How do you find the divergence of a vector field?

andrey21
Messages
475
Reaction score
0
I am just curious how you find the divergence of the following vector field





Heres my example

u = xz^(2)i +y(x^(2)-1)j+zx^(2) y^(3)k



Am I right in thinking

U take the derivative with respect to x for first term derivative with respect to y for second term...

giving me z^(2) + (x^(2) -1) +x^(2)y^(3)
 
Physics news on Phys.org
You have it correct.
 
Thanks LCKurtz just another question I have bin posed.

does div(uxv)

Mean find the divergence of the dot product of vectors u and v.
 
andrey21 said:
Thanks LCKurtz just another question I have bin posed.

does div(uxv)

Mean find the divergence of the dot product of vectors u and v.

No. That wouldn't make any sense because a dot product gives a scalar and divergence applies to vector fields. What it does mean is first take the cross product of a and b, which gives a vector, then take its divergence.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
4
Views
2K
Replies
8
Views
2K
Replies
9
Views
2K
Replies
6
Views
2K
Replies
5
Views
2K
Replies
2
Views
1K
Replies
7
Views
2K
Back
Top