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November 2nd, 2015, 11:18 PM   #1
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(Torus/Volume)

Hello.
I'm really new to this place and hope I can get some help with these two questions.Thank you so much in advance.

1) We are emptying a spherical tank (not completely full) into a tank which is shaped like a torus. What percentage of the torus is filled? Refer to the figure below.


Second question:
A mine shaft can be approximated by the solid of revolution below. The shaft is filled with water.

a) How long would it take to empty the shaft if we are using a ramp that pumps 2500 U.S gallons per hour?



b) If the water in the shaft is pumped into a truck with the shape described below, what is the minimum number of trucks required?


Last edited by skipjack; November 5th, 2015 at 01:53 AM.
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November 4th, 2015, 02:49 PM   #2
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volume of water in the spherical tank ...

$\displaystyle V = \pi \int_{-3}^2 (9 - y^2) \, dy$

volume of the torus (using a theorem of Pappus) ...

V = (cross sectional area)(circumference of centroid rotation) = $\pi \cdot (2\pi \cdot 7)$


mine shaft volume (in $m^3$ which will have to be converted to gals) ...

$\displaystyle V = \frac{1250 \pi^2}{11} \int_{50/11}^{100} \left(\frac{100}{y}+3\right) \, dy$

single truck volume ...

$\displaystyle V = \pi \int_2^4 9^2 - \bigg[(x-3)^4+7\bigg]^2 \, dx$
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November 5th, 2015, 12:17 AM   #3
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Many Thanks. I got them I guess.

Last edited by skipjack; November 5th, 2015 at 01:53 AM.
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November 5th, 2015, 12:19 AM   #4
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Thank you very much
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November 5th, 2015, 07:51 AM   #5
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Mistake in the mine-shaft volume ... forgot my addition sign and multiplied the constants in front of the integral instead of adding them. should be

$\displaystyle V = \frac{1250\pi}{11}+ \pi \int_{50/11}^{100} \left(\frac{100}{y}+3 \right) \, dy$

mea culpa ... it happens.
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