Help with Double Integrals

[ktobrien]

1. The problem statement, all variables and given/known data
Use symmetry to evaluate the given integral.
SS_D 6-x+7y dA
where D is the region bounded by the square with vertices (±5, 0) and (0, ±5).


2. Relevant equations
Integrals


3. The attempt at a solution
I worked it out and got 300. But I did it by integrating one half then the other and adding the two together. I am looking for someone to confirm rather my answer is correct of not and to show me how to do it using symmetry

[Dick]

It think it's 600 isn't it? Hint: can you show that SS_D 6 is the same as your original integral using symmetry?

[ktobrien]

Yes. SS_D 6 is also 300. Does this mean my answer is right?

[HallsofIvy]

So your answer to Dick's question, "can you show that SS_D 6 is the same as your original integral using symmetry?" is yes? How do you show it?

[Dick]

Yes. SS_D 6 is also 300. Does this mean my answer is right?

No, SS_D 6 isn't 300. How are you getting that? D is a 10x10 square. My hint was to show SS_D (-x) and SS_D 7y are zero. And then think about why.

[ktobrien]

SS_D 6 is 300 over the square. x goes from 0 to 5 and y goes from x-5 to 5-x. This is half of the square. This integral equals 150. You then multiply that by 2 to get the whole square. The answer is therefore 300. My answer is correct. Also, D is not a 10X10 square. It is a 5X5 square with its corners at (5,0), (0,5), (-5,0), and (0,-5).

[Dick]

SS_D 6 is 300 over the square. x goes from 0 to 5 and y goes from x-5 to 5-x. This is half of the square. This integral equals 150. You then multiply that by 2 to get the whole square. The answer is therefore 300. My answer is correct. Also, D is not a 10X10 square. It is a 5X5 square with its corners at (5,0), (0,5), (-5,0), and (0,-5).

Ah, you are right. I was drawing the wrong square. Sorry. The sides of the square actually have length 5*sqrt(2). But now can you see why the integral of (-x) or 7*y over the square will automatically vanish?