What is the Last Term in the Expression for \nabla\cdot(\phi\vec{A})?

  • Thread starter Thread starter SAMSAM12
  • Start date Start date
  • Tags Tags
    Curl Grad
SAMSAM12
Messages
2
Reaction score
0

Homework Statement


If \phi= xy^{2}
A=xzi-z^{2}j+xy^{2}k
B=zi+xj+yk

Verify that
\nabla.(\phiA)=A.\nabla\phi+\phi.\nablaA

Homework Equations


The Attempt at a Solution


I have worked out the first two parts of the question:
\phiA = (x^{2}y^{2}z, -xy^{2}z^{2},x^{2}y^{4})
div(\phiA) = 2xy^{2}z-2xyz^{2}

A.grad(\phi) = (xy^{2}z-2xyz^{2})

I'm struggling to work out the last part:
\phi.\nablaA

I tried working out \phi.grad(A)? but the answer sheet has
div(A) = z
\phidiv(A) = xy^{2}z

why?
Any help appreciated.

Merry Christmas and Happy new year
 
Physics news on Phys.org
Ok, so to start compute \nabla A which will just be (\frac{\partial}{\partial x}\vec{A},\frac{\partial}{\partial y}\vec{A},\frac{\partial}{\partial z}\vec{A})
You will end up with a scalar, which you can multiply by your scalar \phi and you should end up with xy^2z.
 
Last edited:
SAMSAM12 said:

Homework Statement


If \phi= xy^{2}
A=xzi-z^{2}j+xy^{2}k
B=zi+xj+yk

Verify that
\nabla.(\phiA)=A.\nabla\phi+\phi.\nablaA
The last term of the expression doesn't make sense. You can't dot a scalar into anything. It should be
$$\nabla\cdot(\phi \vec{A}) = \vec{A}\cdot\nabla \phi + \phi\nabla\cdot\vec{A}$$
 
vela said:
The last term of the expression doesn't make sense. You can't dot a scalar into anything. It should be
$$\nabla\cdot(\phi \vec{A}) = \vec{A}\cdot\nabla \phi + \phi\nabla\cdot\vec{A}$$

Thank you.
 
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...
Back
Top