x. That's actually very handy for this integral, but as you're also learning differentiation at the same time, you might not be far enough along to be able to actually do the integration, which requires a substitution.
     
    Last edited: Jun 14, 2013
  10. Jun 13, 2013 #9
    How would I determine what the y value is in this case? The region is vertical and only the bottom of it touches the curve?
     
  11. Jun 14, 2013 #10

    Mark44

    Staff: Mentor

    The region being rotated is bounded on the left by the y-axis, on the right by the line x = 1, below by the x-axis, and above by the graph of y = ex2. So I don't know what you mean by saying the region is vertical. A sketch of the region would be helpful to you. The curve runs between (0, 1) and (1, e).
     
  12. Jun 14, 2013 #11
    I have drawn a sketch of the curve and shaded in the region. I meant it is a vertical strip. So I'm presuming the limits would be 0 and 1 for this solid?
     
  13. Jun 14, 2013 #12

    CAF123

    User Avatar
    Gold Member

    That is correct - what did you get as your resulting integral?
     
  14. Jun 14, 2013 #13
    I have this:

    2∏* ∫x* (e^(x^2)) dx
     
  15. Jun 14, 2013 #14

    Mark44

    Staff: Mentor

    Yes, with limits of integration being 0 and 1.

    $$ 2\pi \int_0^1 x~e^{x^2}~dx$$
     
  16. Jun 14, 2013 #15
    So when I solve this, I get:

    2∏ * e^(x^2)/2
    2∏*e/(2) - (∏)

    Is this correct?
     
  17. Jun 14, 2013 #16

    Mark44

    Staff: Mentor

    It's correct, but it can be (and should be) simplified.
    The first expression above is the antiderivative, which simplifies to ##\pi~e^{x^2}##.

    The second expression results from evaluating the antiderivative at the limits of integration. IOW, this:
    $$ \left.\pi e^{x^2}\right|_0^1$$
     
  18. Jun 15, 2013 #17
    Okay, I will make sure to simplify. Thank you!
     
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