Describing D is Green's Theorem

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Homework Help Overview

The problem involves applying Green's Theorem to evaluate a line integral over a specified boundary of a rectangle defined by its vertices. The vector field given is F(x, y) = (tan−1(x))i + 3xj, and the boundary C is traversed counterclockwise.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to describe the boundary of the rectangle and set up the integrals accordingly. They express uncertainty about the correctness of their region descriptions and integral setup. Other participants question the accuracy of the region splits and suggest considering the surface integral of a constant function.

Discussion Status

Participants are actively discussing the setup of the problem, with some guidance provided on the interpretation of surface integrals. There is an exploration of the area of the region D and its relation to the integral, but no consensus has been reached regarding the correctness of the original poster's approach.

Contextual Notes

There is a focus on ensuring the described regions accurately match the rectangle's boundaries, and the discussion includes considerations of how to calculate the area of D in relation to the integral.

TranscendArcu
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Describing "D" is Green's Theorem

Homework Statement



Let F(x, y) = (tan−1(x))i+3xj. Find [tex]\int_C F • dr[/tex]where C is the boundary of the rectangle with vertices (0, 1), (1, 0), (3, 2), and (2, 3), traversed counterclockwise.

The Attempt at a Solution



I have Qx = 3 and Py = 0. Therefore Qx - Py = 3 - 0 = 3. Now, what I'm having the most trouble with is just describing this rectangle in terms of x and y.

I think the boundary of my rectangle is described by line segments that follow the equations y=x+1, y=-x+5, y=x-1, and y=-x+1. I rewrite these as x=y-1, x=-y+5, x=y+1, and x=-y+1. I think I have to split up the region somehow, so my integrals are:

[tex]\int_0 ^2 \int_{-y+1} ^{y+1} 3 dxdy + \int_1 ^3 \int_{y-1} ^{-y+5} 3 dxdy[/tex]

At this point, I think I should ask if I'm doing this correctly.
 
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Hi TranscendArcu! :smile:

Your split up regions do not appear to match your rectangle, which you can see if you would plot the outermost points of your boundaries.
2 regions would also not be enough.

But to make it a bit easier, what is in general the surface integral of a constant function, say 1?
 


The surface integral of a constant function is the surface area of the surface (multiplied by the constant, in this case one), right?
 


Yes, but only if the constant function is 1.
So...
 


So I might just write 3*Area(D), right?

So if I find vectors that describe the edges of D, and set them in R3, and take the magnitude of their cross product, I should find the area of D. I have, <-1,1,0> and <2,2,0>. Cross product gives <0,0,-4>

|<0,0,-4>| = 4.

3*4 = 12?
 


Yup.
 

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