Answer to Question #344330 in Calculus for Jane

Question #344330

5. Suppose that a water tank is shaped like a right circular cone with the tip at the bottom, and has height 10 meters and radius 2 meters at the top. If the tank is full, how much work is required to pump all the water out over the top?

1
Expert's answer
2022-06-01T14:34:51-0400


Let's consider the circular cross-section through the tank at depth h and find its radius using similar triangles:

"\\frac{2}{r}=\\frac{10}{10-h},"

"r=\\frac{2(10-h)}{10}=\\frac{10-h}{5}."

The aea of the cross-section: "A=\\pi r^2=\\frac{\\pi (10-h)^2}{25}."

The volume of the section of water at depth h:

"V=\\pi r^2dh=\\frac{\\pi (10-h)^2}{25}dh."


The density of the water: "\\sigma=1000 kg\/m^3," so the mass of water in the section is

"m=\\sigma V=1000\\frac{\\pi (10-h)^2}{25}dh=40\\pi (10-h)^2dh."

To raise this mass of water we need to overcome the force of gravity "F=mg" , "g=9.8 m\/s^2."

"F=9.8\\cdot 40\\pi (10-h)^2dh=392\\pi(10-h)^2dh."

We must lift this water a distance h and require a work

"W=392\\pi(10-h)^2hdh."

The total work:

"W=\\int_0^{10} 392\\pi(10-h)^2hdh=392\\pi \\int_0^{10} (10-h)^2hdh="

"=392\\pi \\int_0^{10} (100h-20h^2+h^3)dh=392\\pi (100\\frac{h\n^2}{2}-20\\frac{h^3}{3}+\\frac{h^4}{4})|_0^{10}="

"=392\\pi (5000 - \\frac{20,000}{3}+\\frac{10,000}{4})=392\\pi\\frac{2500}{3}=\\frac{980,000}{3}\\pi\\approx1,026 kJ."


Answer: 1,026 kJ.


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