# Does my work suggest that the buoyancy force isn't generally just the pressure exerted on a body? [closed]

If I got it right, the buoyancy force is a result of pressure changes exerted by the gravitational force. Archimedes’ law then generally predicts the buoyant force to be $$\rho_\text{fluid} Vg$$. What I fail to understand is why when I calculate the total force due to pressure around a sphere of radius $$R$$ for example, I don't get exactly this force, but the force factored by some constant. Following is an example of what I'm talking about for the sake of making myself clear, I'm not concerned about specific examples:

$$F_\text{buoyancy}=\int_{0}^{2\pi}\int_{0}^{\pi}\rho_\text{fluid} gh\,R^{2}\sin\theta d\theta d\varphi$$ when putting $$h = -R(cos\theta-1)$$ for the depth of some infinitesimal surface area, we get:

$$F_\text{buoyancy}=-2\pi g\rho_\text{fluid} gR^{3}\int_{0}^{\pi}(\cos\theta-1)\sin\theta d\theta=4\pi R^{3}\rho_\text{fluid}g$$

In this calculation for example, I'm missing a $$\frac{1}{3}$$ factor in order to get Archimedes law right. Does that mean the buoyancy force isn't generally just the pressure exerted on a body? Or am I missing something? Seems too close to be a coincidence.

• Have you taken into account the force pushes down on the top and up on the bottom? Oct 23, 2020 at 22:00
• The pressure acts normal to the surface of the sphere at all points on the surface, and this has components both in the horizontal and vertical directions. The buoyancy force is the result only of the vertical component. Oct 23, 2020 at 22:10
• @BioPhysicist : this is regulated by the missing $-\cos\theta$, see Puk's answer. Oct 23, 2020 at 22:18
• Related : My answer as "user82794" here Proof of Archimedes Principle. Oct 23, 2020 at 22:33
• Yoiu forgot to integrate over R and indeed the cos$\theta$. Note that this is not a homework or check my work site. Oct 29, 2020 at 12:48

You are missing a $$-\cos\theta$$ factor: this will account for the fact that you are integrating the vertical (upward) components of the force distribution on the sphere. Without this, you are just integrating the magnitude of this force, which isn't very useful.

• Well, it does work out this way. I'm not sure I understand why I should generally only count the upward component though. Aren't all the pressure components contributing to the general buoyant force? What do all the other components physically do? Oct 23, 2020 at 21:45
• The pressure does not have components. Forces do. And they are vectors and add like vectors. The sum of the magnitudes is not the same as magnitude of the sum.
– nasu
Oct 23, 2020 at 21:53
• @Darkenin The buoyant force is always upward. The other components end up canceling out, so you don't need to include them in your calculation. If you do want to show that they cancel out, you need to integrate the normal force vector per unit area, breaking it down to its components, rather than integrating the (scalar) pressure. In this example, symmetry considerations will reveal immediately that horizontal components cancel (this happens in the $\varphi$ integral).
– Puk
Oct 23, 2020 at 21:54
• @Frobenius For the sake of clarity, in this case just multiplying the integrand by $\sin \theta$ gives the magnitude of the horizontal force per unit area, but the horizontal component is still a vector with a direction that depends on the position along the surface (always pointing toward the vertical axis of symmetry of the sphere). So that won't actually integrate to zero. You can further decompose it into Cartesian components, which will contribute $\cos \varphi$ and $\sin \varphi$ factors that will cause the $\varphi$ integral to yield $0$.
– Puk
Oct 23, 2020 at 22:34
• You are absolutely right. I post a new comment. Oct 23, 2020 at 22:38