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Homework Help: Multiple Integral Challenge Question, no solution in guide

  1. May 18, 2015 #1
    I have what I think is a valid solution, but I'm not sure, and when I try to check the answer approximately in Matlab, I don't get a verified value, and I'm not sure if my analytic solution or my approximation method in Matlab is at fault.

    1. The problem statement, all variables and given/known data

    Evaluate the integral

    [tex]\int_0^1 \int_0^1 e^{max(x^2,y^2)}dydx[/tex]

    where [itex]max(x^2,y^2)[/itex] is the larger of the numbers x2 and y2.

    Assumption: I'm assuming that x2 and y2 are real valued.

    2. Relevant equations

    3. The attempt at a solution

    I consider the region of integration thusly

    There is symmetry about the line y = x. Below this line, the max value is of y squared, since for y < x, y2 > x2. And so above the line y = x, the max value is x squared.

    This is true because 0 <= x <= 1 and 0 <= y <= 1.

    Using Matlab to sketch the 3-D graph gets


    This looks right, and we can see the symmetry.

    Now, if we setup the region is that we just integrate over the triangle created by y=x dividing the square [0,1]x[0,1], knowing that we must multiply the result by 2 at the end, we could evaluate the integral

    [tex]\int_0^1 \int_0^x e^{y^2} dy dx[/tex]

    Now, this is problematic, since we can't eval. this analytically.

    We can change the order of integration, but this results in further evariable squared integrated with respect to said variable situations. For instance, in one arrangement we get

    [tex]\int_0^1\int_x^1 e^{x^2} dy dx[/tex]

    inner integral

    [tex]\int_x^1 e^{x^2} dy = \left[ye^{x^2}\right]_x^1[/tex]

    outer integral

    [tex]\int_0^1 e^{x^2} - xe^{x^2} dx[/tex]

    Which is still problematic, and happens with all changes to the order of integration

    But [itex]e^{x^2} = e^{y^2}[/itex] so we can just replace [itex]e^{y^2}[/itex] with [itex]e^{x^2}[/itex] in the original problematic integral.


    [tex]2 \times \int_0^1 \int_0^x e^{x^2} dy dx = 2 \times \frac{1}{2}\left(e-1\right) = e - 1[/tex]


    [tex]2 \times \int_0^1 \int_0^y e^{y^2} dx dy = 2 \times \frac{1}{2}\left(e-1\right) = e - 1[/tex]

    Anyhow, I wanted to get an approx. verification of this is Matlab.

    The code I used to generate that graph is

    [x,y] = meshgrid(0:0.01:1,0:0.01:1); % 0.01 increments
    z = max(x^2,y^2);

    so, I figure for a good approximation of the volume under that surface should be

    sum(sum(z.*0.0001)); % element-wise multiplication of z by 0.01 x 0.01

    But this gives 0.513433500000000 which is not e - 1.

    Is my analytic solution incorrect or my numeric approximation method? Or both?
  2. jcsd
  3. May 18, 2015 #2
    I'm not familiar with Matlab, but didn't you forgot the exponential in the definition of z ?
  4. May 18, 2015 #3


    User Avatar
    Science Advisor

    There is no exponential. The purpose of the graph was just to see where [itex]x^2[/itex] was larger and where [itex]y^2[/itex] was larger.

    Though that was not really necessary. Since both integrals are from 0 to 1, both x and y are always positive and so [itex]x^2= y^2[/itex] means x= y. [itex]max(x^2, y^2)= x[/itex] if and only if x> y and [itex]max(x^2, y^2)= y[/itex] if and only if x< y. So the integral, [itex]\int_0^1\int_0^1 e^{ax(x^2, y^2)} dxdy[/itex] can be written [itex]\int_0^1 \int_0^x e^{x^2} dydx+ \int_0^1 \int_x^1 e^{y^2} dydx[/itex].

    However, x= y only on the boundary not everywhere in the region of integration so [itex]e^{x^2}[/itex] is NOT always equal to [itex]e^{y^2}[/itex]. Perhaps you simply meant to appeal to symmetry to say that the two integrals are the same?
    Last edited by a moderator: May 18, 2015
  5. May 18, 2015 #4
    I absolutely did..Ok...this is the second time this week that my reading comprehension has brought me to PF...maybe I can build some sort of mechanical hand to slap me in the face if I don't proof read...thanks
  6. May 18, 2015 #5
    Update, making that correction to the code makes the numeric and analytic answers line up. Thanks for your help :)
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