First, we’re going to assume that the top of the circular plate is 6 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate. Setting the axis system up in this way will greatly simplify our work.

Finally, we will again split up the plate into \(n\) horizontal strips each of width \(\Delta y\) and we’ll choose a point \(y_i^*\) from each strip. We’ll also assume that the strips are rectangular again to help with the computations. Here is a sketch of the setup.

The depth below the water surface of each strip is,

\[{d_i} = 8 - y_i^*\]

and that in turn gives us the pressure on the strip,

\[{P_i} = \rho g{d_i} = 9810\left( {8 - y_i^*} \right)\]

The area of each strip is,

\[{A_i} = 2\sqrt {4 - {{\left( {y_i^*} \right)}^2}} \,\,\Delta y\]

The hydrostatic force on each strip is,

\[{F_i} = {P_i}{A_i} = 9810\left( {8 - y_i^*} \right)\left( 2 \right)\sqrt {4 - {{\left( {y_i^*} \right)}^2}} \,\,\Delta y\]

The total force on the plate is,

\[\begin{align*}F & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {19620\left( {8 - y_i^*} \right)\sqrt {4 - {{\left( {y_i^*} \right)}^2}} \,\,\Delta y} \\ & = 19620\int_{{\, - 2}}^{{\,2}}{{\left( {8 - y} \right)\sqrt {4 - {y^2}} \,dy}}\end{align*}\]

To do this integral we’ll need to split it up into two integrals.

\[F = 19620\int_{{\, - 2}}^{{\,2}}{{8\sqrt {4 - {y^2}} \,dy}} - 19620\int_{{\, - 2}}^{{\,2}}{{y\sqrt {4 - {y^2}} \,dy}}\]

The first integral requires the trig substitution \(y = 2\sin \theta \) and the second integral needs the substitution \(v = 4 - {y^2}\). After using these substitutions we get,

\[\begin{align*}F & = 627840\int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{{{\cos }^2}\theta \,d\theta }} + 9810\int_{{\,0}}^{{\,0}}{{\sqrt v \,dv}}\\ & = 313920\int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{1 + \cos \left( {2\theta } \right)\,d\theta }} + 0\\ & = 313920\left. {\left( {\theta + \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}}\\ & = 313920\pi \end{align*}\]

Note that after the substitution we know the second integral will be zero because the upper and lower limit is the same.