Originally Posted by
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