Emptying water out of a Conical Tank? (Calculus)? Please help me with this Calculus question. I'm not asking you to do the whole thing, but I just need help setting up the height function. Here is the question:
A conical tank of radius $6$ feet and height of $15$ feet is $\frac{2}{3}$(of the height) filled with sea water. The tank is located on top of a water tower. The vertex of the cone is 100 ft above ground level (cone is inverted). Find the work required to pump all the water to a point 18ft above the ground. Seawater weights $64$$\frac{lbs}{ft^3}$.
I set up my graph with the cone's vertex at the origin, but I'm not sure if that is right. If so, would $h(x)=(-82+y)$?
I know that $A(x)=\pi(x)^2$ or more specifially $A(x)=(4\pi/15)*y$
When I finally set up my problem it looked like this:
Work$=64*(4pi/15)*(\int_0^{10} (-82y+y^2)dy)$
Which gives me an answer of $-121173.323$ ft/lbs of work, but I don't think that's right.
How do I set up the height function properly?
 A: You have a couple of errors: as you suspected, the height isn’t quite right, but the cross-sectional area is also wrong.
The radius of the tank at height $y$ is $\frac25y$, so the cross-sectional area is $\frac{4\pi}{25}y^2$. To raise the cross-sectional slice of thickness $dy$ at height $y$ to the top of the tank, you must raise it $15-y$ feet, doing $$dW=64\cdot\frac{4\pi}{25}y^2(15-y)dy=\frac{256\pi}{25}(15y^2-y^3)dy$$ foot-pounds of work. Thus, it takes
$$\frac{256\pi}{25}\int_0^{10}(15y^2-y^3)dy$$
foot-pounds of work to raise all of the water to the top of the tank.
There are $64\cdot\frac13\cdot16\pi\cdot10=\frac{10240\pi}3$ pounds of water in the tank; they all fall a distance of $115-18=97$ feet after being raised to the top of the tank, doing $97\cdot\frac{10240\pi}3=\frac{993,280\pi}3$ foot-pounds of work in the process. Thus, the net work required is
$$\frac{256\pi}{25}\int_0^{10}(15y^2-y^3)dy-\frac{993,280\pi}3\text{ foot-pounds}\;.$$
You can of course set it up as a single integral: the slice at height $y$ is first raised $15-y$ feet and then dropped $97$ feet, so it is ‘raised’ a total of $15-y-97=-y-82$; your error was getting the wrong algebraic sign on $y$. If you do this, the answer is
$$\frac{256\pi}{25}\int_0^{10}(-y^3-82y^2)dy$$
foot-pounds.
