Double Integral: Evaluate ∫∫(x^2 + y^2)dx dy in R

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Homework Help Overview

The problem involves evaluating a double integral of the function (x^2 + y^2) over a specified triangular region defined by the vertices (0,0), (2,0), and (1,1). The original poster seeks verification of their integration limits and approach.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster splits the region into two legs for integration, questioning the correctness of their limits. Some participants express concerns about the second leg's limits and suggest writing the equation of the line connecting specific points. Others inquire about the boundaries of the integration region and whether a sketch has been drawn.

Discussion Status

Participants are actively discussing the setup of the integration limits, with some providing feedback on the original poster's approach. There is an exploration of different methods for setting up the double integral, including whether to integrate with respect to y first or x first. No consensus has been reached, and multiple interpretations of the limits are being considered.

Contextual Notes

Participants note the importance of accurately defining the boundaries of the triangular region for integration and question whether the original poster's limits correctly represent the area in question. There is an emphasis on the need for clarity in the integration setup.

mathsdespair
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Evaluate ∫∫(x^2 + y^2)dx dy over the region enclosed within
R

(0,0), (2,0) and (1,1).
I am not asking someone to do the problem but to just verify, have I got the limits right?



I split it up into 2 legs
for the first leg integrate from , x: 0→1 and y :0→x
for the second leg integrate from x:1→2 and y:1→-x
then add them up.
Thanks
 
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First one looks good, I don't like the second one. Write the equation of the lin that goes from 1,1 to 2,0.
 
BiGyElLoWhAt said:
First one looks good, I don't like the second one. Write the equation of the lin that goes from 1,1 to 2,0.

ok, it is y=-x+2
 
Last edited:
so for the second leg is it x goes from 1 to 2 and y goes from 1 to -x+2?
 
mathsdespair said:
so for the second leg is it x goes from 1 to 2 and y goes from 1 to -x+2?
No. The x values go from 1 to 2, as you said, but what do the y values do in the right-most triangle? Have you drawn a sketch of the region over which integration is taking place?
 
mathsdespair said:
Evaluate ∫∫(x^2 + y^2)dx dy over the region enclosed within
R

(0,0), (2,0) and (1,1).
I am not asking someone to do the problem but to just verify, have I got the limits right?



I split it up into 2 legs
for the first leg integrate from , x: 0→1 and y :0→x
for the second leg integrate from x:1→2 and y:1→-x
then add them up.
Thanks

You are doing the y-integration first; that is, for each fixed x you first integrate over y. The alternative is to do the x-integration first; that is, for each fixed y, first integrate over x. You might find this easier.
 
Ray Vickson said:
You are doing the y-integration first; that is, for each fixed x you first integrate over y. The alternative is to do the x-integration first; that is, for each fixed y, first integrate over x. You might find this easier.

Why?
A)you solve for y in terms of x,
B)you solve for x in terms of y,

Seems pretty tomato tomatoe to me.
 
BiGyElLoWhAt said:
Why?
A)you solve for y in terms of x,
B)you solve for x in terms of y,

Seems pretty tomato tomatoe to me.

The best way to know why is to try it for yourself.
 
Mark44 said:
No. The x values go from 1 to 2, as you said, but what do the y values do in the right-most triangle? Have you drawn a sketch of the region over which integration is taking place?

so for the first leg
x:0 to 1
y: 0 to x

2nd leg
x:1 to 2
y:1 to -x+2
is that right?
 
  • #10
BiGyElLoWhAt said:
Why?
A)you solve for y in terms of x,
B)you solve for x in terms of y,

Seems pretty tomato tomatoe to me.
Because you could do it in a single integration rather than needing to break it up into two integrals.
 
  • #11
HallsofIvy said:
Because you could do it in a single integration rather than needing to break it up into two integrals.

is my method right for doing it into steps please?
 
  • #12
mathsdespair said:
is my method right for doing it into steps please?
No. The method may be right, but your implementation of the method for doing it in steps is wrong.

What is the upper boundary of the right hand region? (the one from x=1 to x=2)

What is the lower boundary of the right hand region?
 

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