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'.
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.
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...
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.
It's because your understanding of the Heaviside Step Function is lacking definition. The fact that you have \inte-8s(\frac{5}{s^{2}+36})ds is the same as saying you have a square wave with the function F(s) convoluted with it. When you integrate this, it's like integrating a Dirac-Delta...
For part 2, my personal preference is to substitute for everything in the cos() term. So for me I'd just say 3x+2 = s and then you have 3dx = ds and the ex becomes e(s-2)/3. Then when I have a solution as a function of s I plug 3x+2 back in. I just think that doing the by parts with multiple...
So you have a 50% chance of X being less than or equal to 500, and a 2.27% chance of it being greater than 650. That means there's a 47.73% chance for 500 < X <= 650. Now, knowing this you should be able to divide your integral into three separate integrals, as you now know the PDF values for...
So the general approach is to use the Cauchy Integral Formula and Cauchy Integral Theorem. The theorem has a wiki page, as does the formula: http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula.
You are basically integrating about the boundary of a disk, where the values contained in the...