Surface integrals to calculate the area of this figure

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SUMMARY

This discussion focuses on using surface integrals to calculate the area of geometric figures, specifically squares and triangles. The user expresses difficulty in applying integration to squares of 1/2 unit length, while others confirm that integration is indeed applicable for calculating areas. The coordinates provided indicate that the squares are positioned at (-0.5, 0) and (0.5, 0), leading to a total length of 1 unit between them. The conversation highlights the importance of understanding geometric positioning and integration techniques in solving such problems.

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VVS2000
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Homework Statement
My homework is to find the area of the following figure(leaving those two squares)
Relevant Equations
The distance between those two squares is one unit. Please do tell if the figure is not descriptive enough
20200320_164038.jpg

I can find the area of the triangles but can't solve the squares for some reason
 
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Sry for not including the equations section, I don't know how to type integrals here
 
VVS2000 said:
Sry for not including the equations section, I don't know how to type integrals here
Are you sure you need integration for this?
 
Well the assignmet was clear of me using integration
 
And the squares are of 1/2 unit length
 
VVS2000 said:
Well the assignmet was clear of me using integration
You can, course, use integration to calculate the area of a square or a triangle - just for practice.
 
Please post your image right side up.
 
One edge of one square is at ##(-0.5, 0)## while the other square has an edge at ##(0.5,0)##, and the total length between them is ##1##.
If you can figure out the distance between ##(-0.5, 0)## and ##(0.5,0)##, do you think that you can deduce something?
20200320_164038.jpg
 
I supose all those denominators that look sort of like an ##\alpha## are actually ##2##'s.Your figure shows the inner distance between the squares as ##1## unit and the outer distance between them is ##\frac 1 2 - (-\frac 1 2) = 1## so the squares have sides of length ##0##, so they aren't there.
 
  • #10
VVS2000 said:
Sry for not including the equations section, I don't know how to type integrals here
See the LaTeX tutorial at the top of the page under INFO, Help. :smile:
 

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