Water pushes harder the deeper you go. At depth $h$ the pressure is $p = \rho g h$, where $\rho g$ is the weight density ($9800$ N/m³ for water). On a flat plate lying horizontally, every point has the same depth, so the force is just pressure times area. A dam or an aquarium wall stands vertically, so different parts sit at different depths. Like Work, the fix is to slice.
Why it works
Cut the vertical plate into thin horizontal strips. A strip at depth $y$ with width $w(y)$ and height $\Delta y$ is nearly at a single depth, so the pressure on it is about $\rho g y$ and the force is about $\rho g\,y\,w(y)\,\Delta y$. (Pressure in a fluid pushes equally in every direction, so the same formula holds for a wall as for a floor.) Adding the strips:
$$F = \int_a^b \rho g\,y\,w(y)\,dy.$$
Key idea
$$F = \int_a^b \rho g\cdot(\text{depth})\cdot(\text{width})\,dy,$$
with $y$ measured down from the surface and $a$, $b$ the depths of the top and bottom of the plate.
Worked example: A rectangular plate
A plate 3 m wide and 2 m tall is submerged vertically with its top edge 1 m below the surface. Find the force on one side.
- Depths run from $y = 1$ to $y = 3$, and the width is always 3.
- $$\begin{aligned} F &= \int_1^3 9800\cdot y\cdot 3\,dy \\ &= 29400\cdot\frac{3^{2} - 1^{2}}{2} \\ &= 29400\cdot 4. \end{aligned}$$
- $F = 117{,}600$ N.
Answer: $117{,}600$ N
Worked example: A triangular plate
A triangle with a 4 m base and a height of 3 m is submerged vertically, base along the surface and point down. Find the force on one side.
- By similar triangles the width shrinks from 4 at the surface to 0 at depth 3: $w(y) = 4\left(1 - \frac{y}{3}\right)$.
- $$\begin{aligned} F &= 9800\int_0^3 y\cdot 4\left(1 - \frac{y}{3}\right)dy \\ &= 9800\cdot 4\left[\frac{y^{2}}{2} - \frac{y^{3}}{9}\right]_0^3 \\ &= 9800\cdot 4\cdot\frac{3}{2}. \end{aligned}$$
- $F = 58{,}800$ N.
Answer: $58{,}800$ N
Common mistake
Using one pressure for the whole plate, such as the pressure at the bottom: $9800\cdot 3\cdot 6 = 176{,}400$ N for the triangle, three times too big. Another slip is measuring $y$ up from the bottom of the plate and then using $y$ as the depth. Check: the integrand must be the depth (which is 0 at the surface and grows downward) times the width at that same depth.
A shortcut you can check against: the force equals $$\rho g\cdot(\text{centroid depth})\cdot(\text{area}).$$ For the triangle the centroid is 1 m deep and the area is 6 m², giving $9800\cdot 1\cdot 6 = 58{,}800$ N. Centroids are the next topic, Moments and Centers of Mass.