Calculating Divergence Using the Divergence Theorem

Click For Summary
The discussion focuses on calculating the integral of the divergence of a vector field using the divergence theorem over a specified region. The vector field is defined as F = (x^2 + y^2 + x^2)(x, y, z), and the region is a sphere with radius 5. The user calculates the surface integral and arrives at a result of 125 times the surface area of the sphere, leading to an answer of 1000π. However, the back of the book states the answer is 100π, prompting the user to question the correctness of the book's solution. The user also mentions that integrating the divergence directly yields a different result, reinforcing their belief that the book's answer is incorrect.
fluxions
Messages
50
Reaction score
0

Homework Statement


the problem is to calculate
\int (\nabla \cdot \vec{F}) d\tau
over the region
x^2 + y^2 + x^2 \leq 25
where
\vec{F} = (x^2 + y^2 + x^2)(x\hat{i} +y\hat{j} + z\hat{k})
in the simplest manner possible.

Homework Equations


divergence theorem!

The Attempt at a Solution


Write
\vec{F} = |\vec{r}|^2 \vec{r} = |\vec{r}|^3 \hat{r},
so
\vec{F} \cdot \hat{n} = \vec{F} \cdot \hat{r} = |\vec{r}|^3 \hat{r} \cdot \hat{r} = |\vec{r}|^3 = 125,
since
\hat{n} = \hat{r}
and
|\vec{r}| = 5
along the surface of the sphere.
Then, invoking the divergence theorem, we obtain:
\int (\nabla \cdot \vec{F}) d\tau = \oint_{\partial{\tau}} \vec{F} \cdot \hat{n} d\sigma = \oint_{\partial{\tau}} 125 d\sigma = 125 \cdot 4 \cdot \pi \cdot 5^2

the back of the book gives 100pi as the answer (and I've checked the errata for the book; no correction has been made). am i wrong? or is the book?
 
Last edited:
Physics news on Phys.org
It's also not very hard to integrate the divergence over the interior of the sphere. I get 4*pi*5^5. It sure looks to me like the book answer is wrong.
 
Dick said:
It's also not very hard to integrate the divergence over the interior of the sphere. I get 4*pi*5^5. It sure looks to me like the book answer is wrong.

Great, thanks much!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K