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Volume of Rotation

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data

    Find volume when the area rotates around a line.

    2. Relevant equations

    x=y^2
    x=1-y^2
    about x = 3

    3. The attempt at a solution

    Could someone please check my formula for Area? I have pi[ (3-(y^2))^2 - (2-(y^2))^2 ] which gives me the integral pi[5y-(2/3)y^3]. I think it's wrong though because I plugged it into the wolfram alpha program and it gives me the wrong answer, the answer according to the textbook is (10)(sqrt2)(pi)/3 and the program says according to my integral between the intersection points (-1/sqrt2, 1/sqrt2) the volume would be (14)(sqrt2)(pi)/3. I'm really confused. Any help would be greatly appreciated! Thank you.
     
  2. jcsd
  3. Feb 2, 2015 #2

    Mark44

    Staff: Mentor

    You are missing either ##\Delta x## or ##\Delta y##, which makes it harder for me to tell whether you are using disks or shells. I'm sure I could figure out what you're doing with some effort, but I shouldn't have to.
     
  4. Feb 2, 2015 #3
    Oh, I think I'm using delta y. Like my widths are on the y axis? Is that what is meant? Sorry I'm not very good with terminology at all, haha.
     
  5. Feb 2, 2015 #4
    The radius of your circles run parallel to the x-axis, and the summing of the areas occurs along the y-axis axis (the circles are vertically stacked), which gives us
    [itex]∫\pi(x(y))^2Δy[/itex]
    In this case, x(y) is not directly explicit. You have to use the given functions and the fact that it revolves around the x=3 axis to determine the real form of x(y)
     
  6. Feb 2, 2015 #5

    Mark44

    Staff: Mentor

    I get the same number you got -- ##\frac{14 \pi \sqrt{2}}{3}##. Double-check the problem information and answer. If everything is as you reported, it seems that the book has a typo.
     
    Last edited: Feb 2, 2015
  7. Feb 2, 2015 #6

    Mark44

    Staff: Mentor

    The above is incorrect. tree.lee is using "washers", or disks with holes in them.
    Please read the problem more carefully. Both functions give x explicitly in terms of y.
     
  8. Feb 2, 2015 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    3-(1-y^2) does not give 2-y^2.
     
  9. Feb 2, 2015 #8
    Oh. Wouldn't subtracting the functions from x=3 give the distance from it and thus the radii desired? Then subtracting the inner function from the outer function should give the area I want to integrate, right? That gives me pi[5y-(2/3)y^3], doesn't it?
    But the textbook says (10)(sqrt2)(pi)/3.
     
  10. Feb 2, 2015 #9

    Mark44

    Staff: Mentor

    Yes, I know. It might be a typo in the book. That has been known to happen.

    What book are you using?
     
  11. Feb 2, 2015 #10
    I believe I got the answer. Let me type out my work and someone correct me if I'm wrong
     
  12. Feb 2, 2015 #11
    OH I
    Sorry I posted replies that I didn't mean to I'm new to this forum and there are a lot of brackets saying quotes and stuff and it confuses me and I'm not sure what I"m inputting when I hit post reply. Sorry. Thanks for the help A LOT! I did double check everything.
     
  13. Feb 2, 2015 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No. See my post #7.
     
  14. Feb 2, 2015 #13

    Mark44

    Staff: Mentor

    Yes, good eye, Dick -- you're right. The inner radius is (3 - (1 - y2)2, which is (2 + y2)2. Using that gives the textbook's answer.
     
  15. Feb 2, 2015 #14
    Oh. Wow, this community is so cool. Thanks a billion. It's seriously appreciated. Hope you're all having a wonderful day. Thanks a lot Dick.
     
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