Determine the net pressure force \( \mathbf{F}_p \) on the surface \( A \) bounded by \( C \). The unit normal \( \mathbf{n} \) to the surface \( \eta \) is \( \mathbf{n} = \frac{\nabla \eta}{|\nabla \eta|} = \frac{(-x/R_1, -y/R_2, 1)}{\sqrt{(x/R_1)^2 + (y/R_2)^2 + 1}} \)
and the area element is \( dA = \sqrt{1 + (\partial \eta/\partial x)^2 + (\partial \eta/\partial y)^2} \, dx \, dy = \sqrt{1 + (x/R_1)^2 + (y/R_2)^2} \, dx \, dy \)
\(\therefore \mathbf{F}_p = - \iint_A \Delta p \, \mathbf{n} \, dA = -\Delta p \int_{-\sqrt{2R_1 \zeta}}^{+\sqrt{2R_1 \zeta}} \left[ \int_{-\sqrt{2R_2 \zeta - x^2 R_2/R_1}}^{+\sqrt{2R_2 \zeta - x^2 R_2/R_1}} \left( \frac{-x/R_1, -y/R_2, 1}{1} \right) dy \right] dx\)
The minus sign appears here because greater pressure above the surface (positive \( \Delta p \)) must lead to a downward force and the vertical component of \( \mathbf{n} \) is positive. The x- and y-components of \( \mathbf{F}_p \) are zero because of the symmetry of the situation (odd integrand with even limits)

The remaining double integration for the z-component of \( \mathbf{F}_p \) produces \((\mathbf{F}_p)_z = \mathbf{e}_z \cdot \mathbf{F}_p = -\pi \Delta p \sqrt{2 R_1 \zeta} \sqrt{2 R_2 \zeta}\)
\(\because\) The net surface tension force \( \mathbf{F}_{st} \) on bounding curve \( C \) can be determined from the integral \( \mathbf{F}_{st} = \sigma \oint_C \mathbf{t} \times \mathbf{n} \, ds \), \( ds = dx \sqrt{1 + (dy/dx)^2} \) is an arc length element of \( C \), and \( \mathbf{t} \) is the unit tangent to \( C \)
\(\therefore\) \(\mathbf{t} = \frac{-(1, dy/dx, 0)}{\sqrt{1+(dy/dx)^2}} = \frac{(-y/R_2, x/R_1, 0)}{\sqrt{(y/R_2)^2 + (x/R_1)^2}}\)
\( dy/dx \) is found by differentiating the equation for \( C \), \( \zeta = \left( x^2/2R_1 \right) - \left( y^2/2R_2 \right) \), with \( \zeta \) regarded as constant
\(\because\) \( \mathbf{t} \times \mathbf{n} \) \(\therefore\) the surface tension force acts perpendicular to \( \mathbf{t} \) and tangent to the curved interface
\(\therefore\) Integrand in \(\boxed{ \mathbf{F}_{st} = \sigma \oint_C \mathbf{t} \times \mathbf{n} \, ds} \) is \( \mathbf{t} \times \mathbf{n} \, ds = \frac{(R_2/y) dx}{\sqrt{(x/R_1)^2 + (y/R_2)^2}} \left( \frac{x}{R_1}, \frac{y}{R_2}, \frac{x^2}{R_1^2} + \frac{y^2}{R_2^2} \right) \cong \frac{R_2}{y} \left( \frac{x}{R_1}, \frac{y}{R_2}, \frac{x^2}{R_1^2} + \frac{y^2}{R_2^2} \right) dx \), the approximate equality holds when \( x/R_1 \) and \( y/R_2 \ll 1 \) and the area enclosed by \( C \) approaches zero
\(\because\) Symmetry of the integration path will cause the \( x \)- and \( y \)-components of \( \mathbf{F}_{st} \) to be zero
\(\therefore\) \((\mathbf{F}_{st})_z = \mathbf{e}_z \cdot \mathbf{F}_{st} = 4\sigma \int_0^{\sqrt{2R_1 \zeta}} \frac{R_2}{\sqrt{2R_2 \zeta - (R_2/R_1) x^2}} \left[ 2\zeta + \frac{x^2}{R_1} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \right] dx\) where \( y \) has been eliminated from the integrand using the equation for \( C \), and the factor of four appears because the integral shown only covers one-quarter of the path defined by \( C \)