
#1
Apr2412, 01:45 AM

P: 17

1. The problem statement, all variables and given/known data
Find the area of a square with each side measuring 1 using double integral and change of euclidean coordinates to polar coordinate. 2. Relevant equations x=rcosθ y=rsin0 dA=dxdy=rdrdθ 3. The attempt at a solution int(int(rdr)dθ) 



#2
Apr2412, 06:25 AM

PF Gold
P: 836

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
Apr2412, 06:45 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi chrisy2012!
(have an integral: ∫ ) (sharks, we can choose the origin wherever we like ) 



#4
Apr2412, 06:46 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,885

Double integral over the area of a square
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. 



#5
Apr2412, 08:44 AM

PF Gold
P: 836

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 realise now that it would have been wrong. 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] 



#6
Apr2412, 10:56 AM

Mentor
P: 39,643

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
Apr2412, 03:49 PM

P: 17

Thanks for the reply guys, I asked the TA today and she says you have to divide it up in to 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 :) 


Register to reply 
Related Discussions  
double integral finding the area  Calculus & Beyond Homework  6  
Double Integral (underneath a surface and above a square)  Calculus & Beyond Homework  4  
Surface Area [Double Integral]  Calculus & Beyond Homework  5  
Double integralSurface area  Calculus  3  
Surface Area Double Integral  Calculus  36 