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August 16th, 2018, 07:36 AM  #1 
Member Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6  Is this calculation correct (Double Integration)?
$\displaystyle w = \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} [m \cdot g + T \cdot \frac{x + (h  y)}{h}] \cdot dxdy = \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}}m \cdot g \cdot dxdy + \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} T \cdot \frac{x + (h  y)}{h} \cdot dxdy = m \cdot g \cdot \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} dxdy + \frac{T}{h} \cdot \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} [x + (h  y)] \cdot dxdy $ $\displaystyle I_{1} = m \cdot g \cdot \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} dxdy $ $\displaystyle I_{2} = \frac{T}{h} \cdot \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} [x + (h  y)] \cdot dxdy $ $\displaystyle w = I_{1} + I_{2} $ $\displaystyle I_{1} = m \cdot g \cdot \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} dxdy = m \cdot g \cdot \int_{y_{1}}^{y_{2}} [x]_{x_{1}}^{x_{2}} \cdot dy = m \cdot g \cdot \int_{y_{1}}^{y_{2}} (x_{2}  x_{1}) \cdot dy = m \cdot g \cdot [x_{2} \cdot y  x_{1} \cdot y]_{y_{1}}^{y_{2}} = m \cdot g \cdot ( x_{2} \cdot y_{2}  x_{1} \cdot y_{2}  x_{2} \cdot y_{1} + x_{1} \cdot y_{1}) = m \cdot g \cdot [x_{2}(y_{2}y_{1})  x_{1}(y_{2}y_{1})] \Leftrightarrow I_{1} = m \cdot g \cdot (y_{2}y_{1}) (x_{2}  x_{1}) $ $\displaystyle I_{2} = \frac{T}{h} \cdot \int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} [x + (h  y)] \cdot dxdy = \frac{T}{h} \cdot \int_{y_{1}}^{y_{2}} [\frac{x^{2}}{2} + h \cdot x  y \cdot x]_{x_{1}}^{x{2}} \cdot dy = \frac{T}{h} \cdot \int_{y_{1}}^{y_{2}} [\frac{x_{2}^{2}}{2} + h \cdot x_{2}  y \cdot x_{2} + \frac{x_{1}^{2}}{2}  h \cdot x_{1} + y \cdot x_{1}] \cdot dy = \frac{T}{h} \cdot [\frac{x_{2}^{2}}{2} \cdot y + h \cdot x_{2} \cdot y  \frac{y^{2}}{2} \cdot x_{2} + \frac{x_{1}^{2}}{2} \cdot y  h \cdot x_{1} \cdot y + \frac{y^{2}}{2} \cdot x_{1}]_{y_{1}}^{y_{2}} = \frac{T}{h} \cdot [\frac{x_{2}^{2}}{2} \cdot y_{2} + h \cdot x_{2} \cdot y_{2}  \frac{y_{2}^{2}}{2} \cdot x_{2} + \frac{x_{1}^{2}}{2} \cdot y_{2}  h \cdot x_{1} \cdot y_{2} + \frac{y_{2}^{2}}{2} \cdot x_{1} + \frac{x_{2}^{2}}{2} \cdot y_{1}  h \cdot x_{2} \cdot y_{1} + \frac{y_{1}^{2}}{2} \cdot x_{2}  \frac{x_{1}^{2}}{2} \cdot y_{1} + h \cdot x_{1} \cdot y_{1}  \frac{y_{1}^{2}}{2} \cdot x_{1} ] = \frac{T}{h} \cdot [ \frac{x_{2}^{2}}{2}(y_{2}  y_{1}) + h(x_{2} \cdot y_{2}  x_{1} \cdot y_{2}  x_{2} \cdot y_{1} + x_{1} \cdot y_{1})  \frac{y_{2}^{2}}{2}(x_{2}  x_{1}) + \frac{x_{1}^{2}}{2}(y_{2}  y_{1}) + \frac{y_{1}^{2}}{2}(x_{2}  x_{1})] = \frac{T}{h} \cdot [(y_{2}  y_{1})( \frac{x_{1}^{2}}{2}  \frac{x_{2}^{2}}{2}) + (x_{2}  x_{1})(\frac{y_{1}^{2}}{2}  \frac{y_{2}^{2}}{2}) + h(x_{2} \cdot y_{2}  x_{1} \cdot y_{2}  x_{2} \cdot y_{1} + x_{1} \cdot y_{1})] = \frac{T}{h} \cdot [(y_{2}  y_{1})( \frac{x_{1}^{2}}{2}  \frac{x_{2}^{2}}{2}) + (x_{2}  x_{1})(\frac{y_{1}^{2}}{2}  \frac{y_{2}^{2}}{2}) + h(x_{2}(y_{2}y_{1})  x_{1}(y_{2}y_{1}))] = \frac{T}{h} \cdot [(y_{2}  y_{1})( \frac{x_{1}^{2}}{2}  \frac{x_{2}^{2}}{2}) + (x_{2}  x_{1})(\frac{y_{1}^{2}}{2}  \frac{y_{2}^{2}}{2}) + h \cdot (y_{2}y_{1})(x_{2}x_{1})] = \frac{T}{h} \cdot [\frac{1}{2} \cdot (y_{2}  y_{1})( x_{1}^{2}  x_{2}^{2}) + \frac{1}{2} \cdot (x_{2}  x_{1})(y_{1}^{2}  y_{2}^{2}) + h \cdot (y_{2}y_{1})(x_{2}x_{1})] $ $\displaystyle w = m \cdot g \cdot (y_{2}y_{1}) (x_{2}  x_{1}) + \frac{T}{h} \cdot [\frac{1}{2} \cdot (y_{2}  y_{1})( x_{1}^{2}  x_{2}^{2}) + \frac{1}{2} \cdot (x_{2}  x_{1})(y_{1}^{2}  y_{2}^{2}) + h \cdot (y_{2}y_{1})(x_{2}x_{1})] $ OMG 1 hour to write the latex code, I hope I didn't do anything wrong... Last edited by skipjack; August 16th, 2018 at 02:30 PM. 
August 16th, 2018, 01:25 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622 
It's correct, but unnecessarily tedious. I looked at your final answer and the original statement and was able to see it is correct without writing anything down. Everything in between seems pointless. For example $I_1=mg\int_{y_1}^{y_2}\int_{x_1}^{x_2}dxdy=mg(y_2y_1)(x_2x_1)$. You don't need any intermediate steps. 
August 16th, 2018, 03:09 PM  #3  
Member Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6  Quote:
It literally took me one hour lol. I'm teaching my self by the way and books doesn't show how to do calculations faster. Last edited by babaliaris; August 16th, 2018 at 03:50 PM.  
August 17th, 2018, 01:46 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622 
The main point to simplify is to realize that with fixed limits for x and y you can look at it as a product of single integrals. With $I_2$, you need to break it up into 3 parts, so you can then make the double integral into the product. Once you do that you will have 8 very easy integrals where 6 have a constant integrand.

August 18th, 2018, 01:19 AM  #5  
Member Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6  Quote:
 
August 18th, 2018, 02:56 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622  

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calculation, correct, double, integration 
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