What Does the Notation in This Vector Field Equation Mean?

JaysFan31
Just a quick question about notation.

I was given the vector field

F = r + grad(1/bar(r)) where r= (x)i+(y)j+(z)k.
grad is just written as the upside down delta (gradient) and the bar I wrote in the above equation looks like an absolute value around just the r (although I don't know if it is absolute value). Basically I want to find the gradient of (1/bar(r)).

What would be a simplification of this vector field so that I can solve the rest of the problem?

I want to find its flux across the surface of a sphere.


I think F would be ((x^3)-1)/x^2+((y^3)-1)/y^2+((z^3-1)/z^2, but I'm not sure.
 
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|\vec{r}| is the modulus, or magnitude, of vector \vec{r}.
 
If \vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k} then
\frac{1}{||\vec{r}||}= \frac{1}{\sqrt{x^2+y^2+ z^2}}= (x^2+y^2+ z^2)^{-\frac{1}{2}}
What is the gradient of that function?
 
How can you take the gradient of the function if it doesn't have i, j, and k?

Is it 0?
 
JaysFan31 said:
How can you take the gradient of the function if it doesn't have i, j, and k?

Is it 0?

A better question would be, how could you take the grad of the function if it *did* have i, j, and k? Remember, the gradient acts on a scalar.
 
JaysFan31 said:
How can you take the gradient of the function if it doesn't have i, j, and k?

Is it 0?
Well, you would first have to know what "gradient" actually means!

Given a function f(x,y,z), how would YOU define
\nabla f?
 
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...
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