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Q: Suppose a spherical tank is full of water, calculate the work required to drain it out completely, if the radius 3ft and there is a spout of 1ft. (water density $62.5 Ilbs/ft^{3}$)

Solution

picture of the situation:

(get a generic cross-sectional slice)

$V_{slice} = π·r^{2}·h$

$V_{slice} = π·r^{2}·dy$

So, now I have to find out what 'r' is. 'r' is the same as the isolated 'x' of the equation of a circle. So,

$x^{2} + y^{2} = 9$

$x = +/-\sqrt{9 - y^{2}}$

so,

$V_{slice} = π·(9 - y^{2})·dy$

and for force:

$F_{slice} = (62.5)·π·(9 - y^{2})$

And now displacement, which I think I have wrong.

$displacement = (7 - y)$

Because the slice would have to travel from the bottom of -3 to 0 which is 3, and from o to +3 which is 3, so a distance of 6, +1 for the spout so 7 altogether. I was told it's suppose to be 4 - y but I have no idea why this is.

For the limits of intergation we only integrate where there is water, so limits of integration would range from 0 to 6.

So altogether: $W = \displaystyle\int_{0}^{6} (62.5)·π·(9-y^{2})·(7 - y)dy$

Why is the displacement/distance wrong?

Thank you

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  • $\begingroup$ The spout is at height $4$ in your coordinate system; a slice of water at the bottom of the tank gets lifted $4 - (-3) = 7$ feet. (Separately, it might have been more appropriate to edit one of your other questions instead of asking variations on the same type of calculus problem...?) $\endgroup$ Commented May 7, 2017 at 13:51

1 Answer 1

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Nice drawing and showing of your work so far!

When you're measuring displacement, find the coordinates of the initial and final position and subtract them. You chose coordinates to put $y=0$ in the middle of the tank; that's a good idea because that's the real symmetry in the problem—it makes the equation for the tank simple. The bottom of the tank is at position $y=-3$, and the top of the tank is at position $y=3$. The top of the spout is at position $y=4$. So the slice at position $y$, in moving to position $4$, travels a distance $4-y$.

You can check this against some of what we just pointed out. The water at the bottom of the tank needs to move 7 ft, just like you said. And $4-(-3) = 7$. The water at the top of the tank only moves 1 ft, and $4-3 = 1$. The water in the middle of the tank moves to the top of the tank (3 ft) , then another foot of spout, and $4-0 = 4$.

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  • $\begingroup$ Thank you!! I am not fully conceptually understanding this. I always thought we strictly looked at the total distance, which here is 7. and then each time an integration happens it subtracted off that number from it. I guess I will just accept that it works algebraically. because if we integrated 7 - y at y = -2 that would be 9 wouldn't make sense. But with 4 - y, y = - 2, 6 makes sense for distance traveled. $\endgroup$
    – yre
    Commented May 7, 2017 at 14:02
  • $\begingroup$ The part where you say that the slice at position 'y' is moving towards position 4 (the max) makes sense to me. Is that how these problems are generally set up, the y-value is moving TOWARDS the peak value. And since we have negative values under the x-axis we only consider the positive end of the sphere because the peak value can only occur at the top end and we don't consider what happens at the bottom of the sphere? $\endgroup$
    – yre
    Commented May 7, 2017 at 14:05
  • $\begingroup$ So essentially the bottom end under the sphere under the x-axis has no effect, we only consider the top end above the x-axis and it's maximum or its peak value? $\endgroup$
    – yre
    Commented May 7, 2017 at 14:06
  • $\begingroup$ Can I use initial minus final like this, initial = - 3. We start at the bottom, right? and then the final position = 4, So initial minus final is.... -3 - 4 = - 7. So that would be - 7 - y ? What am I doing wrong? $\endgroup$
    – yre
    Commented May 7, 2017 at 15:07
  • $\begingroup$ I think I get it now, it's just BIG minus SMALL where small gets the variable 'y' to be evaluated later on by the integral? So here the biggest value to be taken on is '4' and then minus - the smallest guy which is -3 but we'll just save that as a variable 'y'. So 4 - y is the same as BIG minus SMALL where small is just substituted as a variable 'y'. Is that right? $\endgroup$
    – yre
    Commented May 7, 2017 at 15:42

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