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Direct Integration vs. Green's Theorem

  • Thread starter wifi
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  • #1
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Problem:

##\int_R (x-y)dx \ dy=-2/3 ## for ##R=\{(x,y):x^2+y^2 \geq 1; y \geq 0\}## by

a.) Direction integration,

b.) Green's theorem.

Attempt at a Solution:

I'm a little confused with part a. Wouldn't the region R be defined by all the points above the y-axis that lie on, in addition to above, the circle of radius 1 centered at the origin?

I'm confused on what the limits of integration would be for the integral ##\iint (x-y) dx \ dy##.
 
Last edited:

Answers and Replies

  • #2
vela
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I'm sure that's a typo. It probably should be ##x^2+y^2 \le 1##.
 
  • #3
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I'm sure that's a typo. It probably should be ##x^2+y^2 \le 1##.
So by direct integration I have,

[tex]\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}}(x-y) dy \ dx = 2/3[/tex] Is it supposed to be negative?
 
  • #4
D H
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You made a mistake somewhere. It is -2/3 rather than +2/3.
 
  • #5
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Dang. My limits are correct, yes?

Also, how could I apply Green's theorem. I thought the idea was to change a line integral into a double integral, right?

EDIT: Never mind, it was an error on my part. Still not sure how to apply Green's theorem.
 
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  • #6
vela
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Well, you have a double integral. You want to change it into a line integral.
 
  • #7
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Well, you have a double integral. You want to change it into a line integral.
So it'd just be a line integral over the boundary of the region, correct?

Since Green's theorem in a plane is given by [tex]\iint_R (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y})dx \ dy \ =\oint_C (F_1 dx +F_2 dy)[/tex]

for this specific problem we'd have ##\frac{\partial F_2}{\partial x}=x## and ##\frac{\partial F_1}{\partial y}=y##.

Thus, ##F_1=\frac{1}{2}y^2## and ##F_2=\frac{1}{2}x^2##. Right?
 
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  • #8
vela
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Looks good.
 

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