Verify Gauss-Divergence theorem for the following vector field
$${\bf{F}} = 4x{\bf{i}} - 2y{\bf{j}} + z{\bf{k}}$$
Over the region bounded by the surfaces $r = 4$, $z = −2$, and $z = 2$. ${\bf{i}}$, $\bf{j}$, $\bf{k}$ are unit vector directions.
The answer of from the gauss divergence theorem is not matching with the double integral method. Can you check what is the problem, in the image attached. OR give me hint on how to go about the problem
Thanks
Verify Gauss’s Divergence Theorem
I just found the same question but i am not able to understand the ending divergence part , as it does not match with the divergence of the given field. Other things match with my answers..Please guide me to understand that answer Solution,latex beginner
Well, obviously, the first thing you need to know is Gauss's divergence theorem! It says that $\int\int_V\int \nabla\cdot \vec{F} dV= \int_S\int \vec{F}\cdot d\vec{S}$. That is, that is, that the integral of the divergence of a differentiable function of three variables over a region, V, in 3 space is equal to the integral of the function itself over the boundary, S, of the region. Here, $\vec{F}= 4x\vec{i}- 2y\vec{j}+ z\vec{k}$ which is linear in all three variables so its divergence is $\nabla\cdot\vec{F}= 4- 2+ 1= 3$. The integral of that over the region, bounded by r= 4 (I assume that refers to polar coordinates so refers to the circular cylinder with axis the z-axis and radius 4) and the planes z= -2 and z= 2. Since the integrand is a constant, the integral is just that constant, 3, times the volume of the cylinder, $\pi(4^2)(2- (-2))= 64\pi$.
So $\int\int_V\int \nabla\cdot\vec{F}dV= 3(64\pi)= 192\pi$