Question

Physics and calculus?

A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (In other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 0.5 meters, its length is 7 meters, and its top is 1 meter under the ground, find the total amount of work needed to pump the gasoline out of the tank. (The density of gasoline is 673 kilograms per cubic meter; use g=9.81m/s^2
work=

i work so many but can't get correct ans
i don't know whether im correct or not that i first calculate dp then find out dw, finally find out W, but don't know why i can't get the ans
so please help me to check out this

Answer 1

Consider that the tank consists of horizontal layers of infinitesimal thickness `dz`. For each layer, situated at depth `=z` below the top of cylinder, force required to overcome the gravity of a layer is

`dF = m g = ρA_ z gdz`
where `A_z` is area of the gasoline layer at specified depth `z`
`=>dF= ρg (2 sqrt(((0.5)^2-(0.5-z)^2) )xx7) dz`
`=>dF= 14ρg sqrt((z-z^2) ) dz`

Work to lift this layer to the ground level is

`dW = dvecF · vecd = 14ρg sqrt((z-z^2) ) dz· (1+ z )`

Total work needed to lift all the gasoline to the ground level is found by integrating both sides with respective variables

`W = int_0^1\ 14ρg sqrt((z-z^2) ) (1 +z ) dz`

Using online integral calculator we get value of integral as

`W=14rhog((3pi)/16)`

Inserting given values we get

`W=14xx673xx9.81((3pi)/16)`
`W=54445.7\ J`

Answer 2

A simpler approach without calculus. Thanks to @ultrilliam

Explanation:

The center of mass of a cylinder lying horizontally on its side would lie on the axis of the cylinder at the center of length `l`.

Depth of center of mass from ground level `=(r+1)\ m`

Now, work done to pump the gasoline out of the tank is equal to the gain in potential energy by gasoline on lifting it from center of mass to the ground level.

`W = Delta U = m g Delta h`
`=>W= (rho V) g Delta h`
`=>W= (rho (pi r^2 l)) \ g \ ( r + 1)`

Inserting given values we get

`W= 673xx pi xx(0.5)^2 xx7xx9.81 ( 1 + 0.5)`
`W=54445.7\ J`