Surface integrals to calculate the area of this figure

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Homework Help Overview

The discussion revolves around calculating the area of a geometric figure that includes triangles and squares, specifically using surface integrals as required by the assignment.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants express difficulty in applying integration to find the area of squares, while some question whether integration is necessary at all. There are attempts to clarify the dimensions of the squares and their relationship to the overall figure.

Discussion Status

The conversation is ongoing, with participants sharing insights about the dimensions of the squares and the requirements of the assignment. Some guidance has been offered regarding the use of integration for area calculation, but there is no clear consensus on the approach or the interpretation of the figure.

Contextual Notes

Participants note constraints related to the assignment's requirement for integration and express confusion over the dimensions of the squares, particularly regarding their lengths and positions.

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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