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: Double Integrals

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img413.imageshack.us/img413/6839/49887281.gif [Broken]

    3. The attempt at a solution



    [tex]\int^b_{a} f(x^4)dx= \int^{1^4}_{(-1)^4} 1+u \frac{du}{4x^3}[/tex]

    So, why does the method break down? Is it because a=b=1 (it should've been a < b)?

    (b) What does it mean for the Jacobian [tex]\frac{du}{dx}[/tex] to "vanish"? And how do we find this point? :confused:
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 2, 2010 #2


    Staff: Mentor

    Both limits of integration in the new integral are 1, so the value of the definite integral is 0.
    An expression "vanishes" if its value becomes zero. For what value of x does your Jacobian become zero?
    Last edited by a moderator: May 4, 2017
  4. May 2, 2010 #3
    So, the first step is to find the Jacobian? Can I get some clues on how to form the Jacobian matrix for this problem? I know that the u=x4, but I can't work out what the four entries in J are supposed to be. :confused: Any help is appreciated.
  5. May 2, 2010 #4


    Staff: Mentor

    There aren't four entries, since u is a function of only one variable, x. The problem even tells you what it is when it asks in part b, at what point does the "Jacobian" du/dx vanish. "Jacobian" is in quotes here because it's normally applied to functions of more than one variable.
  6. May 2, 2010 #5
    Oh, I see. Since u=x4


    It is only zero when x=0. So is this the point we were required to find (the "Jacobian" is zero here)?

    And for part (c) in order to exclude this point, do I need to treat the integral as a "type I region" or a "type II" region? I sketched the function and it is symmetric about the y axis...
    Last edited: May 2, 2010
  7. May 3, 2010 #6


    Staff: Mentor

    I don't remember what "Type I" and "Type II" regions are. When you make the substitution u = x^4, what is the new integral you get? It should not have x or dx in it.
  8. May 3, 2010 #7
    So, the question says I need to separate it into two integrals (using the substitution u=x4) to avoid the point found in part (b) as an interior point. So is the following correct?

    [tex]\frac{du}{dx}=4x^3[/tex], so [tex]dx=\frac{du}{4x^3}[/tex]

    [tex]\int^0_{-1} 1+u \frac{du}{4x^3} + \int^1_0 1+u \frac{du}{4x^3}[/tex]
  9. May 3, 2010 #8


    Staff: Mentor

    No. As I said before, expressions involving x should not appear in the new integrals.
  10. May 5, 2010 #9
    I see, so is the following correct:

    Since 4x3=4u3/4

    [tex]\int^0_{-1} (1+u) \frac{du}{4u^{3/4}} + \int^1_0 (1+u) \frac{du}{4u^{3/4}}[/tex]

    since the graph is even about the y=axis we can write it as

    [tex]2 \int^1_{0} (1+u) \frac{du}{4u^{\frac{3}{4}}}[/tex]

    [tex]=2 \int^1_{0} 1+u 4u^{-\frac{4}{3}}[/tex]

    [tex]=\frac{2}{4} \int^1_{0} 1+u^{-\frac{1}{3}}[/tex]

    [tex]=\frac{1}{2} \int^1_{0} u+\frac{3}{2}u^{\frac{2}{3}}[/tex]

    [tex]=\frac{1}{2} \frac{5}{2} = \frac{5}{4} =1.25[/tex]

    Is this correct now?
  11. May 5, 2010 #10


    Staff: Mentor

    No, not correct. Evaluate the first integral you started with, which is a very easy one. The transformed integral should have the same value.

    You have made several mistakes in your work.
    [tex](1+u) \frac{du}{4u^{\frac{3}{4}}} \neq 1+u 4u^{-\frac{4}{3}}[/tex]

    (1 + u)/(4u^(3/4)) is not symmetric about the y-axis. It's not even defined for u <= 0 . Your substitution was u = x^4, so u must be >= 0. If you solve for x, you need one formula for x >= 0 and another for x <= 0.

    Finally, part b asked you where du/dx was equal to zero, which happens to be in the interval [-1, 1]. You will have to account for the fact that both integrands are undefined at some point in the interval of integration.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook