If I follow the steps illustrated in post 3 but instead solve for d, I obtain -Csin(ak)=Csin(ak), so two possibility is C=0 or sin(ak)=0, so k=n*pi/a for this instance
My reason is as follows - If I plug y = a into (Csin(ky)+Dcos(ky)), I can solve for C. Then if I plug in y = -a and my value for C, I can get -Dcos(ak) = Dcos(ak) (because sin is odd function and cos is even function) I'm not sure if my reasoning is correct so feedback is appreciated. I'm not...
Homework Statement
Hello,
I'm trying to solve laplaces equation to find a solution for the potential in a pipe with the given boundary conditions:
at x=b, V=V_0
at x= -b, V = -V_0
at y=a, V=0
at y=-a, V=0
(Assume this configuration is centered on the origin, pipe as dimensions -b<=x<=b...
No problem man! I initially tried setting the points I found as the limits of integration but then realized that I couldn't do that. After I plotted the points I found I saw that they formed a rhombus. So I found an equation for x in terms of y and set it up as follows...
Why would I use the change of variables theorem to compute an integral in the xy plane? Sorry if this sounded too demanding but I've been staring at this problem for quite some time now.
Homework Statement
Homework Equations
N/A
The Attempt at a Solution
I solved part a. I got an answer of 140. For part b, however, I am stuck. I came up with a set of points for D in the xy plane [(0,3)(0,6)(4,5)(4,8)] giving me a rhombus. How do i integrate this? I tried to split up the...