How Much Work to Pump Liquid from a Hemispherical Tank?

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A hemisphere shaped tank with vertical outlet pipe is full of mysterious liquid. Find the work required to pump the mysterious liquid out of the outlet. The density of the mysterious liquid is 300/98 kg/m^3. [Radius is 2m and vertical pipe is 1m]

W= F*d
=m*a*d
=D*v*a*d
D=300/98 kg/m^3, V=16pi/3 m^3, a=9.8 m/s^2

Im having trouble finding the distance. Can you help me out and correct any of the information I stated above?

 

[HallsofIvy]

Think of the "hemisphere shaped tank" as the bottom of the sphere $x^2+ y^2+ z^2= 4$ or $$z= -\sqrt{4- x^2- y^2}$$. At each z< 0, a "layer" of liquid, of thickness dz, is a circle of radius $$r= \sqrt{x^2+ y^2}= \sqrt{4- z^2}$$. That has area $\pi r^2= \pi(4- z^2)$ and so volume $\pi(4- z^2)dz$. It's mass is
$$m= \frac{300}{98}\pi (4- z^2)dz$$
and so its weight is
[tex]mg= \frac{300}{98}(9.8)\pi (4- z^2)dz= 30\pi(4- z^2)dz[/itex].<br /> (You see the reason for that "mysterious" density!)<br /> <br /> That "layer" has to be lifted to the top of the hemi-sphere, a distance of -z m (z is negative, remember), plus the additional 1 m of pipe, for a total of -z+ 1 m so the work done is<br /> $$mgh= 30\pi (4- z^2)(1-z)dz$$<br /> <br /> To find the total work, integrate that from z= -2 to z= 0.[/tex]