As per Figure 5.8.4(a), measure y positive upwards from the bottom of the tank.  At any value of y the width of a slab of oil of thickness dy is $2 \sqrt{2 r y-y^2}$.  Therefore the volume of that slab of oil is $2 L \sqrt{2 r y -y^2} \mathrm{dy}\.$  Since the density of the oil is $\mathrm{\ρ}$, the weight of the slab is $\mathrm{\ρ} g 2 L \sqrt{2 r y -y^2} \mathrm{dy}\.$  This slab must move a distance of $h-y\=2 r\+2-y$, so the work done moving this slab is $2 \mathrm{\ρ} g L\sqrt{2 r y-y^2}\left(2 r\+2-y\right)\mathrm{dy}$.  Hence, the total work done in emptying the tank is$$

$2 \mathrm{\ρ} g L {\int }_0^{2 r}\left(2 r\+2-y\right)\sqrt{2 r y-y^2} \mathit{\ⅆ}y\= \mathrm{\ρ} g L r^2 \mathrm{\π} \left(2\+r\right)$ 

(For some integrals, Maple interprets "simplify" to mean "evaluate" because the evaluated form is simpler than the unevaluated form. Because the Context Panel provides for assumptions only with the Simplify option, the integral has been so evaluated in order to apply the assumption that $\mathrm{\ρ}\,g\,L$, and $r$ are all positive.)$$