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 August 16th, 2018, 06:36 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 107 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 01:30 PM.
 August 16th, 2018, 12:25 PM #2 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 669 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 $I_1=-mg\int_{y_1}^{y_2}\int_{x_1}^{x_2}dxdy=-mg(y_2-y_1)(x_2-x_1)$. You don't need any intermediate steps. Thanks from topsquark and babaliaris
August 16th, 2018, 02:09 PM   #3
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Quote:
 Originally Posted by mathman 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 $I_1=-mg\int_{y_1}^{y_2}\int_{x_1}^{x_2}dxdy=-mg(y_2-y_1)(x_2-x_1)$. You don't need any intermediate steps.
Well probably because i don't have the experience. I'm new in integration.
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 02:50 PM.

 August 17th, 2018, 12:46 PM #4 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 669 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. Thanks from babaliaris
August 18th, 2018, 12:19 AM   #5
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Quote:
 Originally Posted by mathman 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.
In other words, breaking Integrals to sums its easier than integrating as whole?

August 18th, 2018, 01:56 PM   #6
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Quote:
 Originally Posted by babaliaris In other words, breaking Integrals to sums its easier than integrating as whole?
Yes, also recognizing when a double integral is the product of two single integrals.

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