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!

Real line integral

  1. May 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the line integral [tex] \int\limits_{AB} x^2 dx+ \sqrt{xy}dy [/tex], where AB is a part of a circle in the first quarter of carthesian coordinates system ranging from A(0,R) to b(R,0).


    2. Relevant equations



    3. The attempt at a solution
    Parametrisation of a circle:
    x=Rcos(t), y=Rsin(t)
    dx=-Rsin(t)dt, dy=Rcos(t)dt
    t ranges from [tex]\frac{\pi}{2}[/tex] to 0.

    And so the integral with the parametrisation applied is:
    [tex]\int\limits_{\frac{\pi}{2}}^0( -R^3 cos^2(t) sin(t)+R^2 \sqrt{sin(t)cos(t)}cos(t))dt [/tex]

    Now, the first part of the sum is simple. The latter one, the one with the root, is, according to Mathematica, analytically uncalculable (returns hypergeometric function). Now, when I tried evaluating the definite integral (using Mathematica), it returned [tex]\frac{1}{24} \left(8 R^2-3 \sqrt{2} \pi R \sqrt{R}\right) [/tex], whereas the answer says it's [tex] \frac{1}{15} R^2 (6 \sqrt{R}-5R) [/tex]. Who's right and is it possible to calculate it on paper. Maybe I shuldn't have used the polar coordinates? But I would then end up with a double root!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 29, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Are you sure the original integral is correct? I don't see how you can get a [tex]\sqrt{R}[/tex] in the answer from the integral you started with.

    EDIT: I think the integral is supposed to be

    [tex]\int_{AB} (x^2\,dx + \sqrt{x}y\,dy)[/tex]

    That'll give the answer you're trying to get, I think.
     
  4. May 29, 2010 #3
    That's what I was calculating, I just forgot to write the brackets ;). I wrote it correctly in 3-attempt at a solution. And this is what I evaluated in Mathematica. And integrating this root is the only problem :).
     
  5. May 29, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Did you notice vela's version of the integral is sqrt(x)*y*dy, not sqrt(xy)*dy?
     
  6. May 29, 2010 #5
    My bad, I didn't, sorry. But what it says in the exercise book is sqrt(xy). And the answer came from mathematica, it seems quite wierd to me as well, but i checked it a couple of times

    E: you're right, there is not root of r in the answer, my apologies. But the integral should be the way i wrote it.
     
  7. May 29, 2010 #6


    Mathematica returns:

    [tex]
    \int\limits_{AB} x^2 dx+ \sqrt{xy}dy =
    \frac{1}{24} \left(8 R^3-3 \sqrt{2} \pi R \sqrt{R}\right) [/tex]

    and is consistent with the numerical results.
     
  8. May 29, 2010 #7
    Whatever the result is, given that this exercise was taken from exercise book i would expect it to be calculable on a paper. And it is a book for undergraduates. WTF??
     
  9. May 29, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Could you show us what you entered into Mathematica to get that result? In the second term, [tex]\sqrt{xy}[/tex] is proportional to R as is dy, so that term should be proportional to R2.
     
  10. May 29, 2010 #9
    Integrate[
    Sqrt[r^2*Sin[t]*Cos[t]]*Cos[t] - r^2*Sin[t]*(Cos[t])^2, {t, \[Pi]/2,
    0}]

    You are right, I was wrong. I messed up with those 'r's'. But nevertheless, question remains actual-how the hell am i supposed to calculate it on a paper with the knowledge i've got after first term??
     
  11. May 29, 2010 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I was actually curious to see what jackmell did since you two got the same result in Mathematica, even though you had changed your answer a few hours earlier.

    I think it's just a typo in your book. To get their answer, you need to integrate [tex]\sqrt{x}y[/tex] instead of [tex]\sqrt{xy}[/tex], and you can do that easily with a substitution.
     
  12. May 29, 2010 #11
    Sorry about that. It's r^2:

    Code (Text):
    Clear[r]
    myval = Integrate[(-r^3)*Cos[t]^2*
         Sin[t] + r^2*Sqrt[Cos[t]*Sin[t]]*
         Cos[t], {t, Pi/2, 0}]
     

    [tex]

    \int\limits_{AB} x^2 dx+ \sqrt{xy}dy =
    \frac{1}{24} \left(8 R^3-3 \sqrt{2} \pi R^2 \right)
    [/tex]

    and that is consistent with the numerical results:

    Code (Text):
    In[28]:=
    N[(1/24)*r^2*(-3*Sqrt[2]*Pi + 8*r) /.
       r -> 5]
    r = 5;
    NIntegrate[(-r^3)*Cos[t]^2*Sin[t] +
       r^2*Sqrt[Cos[t]*Sin[t]]*Cos[t],
      {t, Pi/2, 0}]

    Out[28]=
    27.78265748492177

    Out[30]=
    27.78265752218859
     
    Last edited: May 29, 2010
  13. May 30, 2010 #12
    Well, you mey be right, but on the other hand, a couple pages earlier, in a chapter about double integrals (volume, area etc.) they expect you to integrate
    [tex]z=sin(\pi x y)[/tex]. After first integrating, let's say over y, you get [tex]\frac{cos(\pi x y)}{\pi x}[/tex], which again is imposiible to integrate in the meaning of finding an antiderivative. However, mathematica returns the same answer that thay gave in the book.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Real line integral
  1. Line Integral (Replies: 4)

  2. Line Integrals (Replies: 4)

  3. Line Integrals (Replies: 4)

Loading...