Recent content by BerkMath

You apply integration by parts to the RHS, twice. To begin let I=RHS. Then apply integration by parts twice. After the second integration by parts you will end up with another integral, after dividing out the necessary constants, you can make this integral look exactly like I. Stop. Then solve...

That is not the answer. Proceeding from your last step, integration of both sides yields: y*e^t= -e^t (2cos(2t)-sin(2t))+C. Where the right hand side, RHS, was found using integration by parts. Next, dividing both sides by e^t, gives: y= -(2cos(2t)-sin(2t))+Ce^(-t) or y=sin(2t)-2cos(2t)+Ce^(-t)...

You generally use D'Alembert's solution when you are given the 1-D wave equation with intial conditions (t=0), and use separation of variables when you are given both initial condtions and boundary conditions. For the latter method you simply suppose the the solution u(x,t)=X(x)Y(t) and then...

I found I a cool link on a GR brachistrone problem. It becomes an interesting problem in that there are then 2 distinct frames from which to minimize the travel time.

It was a surprise to me that your substitution at the end of the determinant ameliorated any confusion whether to use J(u,v)/J(x,y) or J(x,y)/J(u,v). Sorry! Your integral is good and your limits of integration in terms of u and v are good. You must have made a mistake in calculation 'cuz I get...

Do you mean proving that [0,1]/C is dense?

Just looking at your Jacobian: You calculated J(u,v)/J(x,y) correctly, but unfortunately you want/need J(x,y)/J(u,v). You have int(int(f(x,y))dA. The change of coordiantes given by x=g_1(u,v) and y=g_2(u,v) is a mapping from the domain, U<R2, of f(x,y), to a new domain, V<R2. But you were given...

Why did you change variables? There exist a nice symmetry in the Region R which would make the integral easily solved after breaking the region into two.

You are mistaking the general problem of finding the constant of integration, with the much simpler one you have presented. In general, the constant of integration is not known unless there exist boundary conditions. When integrating m=dy/dx, the answer given is the most general class, or set of...

You must evaluate int(int(int(z)))dV where V is the region enclosed by the positive cone in the xyz plane. In general, if V is defined by: a<x<b, h_1(x)<y<h_2(x), g_1(x,y)<z<g_2(x,y) then int(int(int(f(x,y,z)))dV=int(int(int(f(x,y,z),z,g_1(x,y),g_2(x,y)),y,h_1(x),h_2(x)),x,a,b). In your case...

What is R? What is the power of y in the original integrand?

Assuming your functional is correct, why don't you proceed with calculating the Euler-Lagrange differential equation. If you do it in polar coordiantes with the correct functional, the ELDE can be integrated once giving the equation for a hypocycloid in polar coordiantes. It can then be written...

Carrying out the method I described earlier, you get (p(p-1)^2)/2.

My guess is that Tavi will think things through before typing incorrect answers, next time.

Best of luck to you Graveneworld! Let us know how it went, and if you can bring us some of the questions. My Algebra teacher at Berkeley, Bjorn Poonen, took first place in that competition 4 years in a row! SO INSANE!