Question

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measurement in feet). If the tank is full of a fluid weighing 100 pounds per cubic foot, set up an integral for the work required to pump the contents of the tank to a level 5 feet above the top of the tank.

Answer 1

`100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy` ft-lbs.

Explanation:

Given:

The shape of the tank is obtained by revolving
`y=x^2` about y axis in the interval
`0\leq x\leq 3`.

Density of the fluid in the tank,
`D=100\ lbs/ft^3`

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is,
`y(0)=0^2=0`

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in
`x=3` in the parabolic equation . This gives,

`H=3^2=9\ ft`

So, the height of top of tank is,
`y(3)=H=9\ ft`

Now, 5 ft above 'H' means
`H+5=9+5=14`

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

`\Delta y=(14-y)` ft

Now, area of cross section of the tank is given as:

`A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section`

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have,
`y=x^2`

So,
`x=\sqrt y`

Therefore, radius,
`r=\sqrt y`

Now, area of cross section is,
`A(y)=\pi (\sqrt y)^2`

Work done in pumping the contents to 5 feet above is given as:

`W=D\int\limits^(y(3))_(y(0)) {A(y)(\Delta y)} \, dy`

Plug in all the values. This gives,

`W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}`