My Math Forum what did I do wrong???
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 July 27th, 2017, 05:31 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 152 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 $\displaystyle \int_{-1}^{0} \int_{-y-1}^{y+1} (x^2+y^2) dx dy + \int_{0}^{1} \int_{y-1}^{1-y} (x^2+y^2) dx dy = 2/3$ What did I do wrong???? Any info would be appreciated... Thanks from SenatorArmstrong
 July 27th, 2017, 05:43 PM #2 Senior Member   Joined: Oct 2009 Posts: 142 Thanks: 60 Why do your bounds depend on $y$?? Thanks from zollen
July 27th, 2017, 05:55 PM   #3
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 152
Thanks: 2

Quote:
 Originally Posted by Micrm@ss Why do your bounds depend on $y$??

It should not matter dy dx or dx dy. Is it not?

 July 28th, 2017, 05:10 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,875 Thanks: 766 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. Thanks from zollen and MaxBenjamin
July 30th, 2017, 05:16 PM   #5
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 152
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:
 Originally Posted by Country Boy 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.

July 30th, 2017, 06:03 PM   #6
Senior Member

Joined: Sep 2015
From: USA

Posts: 1,652
Thanks: 839

Quote:
 Originally Posted by zollen 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.

 July 30th, 2017, 07:08 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,660 Thanks: 965 Math Focus: Elementary mathematics and beyond $$\int_{-1}^{1}\int_{-1}^{1}(x^2+y^2)\,dx\,dy=\int_{-1}^{1}\left(\frac23+2y^2\right)\,dy=\frac23+\frac2 3+\frac23+\frac23=\frac83$$ Thanks from MaxBenjamin
July 30th, 2017, 09:36 PM   #8
Senior Member

Joined: Sep 2015
From: USA

Posts: 1,652
Thanks: 839

Quote:
 Originally Posted by greg1313 $$\int_{-1}^{1}\int_{-1}^{1}(x^2+y^2)\,dx\,dy=\int_{-1}^{1}\left(\frac23+2y^2\right)\,dy=\frac23+\frac2 3+\frac23+\frac23=\frac83$$
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

$(-1,0), (0,1), (1,0), (0,-1)$

 July 30th, 2017, 10:27 PM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,660 Thanks: 965 Math Focus: Elementary mathematics and beyond I suggest you make a diagram.
July 30th, 2017, 11:28 PM   #10
Senior Member

Joined: Sep 2015
From: USA

Posts: 1,652
Thanks: 839

$\displaystyle \int \limits_{-1}^{0} \int \limits_{-1-x}^{1+x}~x^2+y^2~dy~dx + \int \limits_{0}^{1} \int \limits_{-1+x}^{1-x}~x^2+y^2~dy~dx =\dfrac 2 3$
Attached Images
 Clipboard01.jpg (12.7 KB, 33 views)

 Tags double integral, wrong

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post shunya Elementary Math 4 February 5th, 2015 09:54 AM Mathbound Calculus 2 August 28th, 2014 07:25 PM Regnes Algebra 2 January 19th, 2012 10:58 PM DanS29 Calculus 5 April 27th, 2010 06:37 PM skarface Algebra 4 January 23rd, 2010 07:03 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2017 My Math Forum. All rights reserved.