Answer
$1.962 \times 10^7 \ N$
Work Step by Step
The cross-sectional-area $A(y)=\dfrac{y^2}{4} \pi$
We need to use the formula for finding the total force on the dam such as: $F=\int_0^a \rho g (a-y) w(y) \ dy$.
Plug in the above formula the given values to obtain:
$W=\int_0^a \rho g (a-y) w(y) \ dy \\= 40 \times \int_0^{10} \rho g (10-y) \ dy\\=40 \rho g (10y-\dfrac{y^2}{2})|_0^{10} \\= (2000) (1000) (9.81) \\=1.962 \times 10^7 \ N$
