# Recent content by jaytech

1. ### Physics Problem: Fnet=ma or Fnet= 0?

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...
2. ### Partial differentiation problem, multiple variables (chain rule?)

You should try the steps I previously mentioned. Then reflect on why it works..
3. ### Partial differentiation problem, multiple variables (chain rule?)

Try saying z = x2 + 2\frac{x^2}{cos^2θ}sin2θ = x2(1 + 2tan2θ) Now with this for z you can perform \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'.
4. ### Evaluate the surface integral

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.
5. ### Evaluate the surface integral

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.
6. ### Evaluate the surface integral

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.
7. ### Evaluate the surface integral

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

You're off by a factor of 2. You're integral for Phi should be 0< Phi < pi.
9. ### Evaluate the surface integral

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.
10. ### Evaluate the surface integral

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...
11. ### General solution to PDE

I think he just made a typo.
12. ### General solution to PDE

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.
13. ### General solution to PDE

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...
14. ### General solution to PDE

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...
15. ### Integration by parts with orthogonality relation

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.