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(1) Is this proof valid ?

Yes, this proof for the cylindrical body is valid.

But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out).

(2) How can I write a proof with any general solid ? (not just cylinder)

Archimedes' principle for an arbitrarily shaped body can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area $\partial V$ to an integral over the enclosed volume $V$. $$\oint_{\partial V} p(\vec{r})\ d\vec{A} = \int_V \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.

Now, as the position-dependent function we choose the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$).

Then the gradient of (2) is $$\vec{\nabla}p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction.

Inserting (3) into (1) we get $$\oint_{\partial V} p(\vec{r}) d\vec{A} = \int_V (-\rho g\hat{z})\ dV. $$

Now on the left side $p\ d\vec{A}$ obviously is the pressure force acting on the surface area element $d\vec{A}$ (except for a minus sign, because the force element $d\vec{F}$ points inside the body, while the area element $d\vec{A}$ points outside). And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint_{\partial V} d\vec{F}=-\rho g \hat{z} \int_V dV$$ or finally $$\vec{F}=\rho g \hat{z} V$$ which is just Archimedes' principle.

1. Is this proof valid ?

Yes, this proof for the cylindrical body is valid.

But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out).

2. How can I write a proof with any general solid ? (not just cylinder)

Archimedes' principle for an arbitrarily shaped body can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area $\partial V$ to an integral over the enclosed volume $V$. $$\oint_{\partial V} p(\vec{r})\ d\vec{A} = \int_V \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.

Now, as the position-dependent function we choose the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$). We need a minus sign here, because pressure increases when going down in the liquid (i.e. in negative $z$-direction).

Then the gradient of (2) is $$\vec{\nabla}p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction (i.e. upwards).

Inserting (3) into (1) we get $$\oint_{\partial V} p(\vec{r}) d\vec{A} = \int_V (-\rho g\hat{z})\ dV. $$

Now on the left side $p\ d\vec{A}$ obviously is the pressure force acting on the surface area element $d\vec{A}$ (except for a minus sign, because the force element $d\vec{F}$ points inside the body, while the area element $d\vec{A}$ points outside). And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint_{\partial V} d\vec{F}=-\rho g \hat{z} \int_V dV$$ or finally $$\vec{F}=\rho g \hat{z} V.$$ This is just Archimedes' principle (Buoyant force is pointing upwards and equal to the weight of the displaced liquid).

(1) Is this proof valid ?

Yes, this proof for the cylindrical body is valid.

But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out).

(2) How can I write a proof with any general solid ? (not just cylinder)

Archimedes' principle for an arbitrarily shaped body can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area to an integral over the enclosed volume. $$\oint p(\vec{r})\ d\vec{A} = \int \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.

Now, as the position-dependent function we take the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$).

Then the gradient of (2) is $$\nabla p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction.

Inserting (3) into (1) we get $$\oint p(\vec{r}) d\vec{A} = \int (-\rho g\hat{z})\ dV. $$

Now on the left side $p\ d\vec{A}$ obviously is the pressure force acting on the surface area element $d\vec{A}$ (except for a minus sign, because the force element $d\vec{F}$ points inside the body, while the area element $d\vec{A}$ points outside). And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint d\vec{F}=-\rho g \hat{z} \int dV$$ or finally $$\vec{F}=\rho g \hat{z} V$$ which is just Archimedes' principle.

(1) Is this proof valid ?

Yes, this proof for the cylindrical body is valid.

But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out).

(2) How can I write a proof with any general solid ? (not just cylinder)

Archimedes' principle for an arbitrarily shaped body can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area $\partial V$ to an integral over the enclosed volume $V$. $$\oint_{\partial V} p(\vec{r})\ d\vec{A} = \int_V \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.

Now, as the position-dependent function we choose the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$).

Then the gradient of (2) is $$\vec{\nabla}p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction.

Inserting (3) into (1) we get $$\oint_{\partial V} p(\vec{r}) d\vec{A} = \int_V (-\rho g\hat{z})\ dV. $$

Now on the left side $p\ d\vec{A}$ obviously is the pressure force acting on the surface area element $d\vec{A}$ (except for a minus sign, because the force element $d\vec{F}$ points inside the body, while the area element $d\vec{A}$ points outside). And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint_{\partial V} d\vec{F}=-\rho g \hat{z} \int_V dV$$ or finally $$\vec{F}=\rho g \hat{z} V$$ which is just Archimedes' principle.

(1) Is this proof valid ?

Yes, this proof for the cylindrical body is valid.

But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out).

(2) How can I write a proof with any general solid ? (not just cylinder)

Archimedes' principle for an arbitrarily shaped body can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area to an integral over the enclosed volume. $$\oint p(\vec{r})\ d\vec{A} = \int \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.

Now, as the position-dependent function we take the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$).

Then the gradient of (2) is $$\nabla p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction.

Inserting (3) into (1) we get $$\oint p(\vec{r}) d\vec{A} = \int (-\rho g\hat{z})\ dV. $$

Now on the left side $p\ d\vec{A}$ obviously is (except for its sign) the pressure force acting on the surface area element $d\vec{A}$. And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint d\vec{F}=-\rho g \hat{z} \int dV$$ or finally $$\vec{F}=\rho g \hat{z} V$$ which is just Archimedes' principle.

(1) Is this proof valid ?

Yes, this proof for the cylindrical body is valid.

But the author should better say "upward and downward pressure force", instead of "upward and downward pressure", because pressure has no direction (as you correctly pointed out).

(2) How can I write a proof with any general solid ? (not just cylinder)

Archimedes' principle for an arbitrarily shaped body can most easily be proved with Gauss' gradient theorem. This theorem relates an integral over a closed surface area to an integral over the enclosed volume. $$\oint p(\vec{r})\ d\vec{A} = \int \vec{\nabla} p(\vec{r})\ dV \tag{1}$$ where $p(\vec{r})$ is any position-dependent function, and $\vec{\nabla}$ is the gradient operator.

Now, as the position-dependent function we take the pressure $$p(\vec{r})=p_0-\rho gz \tag{2}$$ where $z$ is the vertical position coordinate and $p_0$ is the pressure at zero-level ($z=0$).

Then the gradient of (2) is $$\nabla p(\vec{r})=-\rho g\hat{z} \tag{3}$$ where $\hat{z}$ is the unit-vector in $z$-direction.

Inserting (3) into (1) we get $$\oint p(\vec{r}) d\vec{A} = \int (-\rho g\hat{z})\ dV. $$

Now on the left side $p\ d\vec{A}$ obviously is the pressure force acting on the surface area element $d\vec{A}$ (except for a minus sign, because the force element $d\vec{F}$ points inside the body, while the area element $d\vec{A}$ points outside). And on the right side, the constants $(-\rho g\hat{z})$ can be factored out. So we get $$-\oint d\vec{F}=-\rho g \hat{z} \int dV$$ or finally $$\vec{F}=\rho g \hat{z} V$$ which is just Archimedes' principle.