July 27th, 2017, 04:31 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 140 Thanks: 2  what did I do wrong???
Evaluate ∫∫ x2 + y2 dx dy over the square with corners: (1, 0), (0, 1), (1, 0), and (0, 1)?? Answer: 8/3 My solution $\displaystyle \int_{1}^{0} \int_{y1}^{y+1} (x^2+y^2) dx dy + \int_{0}^{1} \int_{y1}^{1y} (x^2+y^2) dx dy = 2/3 $ What did I do wrong???? Any info would be appreciated... 
July 27th, 2017, 04:43 PM  #2 
Senior Member Joined: Oct 2009 Posts: 142 Thanks: 60 
Why do your bounds depend on $y$??

July 27th, 2017, 04:55 PM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 140 Thanks: 2  
July 28th, 2017, 04:10 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,720 Thanks: 699 
It's hard to say what you did wrong as I have no idea what you were trying to do! To integrate f(x, y) over the square with corners (1, 0), (0, 1), (1, 0), and (0, 1), x goes from 1 to 1 and, for each x, y goes from 1 to 1. The integral is $\displaystyle \int_{1}^1\int_{1}^1 f(x, y) dx dy$. In this case, $\displaystyle f(x,y)= x^2+ y^2$ so the integral is $\displaystyle \int_{1}^1\int_{1}^1 x^2+ y^2 dxdy= \int_{1}^1 \int_{1}^1 x^2 dx dy+ \int_{1}^1\int_{1}^1 y^2 dxdy$$\displaystyle = \left(\int_{1}^1 x^2 dx\right)\left(\int_{1}^1 dy\right)+ \left(\int_{1}^1 dx\right)\left(\int_{1}^1 y^2 dy\right)$$\displaystyle = \left(\frac{x^3}{3}\right)_{1}^1\left(2\right)+ \left(2\right)\left(\frac{y^3}{3}\right)_{1}^1= \frac{4}{3}+ \frac{4}{3}= \frac{8}{3}$. Let me see if I can figure out what you were thinking! You have two integrals. The first has y going from 1 to 0 and, for each y, x going from y1 to y 1. x= y 1 is the line from (0, 1) (when t= 1) to (1, 0) (when y= 0). That the line from the midpoint of the lower edge of the square to the midpoint of the left edge of the square. x= y 1 is the line from (2, 1) (when y= 1) to (1, 0) (when y= 0). That is the line from (2,1), which is not in the square, to the midpoint of the left side of the square. The integral between those lines has nothing to do with integrating over the square. The same is true of the second integral its limits of integration have nothing to do with the given square. That was why Microm@ss asked "Why do your bounds depend on y". It has nothing to do with "dx dy" or "dy dx". No, sorry, I still have no idea what you were thinking. 
July 30th, 2017, 04:16 PM  #5  
Senior Member Joined: Jan 2017 From: Toronto Posts: 140 Thanks: 2 
1. I thought I have been evaluating a diamond shape (not square). This is why I was trying to figure out the boundaries of each diagonal lines that make up the diamond. 2. This is an integration of a diamond shape, I would have thought the result of the integral is 0 because positive regions would 'negate' the negative regions. 3. I guess I need more help... Quote:
 
July 30th, 2017, 05:03 PM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749  Quote:
I think your text is wrong or there's some piece of this missing.  
July 30th, 2017, 06:08 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,591 Thanks: 936 Math Focus: Elementary mathematics and beyond 
$$\int_{1}^{1}\int_{1}^{1}(x^2+y^2)\,dx\,dy=\int_{1}^{1}\left(\frac23+2y^2\right)\,dy=\frac23+\frac2 3+\frac23+\frac23=\frac83$$

July 30th, 2017, 08:36 PM  #8  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749  Quote:
$(1,1), (1,1), (1,1), (1,1)$ His integration area is a square with corners $(1,0), (0,1), (1,0), (0,1)$  
July 30th, 2017, 09:27 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,591 Thanks: 936 Math Focus: Elementary mathematics and beyond 
I suggest you make a diagram.

July 30th, 2017, 10:28 PM  #10 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749  $\displaystyle \int \limits_{1}^{0} \int \limits_{1x}^{1+x}~x^2+y^2~dy~dx + \int \limits_{0}^{1} \int \limits_{1+x}^{1x}~x^2+y^2~dy~dx =\dfrac 2 3$ 

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double integral, wrong 
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