# Multiple Integral Challenge Question, no solution in guide

• kostoglotov
In summary, the homework statement asks for the integral of a function over a region where there is symmetry about the line y = x. In Matlab, using the surf function, the author sketches a 3-D graph and finds that the max value is x squared. However, when trying to approximate the volume under the surface, the author finds that the result is not e - 1. The author then tries to verify this by evaluating the integral analytically, but finds that this is problematic. The author then tries to approximate the volume using a numeric approximation method, but this also results in an incorrect answer. Finally, the author solves the problem by realizing that x= y only on the boundary of the integration region and that e^{
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.

## 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 evariable 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?

kostoglotov said:
[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 ?

kostoglotov
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?

Last edited by a moderator:
kostoglotov
Dansuer said:
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

Dansuer said:
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 :)

## 1. What is a multiple integral?

A multiple integral is a type of integral that involves integrating a function of multiple variables over a region in space. It is used to find the volume, surface area, and other properties of a three-dimensional object or region.

## 2. What is the challenge question about?

The challenge question involves solving a multiple integral problem that does not have a solution provided in the guide. It requires critical thinking and problem-solving skills to find a solution.

## 3. Why is there no solution provided in the guide?

The purpose of the challenge question is to test your understanding of multiple integrals and your ability to apply the concepts to a problem without a given solution. It is meant to challenge you and allow you to practice your skills.

## 4. How can I approach solving the multiple integral challenge question?

To solve the challenge question, you can start by breaking down the integral into smaller parts and using known integration techniques to simplify it. You can also try using different coordinate systems or making substitutions to make the integral more manageable.

## 5. What is the importance of being able to solve multiple integral problems?

Multiple integrals are used in many fields of science and engineering, including physics, chemistry, and engineering. Being able to solve these types of problems is crucial for understanding and analyzing complex systems and phenomena in these fields.

Replies
2
Views
798
Replies
1
Views
818
Replies
4
Views
510
Replies
10
Views
1K
Replies
14
Views
970
Replies
2
Views
736
Replies
12
Views
1K
Replies
27
Views
2K
Replies
34
Views
2K
Replies
2
Views
776