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August 16th, 2018, 07:36 AM   #1
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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.
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August 16th, 2018, 01:25 PM   #2
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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.
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August 16th, 2018, 03:09 PM   #3
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Quote:
Originally Posted by mathman View Post
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 03:50 PM.
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August 17th, 2018, 01:46 PM   #4
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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.
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August 18th, 2018, 01:19 AM   #5
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Quote:
Originally Posted by mathman View Post
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?
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August 18th, 2018, 02:56 PM   #6
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Quote:
Originally Posted by babaliaris View Post
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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