Question

A cylindrical shaped gas tank is 12 m tall with base radius to be 3 m. There is a spout on top of the tank with height 0.3 m. Suppose the tank is one-third full. Set up the integral for the work required to pump the gasoline out of the spout. Do NOT compute the integral. Suppose the gasoline density is p= 749 kg/m", and you may use the approximation g 10 m/s2 for gravity. (Requirements: You must show your work to get full credit. In particular, you need to introduce a coordinate axis and show how to set up the Riemann sum approximation, then convert the work to be an integral.)

Answer 1

VTTTTT 12m we know, the woook done is giren by, W = Force x displacement. 1.10.3m Here force is the weight of gasoline. Let us see the given data, Height of tank = 12m Radius of the tank = 3m Height of spout = 0.3m Gasoline density = P = 749kg/m3 Now, we take a small section of height sh in the cylinder, Volume of this section = TTx3 xsh mo ("volume of cylinder is - T8th o Now, weight of this section = P x volume x 3m of Force forth = Px3²nd hxg

Now, it is given that the cylinder is one third full. If we take the base to be at om, gasolene level is at (1273) m = 4m. Total height of cylinder and spout = (12+0. 3) m. = 12, 3 m om the figure, to pump out of the cylinder the dise has to rise up a distance = (12.3-4)m to slice. are up a distance = (12.3-hom for each Now, we set up an integral for work done, Woook = Foree x distance from 4 I I set the as it is limit o d filled] S Px35x g * (12.3-h) dh [We set - 749x3 rx 10xS" (12.8-fi) da = 221770x5" (12.3-6 du So, the integral for the woook required is - W = 221770 x 8(12.3-1)dh. So