# Calculus Work Problem - Inverted Cone - See body

Question: A tank in the shape of a right circular cone is 4 feet tall and 2 feet in diameter at the top, and is half full of water. How much work is done in pumping the water to the top of the tank?

I believe I solved this problem right but my teacher's answer is different from mine. This was my integral:

$$work = 62.4\cdot32.2\int_2^4{x\pi\cdot}{(4-x)^2\over16}\cdot d{\bf r}=2630.141 ftlb$$

Where did I go wrong? I used similar triangles to figure out the v(x) and I believe d(x) is simply x since you are pumping the water up. Any help appreciated. Also, what if the conical tank was completely full of oil weighing 50 lbs/ft^3 and the oil has to be pumped to a spot 2 feet above the top of the cone? How much work is done and how would you do that?

• I assume that $ftlb$ mean foot $\times$ pound (not evident for 90% of people on Earth using metric system) ? Nov 13, 2019 at 18:48

The issue is that you assumed an INVERTED cone, which I can tell because you used $$(4-x)^2$$ instead of $$\displaystyle \frac{x^2}{16}$$.
I'm not sure what you mean by pumping to a spot two feet above the top of the cone, but it would be multiplying the integral (perhaps $$\displaystyle \int_4^6$$cone) by the density of oil $$(50)$$.