Consider a square plate rotated 45 degrees and submerged in water (diagonal of square is perpendicular to surface of water). To find force on one side of square plate, I assume an $x$-$y$ axis passing through the center of the square, with the $y$-axis perpendicular to the water surface.
At a height of $y$ from the center, I assume a rectangular strip of breadth $dy$. The length of this strip is $2(\frac{a}{2^{0.5}}-y)$. Let $\delta$ be the weight density of water. Therefore, the force on one side of the rectangle will be integral of $f(y)$ as $y$ goes from $-a/2^{0.5}$ to $a/2^{0.5}$, where
$$f(y)=2*\delta*(a/2^{0.5} - y)^{2}.$$
The final answer comes out to be: $a^3*\delta*\frac{2^{5/2}}{3}$.
I am told that this answer is wrong. Where am I going wrong in this one?