Science:Math Exam Resources/Courses/MATH101/April 2018/Question 07 (ii)/Solution 1

From the part (i), the volume of a cylindrical pancake at the height $y$ from $y=0$ is

$$dV=\pi (4-(y+1)^{2})dy.$$

Combining with the constant density of the fluid ${\textstyle 1000\,({{\mbox{kg}}/{\mbox{m}}^{3}})}$, the mass of the pancake is

$$dm=\rho dV=1000\pi (4-(y+1)^{2})dy.$$

Then, by Newton's second law of motion with ${\textstyle g=9.8({\mbox{m}}/{\mbox{s}}^{2})}$, we can raise the pancake by applying the compensating force to the pancake

$$dF=gdm=(9.8)\cdot 1000\pi (4-(y+1)^{2})dy.$$

Finally, to move the fluid in the pancake at height $y$ out of the reservoir, we need to raise $-y({\mbox{m}})$, so that the required work is

$$dW=(-y)dF=(9.8)\cdot 1000\pi (-y)(4-(y+1)^{2})dy.$$

Integrating it in the range $(-3,0)$, the work required to pump all the fluid out of the top of the reservoir is

$${\begin{aligned}W&=\int _{-3}^{0}(9.8)\cdot 1000\pi (-y)(4-(y+1)^{2})dy=9800\pi \int _{-3}^{0}y^{3}+2y^{2}-3ydy\\&=9800\pi \left[{\frac {1}{4}}y^{4}+{\frac {2}{3}}y^{3}-{\frac {3}{2}}y^{2}\right]_{-3}^{0}=-9800\pi \left({\frac {1}{4}}(-3)^{4}+{\frac {2}{3}}(-3)^{3}-{\frac {3}{2}}(-3)^{2}\right)\\&=-9800\pi \left({\frac {81}{4}}-18-{\frac {27}{2}}\right)=9800\pi \cdot {\frac {45}{4}}=110250\pi .\end{aligned}}$$

Answer: $\color {blue}110250\pi$