Double integral over the area of a square

In summary, the student attempted to find the area of a square using double integral and euclidean coordinates to polar coordinate, but was not able to do so without knowing the coordinates of at least one corner of the square.
  • #1
chrisy2012
17
0

Homework Statement


Find the area of a square with each side measuring 1 using double integral and change of euclidean coordinates to polar coordinate.


Homework Equations


x=rcosθ
y=rsin0
dA=dxdy=rdrdθ


The Attempt at a Solution


int(int(rdr)dθ)
 
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  • #2
Is that the entire question? Have you been given a graph? Without knowing the coordinates of at least one corner of the square, you can't determine its limits, hence you can't evaluate the area. Unless, the answer is to be given in terms of r and θ?
 
  • #3
hi chrisy2012! :smile:

(have an integral: ∫ :wink:)
chrisy2012 said:
int(int(rdr)dθ)

yup! … but what are the limits? :wink:

(sharks, we can choose the origin wherever we like :redface:)
 
  • #4
A "square" does not automatically have a coordinate system associated with it. So you have to be given the coordinates for at least some of the corners of the square.

For example, if one corner is at (0, 0) and another at (a, 0), then the other two are at (a, a) and (0, a). But you are still not going to have a smooth formula in polar coordinates. The line x= a becomes [itex]rcos(\theta)= a[/itex] or [itex]r= a sec(\theta)[/itex]. Of course, [itex]\theta= 0[/itex] along the line y= 0 and [itex]\theta= \pi/2[/itex] at (a, a) so the first integral would be
[tex]\int_{\theta= 0}^{\theta/4}\int_{r= 0}^{asec(\theta)} f(r,\theta)rdrd\theta[/tex].

The second half would along the top line, y= a so [itex]r sin(\theta)= a[/itex] or [itex]r= a csc(\theta)[/itex] and that extends from [itex]\theta= \pi/4[/itex] to [itex]3\pi/4[/itex].

Or you could choose a coordinate system so that the origin is at the center of the square. That might simplify the values but now, because you have to pass include all four corners so you will need four separate integrals.
 
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  • #5
HallsofIvy said:
A "square" does not automatically have a coordinate system associated with it. So you have to be given the coordinates for at least some of the corners of the square.

For example, if one corner is at (0, 0) and another at (a, 0), then the other two are at (a, a) and (0, a). But you are still not going to have a smooth formula in polar coordinates. The line x= a becomes [itex]rcos(\theta)= a[/itex] or [itex]r= a sec(\theta)[/itex]. Of course, [itex]\theta= 0[/itex] along the line y= 0 and [itex]\theta= \pi/2[/itex] at (a, a) so the first integral would be
[tex]\int_{\theta= 0}^{\theta/4}\int_{r= 0}^{asec(\theta)} f(r,\theta)rdrd\theta[/tex].

The second half would along the top line, y= a so [itex]r sin(\theta)= a[/itex] or [itex]r= a csc(\theta)[/itex] and that extends from [itex]\theta= \pi/4[/itex] to [itex]3\pi/4[/itex].

Or you could choose a coordinate system so that the origin is at the center of the square. That might simplify the values but now, because you have to pass include all four corners so you will need four separate integrals.

I didn't realize that we required the sum of 2 double integrals for a square expressed in terms of polar coordinates.

Suppose the square has the following Euclidean coordinates: (0,0), (1,0), (1,1) and (0,1). I've attached the graph to this post.

The area of the square in terms of Euclidean coordinates is the double integral:
[tex]\int^{y=1}_{y=0} \int^{x=1}_{x=0} dxdy[/tex]
Originally, i thought of converting the limits and dxdy directly into their respective polar forms, but i realize now that it would have been wrong.
HallsofIvy said:
The second half would along the top line, y= a so [itex]r sin(\theta)= a[/itex] or [itex]r= a csc(\theta)[/itex] and that extends from [itex]\theta= \pi/4[/itex] to [itex]3\pi/4[/itex].
I think that's wrong. For the second half, [itex]\theta[/itex] varies from [itex]\frac{\pi}{4}[/itex] to [itex]\frac{\pi}{2}[/itex].

Also, instead of calculating the double integral for the second half and then adding to the double integral for the first half, why not simply write the total area like this:
[tex]2\int_{\theta= 0}^{\theta=\frac{\pi}{4}}\int_{r= 0}^{r=asec(\theta)} f(r,\theta)\,.rdrd\theta[/tex]
 

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  • #6
Please remember that this is a schoolwork questsion, and the OP has shown zero effort so far. We should not be doing his homework for him...
 
  • #7
Thanks for the reply guys, I asked the TA today and she says you have to divide it up into two identical triangles. so the answer would be
∫ 0 to ∏/4 of ( ∫ 0 to secθ (rdr))dθ
which would equal to 1/2.
But since there are two of the triangles the area would equate to 1. So it all works out :)
 

1. What is a double integral over the area of a square?

A double integral over the area of a square is a mathematical operation that calculates the total value of a function over the entire area of a square. It involves integrating a function with respect to two different variables, typically represented by x and y coordinates.

2. How is a double integral over the area of a square different from a single integral?

A single integral calculates the total value of a function over a one-dimensional interval, while a double integral calculates the total value of a function over a two-dimensional region, such as a square. In other words, a double integral involves integrating a function with respect to two variables, while a single integral involves only one variable.

3. What is the purpose of calculating a double integral over the area of a square?

The purpose of calculating a double integral over the area of a square is to find the total value of a function over a two-dimensional region, which can be useful in many fields of science and engineering. It can also be used to solve problems involving finding the average value of a function, finding the center of mass of an object, and calculating volumes and areas of complex shapes.

4. What are the steps involved in calculating a double integral over the area of a square?

The steps for calculating a double integral over the area of a square are as follows:

  • 1. Determine the limits of integration for both variables, typically represented by x and y coordinates.
  • 2. Set up the integral by writing the function to be integrated and the limits of integration in the correct order.
  • 3. Solve the inner integral first by treating the outer variable as a constant.
  • 4. Solve the outer integral by plugging in the result from the inner integral.
  • 5. Simplify the expression and evaluate the integral to get the final result.

5. Are there any special cases when calculating a double integral over the area of a square?

Yes, there are some special cases that may require certain adjustments when calculating a double integral over the area of a square. These include cases where the function being integrated is not continuous or when the region of integration is not a simple square. In these cases, additional techniques may be needed to accurately calculate the integral.

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