Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Having trouble graphing a Volume problem

  1. Oct 6, 2006 #1
    Hello again,

    Once I get this graphed correctly, I don't see any problems as to why I couldn't solve it. But for now I have a question about graphing this volume problem.

    The problem says to find the volume generated by rotating the region bounded by [itex]y=e^x[/itex], [itex]x=0[/itex], and [itex]y=\pi[/itex] about the x-axis.

    I have a picture of what I my graph would look like. The green curve is the line [itex]y=e^x[/itex], and the rust colored line is, [itex]x=0[/itex]. My question is where do I graph [itex]y=\pi[/itex] ? Is it as simple as placing the line at y = 3.14?

    Thanks for taking a look. It is greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 7, 2006 #2
    I still am unsure about this. I could use the help. Thanks.

    Actually I think I got it. But if you could look at confirm or deny what I have come up with I would appreciate it.
     

    Attached Files:

    • scan.gif
      scan.gif
      File size:
      24.6 KB
      Views:
      155
    Last edited: Oct 7, 2006
  4. Oct 7, 2006 #3
    I'd love to confirm what you have, however, for some reason the attachment is "pending approval." I can't view it.

    Left boundary is a vertical line at x=0. Right and bottom boundary is y=e^x. Top boundary is y=pi. There's an intersection point where y=e^x and y=pi meet. It should look like a triangle. Then you rotate about the x-axis.

    I didn't bother attaching a picture since attachments on this forum need some approval.
     
  5. Oct 7, 2006 #4

    radou

    User Avatar
    Homework Helper

    Yes, it is as simple as placing a line parallel to the x-axis at y = 3.14.
     
  6. Oct 7, 2006 #5
    Thanks fire and radou.

    Fire ~ Your description is what my graph looks like so thanks again.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook