Determining hydrostatic force on a sphere

[thebrennus]

Homework Statement

A solid sphere with radius $R$ is submerged on water at a depth $h_1$. Water's specific weight is $\gamma$.
Determine both horizontal and vertical forces exerted by the water on the surface of the sphere, in terms of $R$ and $\gamma$.

Homework Equations

My lecturer wants the solution to be achieved by integrating the differential pressure of the sphere, so I believe the only relevant equations are $F = p A$ and $p = (p_0 + \gamma h)$. I am also assuming $p_0 = 0$

The Attempt at a Solution

I started by decomposing the resultant force on the surface of the sphere into its vertical and horizontal components, allowing me to write: $F_V = F \sin\theta$ and $F_H = F \cos\theta$

3.1. The vertical force
So I divide the sphere into differential rings with area $dA = 2\pi R^2 \cos\theta dθ$
If $F_V = F \sin\theta$, then $dF_V = dF \sin\theta$, but $dF = p dA$ and $p = (p_0 + \gamma h)$, which yields:
$$dF_V = (p_0 + \gamma h) (2\pi R^2 \cos\theta dθ) \sin\theta$$
The point where I'm evaluating $h$ is $h = h_1 + R - R\sin\theta$, so we have:
$$dF_V = \gamma (h_1 + R - R\sin\theta) (2\pi R^2 \cos\theta dθ) \sin\theta$$
Finally, the vertical force is given by:
$$F_V = \gamma 2\pi R^2 \left[h_1 \int_{\frac \pi 2}^{\frac {3\pi} 2} \sin\theta \cos\theta\, d\theta + R \int_{\frac \pi 2}^{\frac {3\pi} 2} \sin\theta \cos\theta\, d\theta - R \int_{\frac \pi 2}^{\frac {3\pi} 2} \sin^2\theta \cos\theta\, d\theta\right]$$
Solving the integral we have:
$$F_V = \gamma 2\pi R^2 * (2 R /3) \rightarrow F_V = \frac 4 3 \pi R^3 \gamma$$
Which I believe is the correct answer since it is the buoyancy of a sphere.

3.2 The horizontal force
As far as I know, the method for solving this should be the same as before. The only difference I can see is that the horizontal force is given by $F_H = F \cos\theta$, so the equation is:
$$F_H = \gamma 2\pi R^2 \left[h_1 \int_{\frac \pi 2}^{\frac {3\pi} 2} \cos^2\theta\, d\theta + R \int_{\frac \pi 2}^{\frac {3\pi} 2} \cos^2\theta\, d\theta - R \int_{\frac \pi 2}^{\frac {3\pi} 2} \sin\theta \cos^2\theta\, d\theta\right]$$
Solving the above integral yields:
$$F_V = \gamma 2\pi R^2 * (h_1\frac \pi 2 + R\frac \pi 2) \rightarrow F_V = \pi^2 R^2 \gamma[h_1 + R]$$
However I am pretty sure that this is wrong. Since the whole sphere is submerged there should be forces cancelling themselves out so that the horizontal force should be just zero. Besides, this solution is in terms of $h_1$, and my lecturer explicitly said the answer shouldn't be in therms of any variables other than $R$ and $h_1$

 

[Charles Link]

I suggest you redo your coordinate system and use a standard spherical with $\theta$ being the polar angle measured from the vertical (upward z) direction and $dA=2 \pi R^2sin(\theta) \, d \theta$ because of the $\phi$ symmetry. The angle $\theta$ will then vary from $0$ to $\pi$. The vertical force is the z-component. For the horizontal force, you will need to consider the $x$ and $y$ components and integrate over $\phi$. The horizontal component will have a $sin(\theta)$ just as the vertical has a $cos(\theta)$, but it is the $\phi$ integrals that will give you the zero answer for the horizontal (x and y) components. For the vertical (z component), the integral over $d \phi$ is just $2 \pi$. For the x and y components, the $sin(\phi)$ or $cos(\phi)$ terms integrate to $0$ when integrated from $0$ to $2 \pi$.

 

[thebrennus]

I suggest you redo your coordinate system and use a standard spherical with $\theta$ being the polar angle measured from the vertical (upward z) direction and $dA=2 \pi R^2sin(\theta) \, d \theta$ because of the $\phi$ symmetry. The angle $\theta$ will then vary from $0$ to $\pi$. The vertical force is the z-component. For the horizontal force, you will need to consider the $x$ and $y$ components and integrate over $\phi$. The horizontal component will have a $sin(\theta)$ just as the vertical has a $cos(\theta)$, but it is the $\phi$ integrals that will give you the zero answer for the horizontal (x and y) components. For the vertical (z component), the integral over $d \phi$ is just $2 \pi$. For the x and y components, the $sin(\phi)$ or $cos(\phi)$ terms integrate to $0$ when integrated from $0$ to $2 \pi$.

Thank you so much for your answer, it does make sense. I didn't really think about the coordinate system before