Homework Help: Intergal Application(Work)

1. Jul 19, 2007

Winzer

1. The problem statement, all variables and given/known data
A tank in the shape of the bottom half of a sphere of radius 10ft. is buried so that the top of the tank is 5ft. below the surface of the ground. If the tank is intially filled woth oil ( with wieght density $$\delta=40\frac{lbs}{ft^3}$$ determine how much work is required to empty the tank through a valve 1ft above the ground.

2. Relevant equations
$$W=FD$$
$$x^2+y^2=100$$

3. The attempt at a solution
Ok so I drew a picture with a half circle with the top at -5 and the valve at 1. So I decided to slice out an element $$x_{i}$$ which looks like a disk . The ith volume of that disk is $$V_{i}\approx \pi (r_{i})^2\Delta Y$$ I let $$x_{i}=-\sqrt(100-(y_{i})^2)$$ be equall to $$r_{i}$$.
$$V_{i}\approx \pi (100-(y_{i})^2) \Delta Y$$. Multiplying that quantity by $$/delta$$ I get $$m_{i}$$. I then multiply that by 32ft/sec^2 which is my g. Then my $$D_{i}=1-y_{i}$$ since each element will be traveling this distance. So $$W_{i}\approx 32\delta\pi\ (100-y_{i})^2(1-y_{i}) \Delta Y$$ As $$\Delta Y \rightarrow 0$$
I finally have $$W=32\pi\delta\int_{5}^{15} (100-y^2)(1-y)dy$$
Sound reasonable?

Last edited: Jul 19, 2007
2. Jul 19, 2007

Winzer

mmm ... maybe that $$D_{i}=1-y_{i}$$ should be $$D_{i}=1+y_{i}$$?

3. Jul 19, 2007

Winzer

4. Jul 19, 2007

Dick

Nope, doesn't sound reasonable. And no picture necessary, you described it pretty well. And you've got all the right parts, but the solution is all jumbled up. You don't seem to be really clear on what y or y_i is. The clue something is wrong is that the integrand changes sign. So parts of it contribute negative work and parts positive? I don't think so. And you are already given a 'weight density' which is a 'force density' so there is no need for a g in the problem. Tell us what is the y you are integrating over, is it distance from the ground to the sphere section, distance from the sphere section to the exit point, distance from the sphere section to the top of the sphere? I think you may have just used all of those for the same variable.

Last edited: Jul 19, 2007
5. Jul 19, 2007

Winzer

Oh! ok thats true I definatley do not need g, lol. $$y_{i}$$ is just an element, a slice takin out, it doesn't become y till I take the reimman some and n->infinity. Yes so I am intergrating y over the bottom of the of the sphere, to the top of the sphere because that is the section containing the oil.
So my limits of intergrations should be negitive right and switched around?
$$W=\pi\delta\int_{-15}^{-5} (100-y^2)(1-y)dy$$ which becomes
$$W=-\pi\delta\int_{5}^{15} (100-y^2)(1-y)dy$$

I know I am close.
Edit: is $$D_{i}=1-y_{i}$$ right?

6. Jul 19, 2007

Dick

y is the distance from what to what?

7. Jul 19, 2007

Dick

Ok, let's say, judging by the (100-y^2) term that y is the distance from the top of the hemisphere to the disk you are lifting. The range on your integral is 0-10. Then how far does a disk at y have to be lifted? As I read the question a disk at y=0 goes up 6ft. A disk at y=10ft goes up 16ft. What does this tell you about the linear factor in your integral?

Last edited: Jul 19, 2007
8. Jul 20, 2007

Winzer

No.! $$x=-\sqrt(100-y^2)$$ is the raduis of the sphere. Then
$$y=\sqrt(100-x^2)$$ would be the hieght.
But I chose $$x_{i}=-\sqrt(100-y^2)$$ because $$r_{i}=x_{i}$$. This $$r_{i}$$ is the radius of the partioned disk so: $$r_{i}=-\sqrt(100-y^2)$$
Yes I see now: so $$D_{i}=1+y_{i}$$.
I don't think it would be on the interval 0-10. The half sphere is 5ft below the surface with a radius of 10ft so the bottom of the tank would be -15, and the flat top is -5.

Last edited: Jul 20, 2007
9. Jul 20, 2007

Winzer

This attachment should clear things up

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• calc.jpg
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10. Jul 20, 2007

Dick

Ok so r^2=(100-y^2). If y=(-15), which you say is the bottom of the tank, then r^2 is negative. You haven't convinced me this is correct. Even without looking at the attachment.

11. Jul 20, 2007

Winzer

correct. R^2 is the raduis of the tank we can chose it to be +/-, either way when we plug it into our volume equation and square we get $$V_{i}=\pi (r_{i})^2\Delta y\longrightarrow V_{i}=\pi (-/+\sqrt(100-y^2))^2\Delta Y\longrightarrow V_{i}=\pi (100-y^2)\Delta Y$$

12. Jul 20, 2007

Dick

You aren't listening. You may be able to choose +/- but for y=-15 sqrt(100-225) is neither. It's IMAGINARY. If you mean to say something like sqrt(|100-225|), then that's larger than 10. An unusual radius of a hemisphere of radius 10.

13. Jul 20, 2007

Winzer

mmmmm......

14. Jul 20, 2007

Winzer

My picture I drew looks correct right? yah I see where you are saying it is imaginary.

15. Jul 20, 2007

Winzer

Ah...now I get it
how about $$x_{i}=+/-\sqrt(100-(y-5)^2)=r_{i}$$
That looks way better.

So instead of $$x^2+y^2=100$$
it should be $$x^2+(y+5)^2=100$$

Last edited: Jul 20, 2007
16. Jul 20, 2007

Winzer

I also think my $$D_{i}$$ is wrong. since each element will have to travel between my partition interval of [-15,-5], to 1,
So i should be $$D_{i}=6+y_{i}$$

17. Jul 20, 2007

Dick

If you are going from y=-15 (bottom of tank) to y=-5 top of tank) I'm ok with r=sqrt(100-(y+5)^2). y=-15, r=0 and y=-5, r=10. Seems ok. But now look at your D=6+y. y=-5, D=1??? Shouldn't that be 6?

18. Jul 20, 2007

Winzer

I guess I am confused finding $$D_{i}$$

19. Jul 20, 2007

Dick

You've picked y=-5 to be the coordinate of the top of the tank. D should be the distance the oil has to travel to get out. At y=-5 that is D=6ft. At y=-15 that is D=16ft. So you want a linear expression in y that evaluates to those D's at those values of y.

20. Jul 20, 2007

Winzer

But doesn't each slab have to travel so shouldn't it be D= 6+Y.?
If it was d=6, then that would just be the top slab.