Solving Double Integrals: Finding J in R2

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Homework Help Overview

The discussion revolves around evaluating a double integral J defined over a specific region U in R². The integral involves the function y cos(x + y) and the region is described by certain inequalities that participants are interpreting geometrically.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the geometric interpretation of the region U, with one participant suggesting that it consists of two triangles, while another believes it should only be one triangle. There is also mention of using a graphing tool to verify the shape.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the region U and its implications for setting up the integral. Some guidance has been offered regarding the limits of integration based on the identified geometric shapes.

Contextual Notes

There is a potential discrepancy in the interpretation of the region U, as participants are not in agreement on the number of triangles formed. This may affect the limits of integration and the overall approach to solving the integral.

squenshl
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How do I find
J := \int\int_U y cos(x + y) dA
where U = {(x,y) \in R2 : 0 \leq x \leq 2, 1-x \leq x \leq -2 + 2x}.

I drew the picture & i get 2 triangles but not sure what to do with them. Help please.
 
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in rectangular coordinates, dA=dx dy, so the region U describes will give you the limits for the integral.
 
I think you should only be getting one triangle. Try checking your picture against one generated by a graphing tool.

Cheers,
W. =)
 
Very true. I got a triangle & J = -0.5717341110 using maple.
 

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