# Simple volume calculation problem (double integrals)

• xio
In summary, the attempt at a solution to the homework statement is to find the volume at the first quadrant and then multiply by four. Apparently so simple, but I just can't see the mistake.
xio
[EDIT]: Found the mistake, see the next post.

## Homework Statement

Evaluate $$\iint_{S}{\rm e}^{x+y}dx\, dy,S=\{(x,y):\left|x\right|+\left|y\right|\leq1\}$$

2. The attempt at a solution
##\left|x\right|+\left|y\right|## is the rhombus with the center at the origin, symmetrical about both axes, so we find the volume at the first quadrant and then multiply by four.

##\varphi(x)=1-x## limits the region of integration and so we have, first integrating over ##y## : $$\iint_{S}{\rm e}^{x+y}dx\, dy=4\int_{0}^{1}\left[\int_{0}^{1-x}{\rm e}^{x}{\rm e}^{y}dy\right]dx=1\neq {\rm e}-{\rm e}^{-1}$$ (I know the correct answer which is ##{\rm e}-{\rm e}^{-1}## )

Apparently so simple, but I just can't see the mistake.

Last edited:
Oh Lord, ##{\rm e}^{x+y}## obviously isn't symmetric.

xio said:
Oh Lord, ##{\rm e}^{x+y}## obviously isn't symmetric.

True. Did you work the problem using the "obvious" change of variables?

LCKurtz said:
True. Did you work the problem using the "obvious" change of variables?

Hrm... I guess I cheated: I took the total volume to be the sum of volumes over the triangles formed by the axes and four diagonal lines (##x+1##, ##1-x##, ##-x-1##, ##x-1##): $$\int_{-1}^{0}\left[\int_{0}^{x+1}{\rm e}^{x+y}dy\right]dx+\int_{-1}^{0}\left[\int_{-x-1}^{0}{\rm e}^{x+y}dy\right]dx+\ldots,$$ took the first one $$\int_{1}^{0}\left[\int_{0}^{x+1}{\rm e}^{x+y}dy\right]dx=\int_{-1}^{0}{\rm e}^{2x+1}-{\rm e}^{x}dx=\frac{1}{2}({\rm e}+{\rm e}^{-1})-1$$ and then asked sage to do the rest because they are essentially the same and evaluating them all has no didactic value.

By the way, I'm learning this stuff on my own, so I might not know the standard way of approaching problems like this. My book (Apostol) says nothing about the change of variables if I understand correctly.

What I dislike about this sort of examples is that I have to plot some of the functions first to actually understand what's going; TBH I'd enjoyed more analytic (or algorithmic) approach, where the peculiarities of a particular example wouldn't matter.

(too many "I"'s in this post)

No, that's not what LCKurtz meant by "the obvious change of variables".

If you let u= x+ y and v= x- y, then x= (u+ v)/2 and y= (u- v)/2. The figure is then $-1\le u\le 1$, $-1\le v\le 1$.The Jacobean is 1/2 so the integral becomes
$$\frac{1}{2}\int_{u=-1}^1\int_{v=-1}^1 e^u dvdu$$

HallsofIvy said:
No, that's not what LCKurtz meant by "the obvious change of variables".

I never suggested that.

Change of variables is some ten subsections below my current position in Apostol, but I liked the technique, thank you.

## What is a double integral?

A double integral is a mathematical concept that involves calculating the volume of a three-dimensional region by integrating a function over a two-dimensional region. It is represented by two nested integrals and is commonly used in science and engineering to solve problems related to volume, mass, and other physical quantities.

## How is a double integral different from a single integral?

A single integral calculates the area under a curve in a two-dimensional plane, while a double integral calculates the volume under a surface in a three-dimensional space. In other words, a single integral is a one-dimensional operation, and a double integral is a two-dimensional operation.

## What is the process for solving a simple volume calculation problem using double integrals?

The first step is to determine the bounds of integration, which define the limits of the two-dimensional region. Next, the function to be integrated is determined, which represents the height of the three-dimensional region at each point in the two-dimensional region. The double integral is then evaluated by integrating the function over the two-dimensional region, resulting in the volume of the three-dimensional region.

## What are some real-world applications of double integrals?

Double integrals are used in a wide range of scientific and engineering fields, including physics, chemistry, biology, and economics. Some common applications include calculating the volume of a chemical solution in a container, determining the mass of an object with varying density, and finding the average temperature of a three-dimensional object.

## Are there any limitations or drawbacks to using double integrals?

One limitation of double integrals is that they can only be used to calculate the volume of a region that is well-defined by a mathematical function. Additionally, the process of setting up and evaluating double integrals can be complex and time-consuming, especially for more complicated three-dimensional regions. It is also important to ensure that the bounds of integration are correctly determined to avoid incorrect results.

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