# Gauss Divergence Theorem

#### swinburneguy2009

Verify the Gauss Divergence Theorem:​
∫∫∫∇ ⋅F dxdydz= ∫∫ F .nˆ dA , where

S​

the closed surface
S is the sphere x2 + y2 + z2 = 9 and the vector field

F = xz2i + x2 y j + y2 z k .

#### HallsofIvy

MHF Helper
If you are taking a course in which you are expected to be able to do these problems you should know something about 3 dimensional Calculus. What do you know and what have you tried on this problem?

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#### swinburneguy2009

I am sorry ...clearly i do not know too much about the subject. I am almost 40 years old, and have taken on mech. engineering part time via correspondance while raising a family and working full time. It is very hard, and I am having great difficulty getting a grasp on this. I have work due tomorrow, which is a "must pass" for me. due to being via correspondance, i have no contact with lecturers or tutors.
I plan to take the next 2 weeks off work annual leave to study for the exam. unfortunately, I have this work due tomorrow and am stressed out. call me unprepared and say what you will, but i am in real need of help right now! I have done all but 3 of the questions needed for my assignment. It is very hard to do this work when the last maths i did was over 20 years ago...

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#### AllanCuz

Verify the Gauss Divergence Theorem:​
∫∫∫∇ ⋅F dxdydz= ∫∫ F .nˆ dA , where

S​

the closed surface
S is the sphere x2 + y2 + z2 = 9 and the vector field

F = xz2i + x2 y j + y2 z k .

I'm confused here. Are we verifying the divergence theorem in general (i.e. prove it is always true) or prove that it is true in this case?

I'll assume we're doing the latter.

$$\displaystyle \iiint_S \nabla \cdot F dV$$

What does the above represent? Well...It represents the entire flux out of our surface. Can we compute this? Of course!

$$\displaystyle \nabla \cdot F = divF$$

Remember that $$\displaystyle divF = \frac{ \partial F1 }{\partial x } + \frac{ \partial F2 }{\partial y } + \frac{ \partial F3 }{\partial z }$$

Assuming you mean

$$\displaystyle F = xz^2 \hat i + x^2 y \hat j + y^2 z \hat k$$

We arrive at

$$\displaystyle \iiint_S \nabla \cdot F dV = \iiint_S (z^2 + x^2 + y^2) dV$$

Of course our surface is defined as $$\displaystyle x^2 + y^2 + z^2 = 3^3$$ which is a sphere! So we have 2 options here, either use cylindrical co-ordinates or spherical. I personally like to use cylindrical, so let's go with that and see if that's easy enough to compute.

$$\displaystyle \iiint_S (z^2 + x^2 + y^2) dV$$

$$\displaystyle \iiint_S r(z^2 + r^2) d \theta dr dz = \int_0^{2 \pi } d \theta \int_0^3 dr \int_0^{ \sqrt{ 3^2 - x^2 - y^2 } } r(z^2 + r^2) dz$$

Compute the above and we will then compare to

$$\displaystyle \iint_D F \cdot \hat N dA$$

Let's remember that the above represents the flux out of the surface. So draw our shape (which is a sphere) and we will note that it is smooth everywhere!!!!

What does that mean? Well...it makes the above integral easy because we only have 1 surface to compute for total flux.

Knowing,

$$\displaystyle \hat N = +- ( - \frac { \partial D}{\partial x } \hat i - \frac { \partial D}{\partial y } \hat j + \hat k )$$

We can easily compute the above. Do so and compare 