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July 27th, 2017, 04:31 PM   #1
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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_{-y-1}^{y+1} (x^2+y^2) dx dy + \int_{0}^{1} \int_{y-1}^{1-y} (x^2+y^2) dx dy = 2/3
$

What did I do wrong???? Any info would be appreciated...
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July 27th, 2017, 04:43 PM   #2
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Why do your bounds depend on $y$??
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July 27th, 2017, 04:55 PM   #3
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Quote:
Originally Posted by Micrm@ss View Post
Why do your bounds depend on $y$??

It should not matter dy dx or dx dy. Is it not?
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July 28th, 2017, 04:10 AM   #4
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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 -y-1 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.
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July 30th, 2017, 04:16 PM   #5
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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:
Originally Posted by Country Boy View Post
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 -y-1 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.
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July 30th, 2017, 05:03 PM   #6
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Quote:
Originally Posted by zollen View Post
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...
I got 2/3 as well and I did note it was a diamond shaped area of integration.

I think your text is wrong or there's some piece of this missing.
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July 30th, 2017, 06:08 PM   #7
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$$\int_{-1}^{1}\int_{-1}^{1}(x^2+y^2)\,dx\,dy=\int_{-1}^{1}\left(\frac23+2y^2\right)\,dy=\frac23+\frac2 3+\frac23+\frac23=\frac83$$
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July 30th, 2017, 08:36 PM   #8
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Quote:
Originally Posted by greg1313 View Post
$$\int_{-1}^{1}\int_{-1}^{1}(x^2+y^2)\,dx\,dy=\int_{-1}^{1}\left(\frac23+2y^2\right)\,dy=\frac23+\frac2 3+\frac23+\frac23=\frac83$$
that's an integration of the square with corners at

$(-1,-1), (-1,1), (1,1), (1,-1)$

His integration area is a square with corners

$(-1,0), (0,1), (1,0), (0,-1)$
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July 30th, 2017, 09:27 PM   #9
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I suggest you make a diagram.
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July 30th, 2017, 10:28 PM   #10
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$\displaystyle \int \limits_{-1}^{0} \int \limits_{-1-x}^{1+x}~x^2+y^2~dy~dx +

\int \limits_{0}^{1} \int \limits_{-1+x}^{1-x}~x^2+y^2~dy~dx =\dfrac 2 3$
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