Work — Question 9

Question 9 – Solution

1. Locate the slices and outlet.

Let $y$ be height above the bottom. Oil occupies $0\le y\le5$, and the spout is at height $5+2=7$ m. The lift distance is $7-y$.

2. Find the slice volume and weight.

The tank radius is $2$ m, so$$dV=\pi(2)^2dy=4\pi\,dy.$$$$dF=w(y)\,dV=9000(1+0.1y)(4\pi)\,dy=36000\pi(1+0.1y)\,dy.$$

3. Set up the work integral.

Multiply each slice weight by its lifting distance:$$W=36000\pi\int_0^5(1+0.1y)(7-y)\,dy.$$

4. Expand and integrate exactly.

$$(1+0.1y)(7-y)=7-\frac3{10}y-\frac1{10}y^2.$$$$W=36000\pi\left[7y-\frac3{20}y^2-\frac1{30}y^3\right]_0^5.$$

5. Apply the bounds and simplify.

$$\begin{align*} W&=36000\pi\left(35-\frac{15}{4}-\frac{25}{6}\right)\\&=36000\pi\left(\frac{420-45-50}{12}\right)=\boxed{975000\pi\ \mathrm{J}}. \end{align*}$$