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**DBA** Hi, given is a sphere filled with water. The radius is 3m

The sphere has a spout on top and the hight of the spout is 1m.

What I did

I did draw a circle with r=3 in (0,0)

Then I made a horizontal "cut" somewhere above the x-axis.

The cut gives me a disk with A=pi*r^2

__The formula of work we need to use is__

Work = weight * displacement

= Lambda * Volume * displacement

= Lambda *Area * thickness * displacement

__For the area:__

A= pi*x^2 I need to find x for the disk I cut.

The x lies on the circle and the formula of this circel is

x^2+y^2 = 9 because radius of the sphere is 3

then x^2 = 9-y^2

So A = pi * (9-y^2)

__Displacement__

I am not sure here.

My distance from my cut on the y-axis is Y

The radius of the sphare and therefore of the circle I did draw is 3 also.

Then the distance the water of my disk has to move (3-y) upwards.

But I also have the spout of 1m

Is the displacement then (4-y)?

__Thickness__

dy, because the thickness of my disk is an "up-down" value or y-value

__Integral interval__

I am not sure here also, I would say from -3 to 3. Because that is where the water is

Is that right?

__Integral__ I use S for "Integral from -3 to 3"

Work = Lambda * pi S (9-y^2)(4-y) dy

Can someone let me know if I did that right? I do not understand how I get the displacement when I have a spout and what the interval of the integral should be.

Thanks