Question

Find the work in ft-lbs required to empty a full hemispherical tank with a radius of 6 ft by pumping the water to 2 ft above the top of the tank

Answer 1

@KingGeorge

Answer 2

I'll be totally honest. I forgotten all the details of how to do this kind of problem. I do know that\[W=\int F\,dx\]where W is work, F is force, and x is some distance. I also remember that in your case, the integral will be from 2 (2 feet above the tank) to 8 (radius + 2).

Answer 3

I have some notes

Answer 4

http://assets.openstudy.com/updates/attachments/537d3749e4b015d57c6438e6-hero-1400719958596-workftlbs.png

Answer 5

this is for a similar problem

Answer 6

What's in front of the integrals? A "D"?

Answer 7

Yeah, those are Ds

Answer 8

What do they stand for?

Answer 9

hero made the notes

Answer 10

my guess is distance

Answer 11

Alright, I think I've got a little better idea of what to do know. D stands for density, and g is the gravitational constant. The density is required because we also need to factor in the mass, and mass=density*volume.

So the only thing I'm not sure about at this point is the volume.

Answer 12

One minute.

Answer 13

v = The Integral of f(x) dx from b to a. I think... I'm on my phone right now. The formula was drawn out in another post

Answer 14

I feel a little bad about posting a link, but I think there is a very clear explanation of a very similar problem here. Simply change the units/constants, and the only thing left, is that you have to lift everything an extra 2 feet.
http://www.math.washington.edu/~yumiko88/Final/Work/F%20Work%20Sol.pdf

Answer 15

Will you help for this one so I have an example to go off of?

Answer 16

Sure. I'll be basing my explanation off of what's in the link though.

Answer 17

Thank you. I can't even click the link on mobile. I'll get on my computer soon though

Answer 18

First, I'm going to model it with a graph of a semi-circle that's centered at (0,-2) (so we have to lift an extra 2 units). The equation for the circle is\[x^2+(y+2)^2=36\]and we're taking the lower half, so the equation reduces to\[x=\sqrt{32-4y-y^2}\]Now we find the volume.

Answer 19

Since \(x\) is modeling the radius, the volume at some point \(y\) is simply\[\pi x^2\,\Delta y=\pi(32-4y-y^2)\Delta y\;ft.^3.\]Where the \(\Delta y\) is the height. From this, we can find the mass. The mass is simply density*volume, and for water, we know the density to be \(62.4 \;lbs./ft.^3\). So the mass is\[62.4\pi(32-4y-y^2)\Delta y\;lbs.\]
Next is the force term.

Answer 20

But this is also easy, since the force is merely mass*acceleration, and in our case, I think the constant is usually just \(32 \;ft/s^2.\) So the force is\[32\cdot62.4\pi(32-4y-y^2)\Delta y\;lbs\cdot ft/s^2\]

Answer 21

Lastly, we can't forget that we have to move the water up \(-y\) units (since we're assuming everything is below the x-axis). So we have to multiply by \(-y\) as well.

Finally, we integrate from -8 to -2. So if everything works out, it should be that the work is\[-\int_{-8}^{-2}32\cdot62.4\cdot\pi\cdot y\,(32-4y-y^2)\,dy\]

Answer 22

I will test that on my computer and give you what I got

Answer 23

I really hope that's right...

Answer 24

So I just discovered a slight error (it doesn't effect the actual integral though). When I first said the equation reduced to \[x=\sqrt{32-4y-y^2}\]it should have been\[x=-\sqrt{32-4y-y^2}\]instead. But since we immediately squared this to get the volume, the negative sign doesn't matter.

Answer 25

This is odd. From my knowledge, the final answer is in Pi. Ex.: 9pi

Answer 26

There is a \(\pi\) in my integral. Which you can factor out if necessary since it's a constant.

Answer 27

So if you want, you could write it as\[-\,\pi\int_{-8}^{-2}32\cdot62.4\cdot y\,(32-4y-y^2)\,dy\]

Answer 28

I'm on my computer. I'll try that.