1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface Area rotated about an axis which is not the x or y axis

  1. Apr 17, 2014 #1
    Hi. I understand how to solve surface Area using integration when it is to be revolved about the x or y axis. But when the axis is not x or y I have a difficult time solving it. Please help me. Here is the equation

    sqrt(x+1) rotated at x=-1 and y=5.
    the bounds are 1 to 5.
    since y=sqrt(x+1)
    x = y^2 -1

    I solved for x=-1 and I got ∫ (from 1 to 5) 2pi ((y^2 -1) +1) √(1+4y) dy

    and for y=5 ∫ (from 1 to 5) 2pi ((√x+1)+5) √(1+ [itex]\frac{1}{4x+4}[/itex])


    pls help me. I really do understand on how to rotate it about the x or y axis. I tried searching online for some tutorials but I can't find any.
     
  2. jcsd
  3. Apr 17, 2014 #2
    Are the axes you're rotating about x = -1 and y = 5? As in, these are two separate parts of one question?

    I suggest a change of variables here. You can redefine the axes (shifting everything to the left, right, up or down) and then use the formulae you know to find the surface area of revolution when the axis of revolution is the x or y axis. For example, if you make the substitution x' = x - a, then the line x = a is now the x' = 0 axis, or the y' axis.
     
  4. Apr 17, 2014 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I assume you have a typo under that radical and you mean$$
    2\pi\int_1^5(y^2-1+1)\sqrt {1+4y^\color{red}2}~dy$$Although that isn't the way I would have set it up, at least the integrand is now correct, so you are doing something right. But your limits aren't. When ##x## goes from ##1## to ##5##, what are the corresponding ##y## values? That's what you want for your limits.

    In your second one$$
    2\pi \int_1^5 (\color{red}{\sqrt{x+1}+5})\sqrt{1+\frac 1 {4x+4}}~dx$$you have the radius of rotation wrong. You want ##y_{upper} - y_{lower} = 5 -\sqrt{x+1}##.

    Actually, you seem to understand pretty well what you are doing. Personally I would have done the first integral in terms of ##x## and the second one in terms of ##y## but once you fix the mistakes, you are OK.
     
  5. Apr 17, 2014 #4

    BiGyElLoWhAt

    User Avatar
    Gold Member

    How I look at it is like this:

    If you rotate your function around an axis, what is your radius? Well look at it carefully, what are you really doing?

    Let's say our function is [itex]y=\sqrt{x}[/itex] and we want to find the surface area. Well what's the radius if we rotate about the x-axis? Well at the x-axis [itex]y=0[/itex] and thus the distance from our function to our axis we're rotating about: [itex]r=\sqrt{x}-0[/itex]

    Now let's say we want to rotate our function about [itex]y=-3[/itex]
    So what's r? Well it's the distance from our function to our axis.
    Or: [itex]|f(x)-a_{rotation}|[/itex], which in our case is [itex]|\sqrt{x}-(-3)|=|\sqrt{x}+3|[/itex]

    I would recommend drawing it out.

    Then you just plug it into your [itex]∫_{a}^b2\pi r\sqrt{1+f'{x}^2}dx[/itex]
    Mod note: The above should be
    ##\int_a^b 2\pi r\sqrt{1+(f'(x))^2}dx ##
     
    Last edited by a moderator: Apr 17, 2014
  6. Apr 17, 2014 #5

    BiGyElLoWhAt

    User Avatar
    Gold Member

    oops. in the eq. it's f'(x), not f'x
     
  7. Apr 18, 2014 #6
    I knew I was wrong there. So I solved for the values of y and I got √2 and √6 for my bounds in my second equation. Am I correct? OH! And thanks for noticing the 4y^2. I didn't notice that I did not square the Y



    as for this one, yeah, I get what your trying to say. I changed my answer.

    THIS LAST ONE ACTUALLY MADE ME SMILE. Thanks. It means a lot to hear someone say I'm actually right! I took your advice and here are my answers

    For equation 1: 2pi ∫(from 1 to 5) ([itex]\sqrt{x+1}[/itex]-(-1)) [itex]\sqrt{1+\frac{1}{4x+4}}[/itex] dx

    For equation 2: 2pi ∫(from √2 to √6) (5-[itex]\sqrt{y^2 -1}[/itex]) [itex]\sqrt{1+4y^2}[/itex] dy


    SO! Am I right?
     
  8. Apr 18, 2014 #7
    Also guys...I have a problem. I solved for the both of them and i got different answers. Aren't they supposed to have the same answers?
     
  9. Apr 18, 2014 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You should have two different answers. One for revolving about x=-1 and another for revolving about y = 5. The dx and dy versions should agree for each axis.
     
  10. Apr 18, 2014 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Correct.

    Almost. ##y_{upper} - y_{lower} = 5-y##.
     
    Last edited: Apr 18, 2014
  11. Apr 18, 2014 #10

    you lost me there. May i ask why is it y instead of [itex]\sqrt{y^2 - 1}[/itex]
     
  12. Apr 18, 2014 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can express a point on the curve ##y=\sqrt{x+1}## in terms of ##x## as ##(x,\sqrt{x+1})## or in terms of ##y## as ##(y^2-1,y)##. You want the ##y## value (the second coordinate) in terms of ##y## for the dy integral.
     
  13. Apr 19, 2014 #12

    STUPID ME. To be honest I looked at my notes and I was looking for where I got [itex]\sqrt{y^{2}-1}[/itex] and I can't find it anywhere. Maybe I got it from my original integral which was in terms of x then I just removed the inside of the radical. It's things like that which I need to watch out for. Thanks so so so much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Surface Area rotated about an axis which is not the x or y axis
Loading...