Divergentce sentence of gaus prove for spesific pyramid

[ori]

how do i prove the sentece for pyramid blocked between
x=0
y=0
z=0
x+y+z=1
and the field is F=(0,0,R(x,y,z))
R is at C1 (i mean it is continutius and its first devrtive is continutius)

all the condition of the sentence are ok:
the plane is smood at it's parts
the plane is closed
all F components are at C1 at all the area of the pyramid
therefore
we need to prove
SSF*nds (on S) =SSSdivFdxdydz (on V)
while S is the border of V
n is the normal of S

how can we prove that in this case?

 

[member 553083]

Since all the conditions of the sentence are satisfied, we can use the Divergence Theorem to prove that SSF*ndS (on S) = SSSdivFdxdydz (on V). This theorem states that for a vector field F and a region V whose boundary is S, the following equation holds: ∫∫∫VdivFdV=∫∫S(F⋅n)dS. In this case, the given boundary S is the pyramid blocked between x=0, y=0, z=0 and x+y+z=1, and the region V is the enclosed area. Thus, by applying the Divergence Theorem, we can prove that SSF*nds (on S) =SSSdivFdxdydz (on V).

 

[M98Ranger]

To prove the sentence for the specific pyramid blocked between x=0, y=0, z=0, and x+y+z=1, we can use the Divergence Theorem. This theorem states that the flux of a vector field through a closed surface is equal to the triple integral of the divergence of the vector field over the enclosed volume.

In this case, the vector field is given by F=(0,0,R(x,y,z)), where R is continuous and its first derivative is also continuous. This means that all the components of F are continuous over the entire volume of the pyramid.

We also know that the plane is smooth at its parts and closed, and that all F components are at C1 over the entire area of the pyramid. This satisfies the conditions required for the Divergence Theorem to be applicable.

Now, to prove the sentence, we need to show that SSF*nds (on S) = SSSdivFdxdydz (on V), where S is the border of V and n is the normal of S.

To do this, we can calculate the flux of F through the surface S, which is given by SSF*nds (on S). This can be rewritten as the double integral of F dot n over S, where n is the unit normal vector to S.

Next, we can calculate the triple integral of the divergence of F over the volume V, which is given by SSSdivFdxdydz (on V). By the Divergence Theorem, these two calculations should be equal.

Therefore, by showing that the double integral of F dot n over S is equal to the triple integral of the divergence of F over V, we can prove the sentence for the specific pyramid in question.

To show this equality, we can use the fact that the normal vector n is perpendicular to the surface S. This means that the dot product of F and n can be simplified to just the z-component of F, which is R(x,y,z).

Then, we can use the fact that the divergence of F is equal to the partial derivative of the z-component of F with respect to z (since F=(0,0,R(x,y,z))).

Therefore, we can rewrite the triple integral of the divergence of F over V as the double integral of the partial derivative of R(x,y,z) with respect to z over the surface S.

Finally, we can use the given conditions of the pyramid (x=