Double Integral volumes: Triangular base

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Discussion Overview

The discussion revolves around evaluating a double integral over a triangular region defined by the vertices (1,1), (2,0), and (2,2). Participants explore the correct setup of limits for integration and the subsequent calculations involved in solving the integral.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes the limits of integration as y=x to y=2-x for y and from x=1 to x=2.
  • Another participant agrees with the proposed limits of integration.
  • A participant expresses difficulty in simplifying the resulting expression after integrating with respect to y, leading to a term that does not simplify well.
  • Another participant suggests leaving the problematic term in its original form and integrating it with respect to x, indicating that it should simplify correctly using the power rule.
  • A participant shares their integration steps and expresses uncertainty about obtaining the correct final answer, suspecting that the original setup of limits might be incorrect.
  • Another participant reports consistently arriving at a different result (75/2) and expresses confidence in the limits of integration, indicating a potential disagreement on the calculations or setup.

Areas of Agreement / Disagreement

Participants generally disagree on the correctness of the calculations and the final results, with multiple competing views on the setup and execution of the integral.

Contextual Notes

Participants have not resolved the discrepancies in their results, and there are indications of missing steps or assumptions in the integration process.

dan5
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Evaluate

∫∫R 5x2 + 2y2

where R is triangle (1,1) (2,0) (2,2)

I see the lines bounding the triangle are y=x y=2-x and x=2, and have tried many attempts at setting up the correct limits.

Would it be correct to split this into 2 triangles, or are the limits y=xy=2-x for y and 21 for x, correct
 
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y=x∫y=2-x for y and 2∫1 for x

This.
 
Thanks for the help... but

when I then integrate WRT y I get

2∫1 [5x2y+(2/3) y3] which I have to substitute in my limits of y=x and y=2-x

but then this leaves me with a (2/3) (2-x)3 term, which does not simplify well, and does not lead me to the correct answer.

Any ideas where I have gone wrong, thanks!
 
Well I would just leave that term in the form (2/3) (2-x)^3 and then integrate it with respect to x without factoring it out, that term should just become (-1/6) (2-x)^4, by the power rule. Beyond that I can't see where you went wrong without seeing your work.
 
Right, here goes...

Integrating WRT y

2∫1 [5x2y+(2/3)y3 ]

then entering limits

2∫1 [5x3+(2/3)x3] - [5x2(2-x) + (2/3)(2-x)3]

simplifying to...

2∫1 (32/3)x3 - 10x2 - (2/3)(2-x)3

applying the second integral WRT x

[ (32/12) x4 - (10/3) x3 + (1/6)(2-x)4 ]

then entering limits of 2 and 1 leaves me with 14, where I am meant to find 33/2.

Again thanks in advance! I can't help thinking my original set up of the limits in the problem is wrong
 
hmm, I keep getting 75/2, I'm quite confident the limits of integration are correct, I'm not sure what is going wrong..
 

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