Multiple Integral Challenge Question, no solution in guide

[kostoglotov]

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. Homework Statement

Evaluate the integral

\int_0^1 \int_0^1 e^{max(x^2,y^2)}dydx

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

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

Homework Equations

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

\int_0^1 \int_0^x e^{y^2} dy dx

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

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

\int_0^1\int_x^1 e^{x^2} dy dx

inner integral

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

outer integral

\int_0^1 e^{x^2} - xe^{x^2} dx

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

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

So

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

Or

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

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);
surf(x,y,z)

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?

 

[Dansuer]

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

I'm not familiar with Matlab, but didn't you forgot the exponential in the definition of z ?

 

[HallsofIvy]

There is no exponential. The purpose of the graph was just to see where x^2 was larger and where y^2 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 x^2= y^2 means x= y. max(x^2, y^2)= x if and only if x> y and max(x^2, y^2)= y if and only if x< y. So the integral, \int_0^1\int_0^1 e^{ax(x^2, y^2)} dxdy can be written \int_0^1 \int_0^x e^{x^2} dydx+ \int_0^1 \int_x^1 e^{y^2} dydx.

However, x= y only on the boundary not everywhere in the region of integration so e^{x^2} is NOT always equal to e^{y^2}. Perhaps you simply meant to appeal to symmetry to say that the two integrals are the same?

 

[kostoglotov]

I'm not familiar with Matlab, but didn't you forgot the exponential in the definition of z ?

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

 

[kostoglotov]

I'm not familiar with Matlab, but didn't you forgot the exponential in the definition of z ?

Update, making that correction to the code makes the numeric and analytic answers line up. Thanks for your help :)