Question

A sphere tank is full of water. Find the work required to pump the water out of the spout. The sphere's radius is 3 meters and the spout is 1 meter tall.

Answer 1

i think what there asking for is to find the volume which in this case is the work required to pump the water

Answer 2

I don't know how much the water weighs, so I can't determine the work. I am unsure of how to make the integral for work since there is a 1 meter tube at the top of the sphere. The sphere's volume is (4/3)pi(27) which simpiflifed is 36pi. The volume of the spout is pi. So How would I set up the integral for the work required to pump the water?

Answer 3

Use the fact that the density of water is 10^3 kg/m^3, use the volum,e of the sphere, then you can calculate the work required because you will have a mass, can then use gravitiy to find the force, and then you can calculate work.

Answer 4

Thank you!!!

Answer 5

What would the integral values be for "a" and "b"

Answer 6

How far does the water have to travel to get completely out of the tank?

Answer 7

it has to go from the bottom of the sphere which has a radius of 3 meters and a 1 meter tube. Would it be 0 to 1?

Answer 8

If it has a radius of three, how far does the strip of water at the bottom of the sphere have to travel?

Answer 9

Oh so it would be from 0 to 6 for the sphere and plus one more for the tube, so 0-7?

Answer 10

There ya go.

Answer 11

Thanks so much!!!

Answer 12

yw

Answer 13

Oh last thing just to make sure I have everything set up correctly, Integral 0-7 of (36pi+pi)*9.81*1000?

Answer 14

I am going to check for you, hang on.

Answer 15

Alright thank you!!!

Answer 16

ok, not getting the same thing. What did you use as your radius and height for the strip of water?

Answer 17

I used the radius of the sphere as 3 and the cylinder is 1 meter so wouldnt it still be 36pi for the cylinder and pi for the tube?

Answer 18

The radius of the sphere is constant, but the radius of each strip of water being pumped out varies depending on the depth it is.

Answer 19

Okay, so the less water there is the smaller each strip of water will be and to account for this do I need to divide the sphere into sections and integrate them?

Answer 20

You need to derive a formula for the radius of the strip with respect to the position it is in the sphere. Recall that the height also varies, as each strip is not being lifted the same distance.

Answer 21

I mean displacement, not height.

Answer 22

I get 9810pi(36-y^2)(6-y)dy as my work integral.

Answer 23

I see where you get the 9810 from 9.81x1000 and did you get the 36-y^2 because the diameter of the sphere squared is 36 and y^2 is the water's change squared? Since it's 6-y is the change of the water's height you multiply that by the 36-y^2?

Answer 24

correct

Answer 25

sorry, was called away by my wife

Answer 26

that is quite alright, I appreciate your help! I was completely stumped on this problem!

Answer 27

oops shouldn't be 36 as it's dealing with radius squared not diameter squared. I was thinkning r=6 not r=3

Answer 28

so 9

Answer 29

but the 6-y is correct because the bottom of the sphere is at a depth of 6 feet so the work required to move that bottom strip would be the entire 6 feet. What I am trying to determine is if you go from 0 to 7, or from -3 to 4. If we center the sphere at the origin, the water goes from a depth of -3 to 3+1 to get out of the spigot

Answer 30

I guess you could try both and see if they yield identical solutions.

Answer 31

@collegestudent15 I think if we replace 6-y with 7-y and integrate from -3 to 3 you should be able to get the work dersired.

Answer 32

desired*

Answer 33

Good luck. Let me know if that works!

Answer 34

Okay thank you and sorry about the delay i was helping someone on here with trig sub integrals! :D! I will try it right now and let you know in a few!! Thanks again

Answer 35

Oh quick question, do I still need the 9-y^2?

Answer 36

Yes, because the radius of each strop of water changes with the depth y.

Answer 37

strip*

Answer 38

Okay, that makes so much more sense now!!! and yes it worked thank you so much!!!!!!

Answer 39

You are welcome. I was stumped for awhile there myself. Had to recall everything.

Answer 40

Thank you so so much!!! I've been stuck on this problem forever and now it makes complete sense!