Double integral over the area of a square

[chrisy2012]

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)

 

[DryRun]

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?

 

[tiny-tim]

hi chrisy2012!

(have an integral: ∫ )

int(int(rdr)dθ)

yup! … but what are the limits?

(sharks, we can choose the origin wherever we like )

 

[HallsofIvy]

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 $rcos(\theta)= a$ or $r= a sec(\theta)$. Of course, $\theta= 0$ along the line y= 0 and $\theta= \pi/2$ at (a, a) so the first integral would be
$$\int_{\theta= 0}^{\theta/4}\int_{r= 0}^{asec(\theta)} f(r,\theta)rdrd\theta$$.

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

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.

 

[DryRun]

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 $rcos(\theta)= a$ or $r= a sec(\theta)$. Of course, $\theta= 0$ along the line y= 0 and $\theta= \pi/2$ at (a, a) so the first integral would be
$$\int_{\theta= 0}^{\theta/4}\int_{r= 0}^{asec(\theta)} f(r,\theta)rdrd\theta$$.

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

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:
$$\int^{y=1}_{y=0} \int^{x=1}_{x=0} dxdy$$
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.

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

I think that's wrong. For the second half, $\theta$ varies from $\frac{\pi}{4}$ to $\frac{\pi}{2}$.

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:
$$2\int_{\theta= 0}^{\theta=\frac{\pi}{4}}\int_{r= 0}^{r=asec(\theta)} f(r,\theta)\,.rdrd\theta$$

 

[berkeman]

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...

 

[chrisy2012]

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 :)