Recent content by jaytech

My suggestion to you is to remember that according to Newton's Laws of Force, a body in motion stays in motion and an object at rest stays at rest, unless acted upon by an external force. F = ma, and F = 0 are the mathematical representations of this statement. F = 0 implies the system's...

You should try the steps I previously mentioned. Then reflect on why it works..

Try saying z = x2 + 2\frac{x^2}{cos^2θ}sin2θ = x2(1 + 2tan2θ) Now with this for z you can perform [SIZE="4"]\left(\frac{\partial z}{\partial \theta}\right)_{x} quite easily. Hint* Remember, that after you perform the derivation to look for anywhere you can make a substitution to remove 'x'.

Now I would like some clarification...because I don't see how the projection on the surface could be zero when there is no symmetry argument for the point at the top of the hemispherical bowl.

My cylinder example was derived from a line charge along an axis. I hadn't read the problem until he attempted the integration. A disk is still applicable however, as to close the surface on the plane intersecting the normal would be the same.

It seems to me he did account for the surface? I picked a disk and told him to use the divergence theorem because I assumed it an easier route to take.

My fault, I didn't read the problem. You're right, as that is the volume of a single hemisphere.

You're off by a factor of 2. You're integral for Phi should be 0< Phi < pi.

No, that's right. With the disk, the volume would be that of a cylinder. So you're divergence = 1 implies "a" is growing linearly in a single direction. So the RHS of the equation would be something like (using the disk analogy), z*2pi*R, which is the volume of a cylinder. Both sides hold.

Picturing it in 2D with a disk may be the best way to learn this, then apply the concepts to 3D. Imagine a flat disk in the XY plane. dS would be a vector along the z-axis whose magnitude is equal to a small area of the disk. If integrating this dS, to just S (the surface), one would have a...

I think he just made a typo.

I found a mistake here, \frac{∂f}{∂y} = \frac{∂f}{∂u}\frac{∂u}{∂y}+\frac{∂f}{∂v}\frac{∂v}{∂y}, not what it says in the quote.

Since you factored out a \frac{∂}{∂x}, this implies the right hand side is constant with respect to a change in x. If this were an ODE, it's implied the value is constant, but because you have a function of two variables, it is implied to be a function that is constant with respect to a change...

First look at the equation: \frac{\partial^2 f}{\partial x^2}+4\frac{\partial^2 f}{\partial x \partial y}+\frac{\partial f}{\partial x}=0 We can factor a \frac{∂}{∂x} out and write it as such: \frac{∂}{∂x}(\frac{∂f}{∂x}+4\frac{∂f}{∂y}+f(x,y)) = 0. Now what this implies is...

You're answer is wrong because \int_{0}^{a} sin\frac{n\pi x}{a}sin\frac{m\pi x}{a}dx = \begin{cases} \frac{x}{2} &\mbox{if } n = m; \\ 0 & \mbox{otherwise.} \end{cases} The \frac{a}{2} is after evaluation of the integral from 0 to a. So your \intvdu is actually \int\frac{x}{2}dx.