Ellipsoidal Shell Gravity

Let the position vector of a point on the surface be:

$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

The position vector of the point of interest (at which force needs to be computed) is:

$\vec{r}_p = 1 \hat{i} + 0 \hat{j} + 0 \hat{k}$

It is given that the mass per unit area of this ellipsoid is unity. Let us consider a small area element which is a part of the surface. The surface area of this is computed as follows:

$z = \sqrt{1 - \frac{x^2}{4} - y^2}$

or,

$z = f(x,y)$

Which implies: $z - f(x,y) = \phi = 0$ . The gradient of this expression is:

$\nabla(\phi)= - f_x \hat{i} - f_y \hat{j} + 1 \hat{k}$

Where $f_x$ and $f_y$ are partial derivatives of $f(x,y)$ with respect to $x$ and $y$ respectively. Therefore, the mass of this element is:

$dM = \sigma \mid\nabla(\phi)\mid dydx$

The gravitational force can be computed by using the inverse square law:

$d\vec{F} = dF_x \hat{i} + dF_y \hat{j} + dF_z \hat{k}$

or:

$d\vec{F} = \frac{G(dM)m}{\mid\vec{r} - \vec{r}_p\mid^2}\bigg(\frac{\vec{r} - \vec{r}_p}{\mid\vec{r} - \vec{r}_p\mid}\bigg)$

Integrating this over the elliptical region $\frac{x^2}{4} + y^2 = 1$ , multiplying the answer by 2 (since only the top half of the surface is considered here), and taking the magnitude of the resultant, gives us the answer of 1.085 . This is different from 1.077. I got lucky. Posting this not so detailed solution to know if I made a mistake.

Since the ellipsoid is a solid of revolution, the gravitational force at the point $(1,0,0)$ acts along the $x$ -axis, and has magnitude $\int_0^\pi \frac{2\cos t - 1}{\sqrt{(2\cos t - 1)^2 + \sin^2t}} \times \frac{Gm\sigma 2\pi \sin t \sqrt{4\sin^2t + \cos^2t}\,dt}{(2\cos t - 1)^2 + \cos^2t} \; = \; Gm\sigma \int_0^\pi \frac{2\pi \sin t \sqrt{1 + 3\sin^2t}}{(2 -4\cos t + 3\cos^2t)^{\frac32}}\,dt$ which equals $1.085033MG\sigma$ after numerical integration. This makes the answer $\boxed{1.085033}$ .