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!

Application of integration

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the area of the plane bounded by the hyperbola (x)^1/2+(y)^1/2=(a)^1/2 and the axes of the coordinates is a^2/6

    2. Relevant equations
    ∫yd x


    3. The attempt at a solution
    I just can't integrate the function (a)^1/2-(x)^1/2 considering the fact that there are x as well as a and should I take the limit from 0 to a for x??????
     
  2. jcsd
  3. Oct 20, 2013 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    There's nothing stopping you integrating that, but it would not be ∫ydx. What is y equal to?
    The limits should be at the axes, i.e. where x or y is zero. Does that correspond to x = 0 or a?
     
  4. Oct 20, 2013 #3
    y={ (a)^1/2- (x)^1/2}^2. The integration should be done with respect to x where y is just a function of x. Can you drop a hint as to how I should integrate it and is 0≤ x≤ a or 0≤ x≤ (a)^1/2 true
     
    Last edited: Oct 20, 2013
  5. Oct 20, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Start by expanding the 'squared'.
     
  6. Oct 22, 2013 #5
    Okay, then it becomes y= a-2(a)^1/2(x)^1/2+x. Now, integrating both sides we get y= - 4/3x^3/2(a)^1/2+x^2/2. But, what limits should I take??? Is 0≤x≤a valid in this case??
     
    Last edited: Oct 22, 2013
  7. Oct 22, 2013 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I already answered that. The other bounds are the axes, so the limits correspond to x = 0 for one and y = 0 for the other.
     
  8. Oct 23, 2013 #7
    Taking the limit from 0 to a the integration with respect to x becomes y= -4a^2/3+a^2/2= -8a^2+3a^2/6= -5a^2/6 How come I have a negative sign and the area is not negative. Besides the answer is a^2/6. What mistakes am i making???
     
  9. Oct 23, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You missed a term when you integrated.
     
  10. Oct 23, 2013 #9
    How come I missed a term when I integrated??? Besides the integration of a becomes 0 with respect to x. Hence, the negative sign should also be taken into account because I expanded {(a)^1/2-(x)^1/2}^2
     
  11. Oct 23, 2013 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, the derivative of a constant would be zero, but not the integral.
     
  12. Oct 26, 2013 #11
    Sorry, I made a ridiculously silly mistake. Don't know what happened as I solved harder integration problems like this. How do I solve the second part of the question?? I have no clue as to how I can find out the axes of the coordinates to be a^2/6
     
  13. Oct 26, 2013 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I don't see a second part. It asks you to find an area, which you have now done.
     
  14. Oct 26, 2013 #13
    Sorry for my foolishness. I completely misread the question at first I thought that I had to show that the area and the axes of the coordinates are a^2/6. Anyway, Thank you haruspex for bearing my ridiculousness and showing me the extent of silliness that I exhibit in solving the simplest problems. My brain malfunctions at times especially when I am doing a lot of math per day...
     
    Last edited: Oct 26, 2013
  15. Oct 26, 2013 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Let's not start a competition over who does the stupidest things. I might win.
     
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: Application of integration
Loading...