I have a tank of oil with a density of $900\frac{kg}{m^3}$. My tank has a spout that is $2$ meters tall and the general radius of the tank is $6$ meters. It is half full of oil and I want to find the work required to pump all that oil out. In this case we use $9.8\frac{m}{s^2}$ for the force of gravity.
My initial approach to this problem was to find the work required to pump oil out of the sphere and then add that work to the work required to pump the oil out of the spout. But I realized I could not determine the radius of the spout to do so.
So so far I said that to find the $ x = r \ $ I must take $c^2 = 6^2 + 6^2$ which equals $c = 6\sqrt{2}$ then find x by saying: $$ 72 = x^2+y^2 \\ x^2 = 72 - y^2 \\ x = \sqrt{72-y^2} \\ x = 6\sqrt{2} - y $$
So I have my change in radius, $x$. Now I find the volume of the oil, which is $8820\pi(72-12\sqrt{2}+y^2)\Delta y$. I have my force and my distance I say is $(8-y)$ which is $6 + 2$ if I say the center of the sphere is the point of origin.
I then integrate this by $W = \int F*d$. So: $$ W = \int_{-6}^0 8820\pi(72-12\sqrt{2}+y^2)(8-y)\Delta y \\ W = 8820\pi\int_{-6}^0 (72-12\sqrt{2}+y^2)(8-y) \ dy \\ W = 8820\pi\int_{-6}^0 (576-72y-96\sqrt{2}+12\sqrt{2}y+8y^2-y^3) \ dy $$
I integrated it and plugged in values but kept getting the wrong answer. So my logic has been to push the oil from its floor to its half point. And also my distance in the work formula has been from the top of the spout to the surface of the oil which is $y$. What am I doing wrong?
The radius of the spout doesn't matter at all. You just need to lift all the oil to the height of the top of the spout. Let $z$ be the vertical axis, positive upward, with $0$ at the center of the tank. The work required is then $$W=\int_V 9.8\rho(8-z)dV$$because a small element of volume $dV$ needs to be lifted $8-z$ against gravity. Now you can use the disk method for a solid of revolution to do the integral.
Using the disk method, at a particular $z$ we have a circle of radius $\sqrt{36-z^2}$ which needs to be lifted $8-z$ to get out of the tank. The work is then $$\int_{-6}^0 9.8\rho (8-z) \pi(36-z^2)dz$$