Hm, I was able to work through the problem, correcting for the period, and it looks like I'm now only off by a factor of 2. A friend of mine is also having the same problem. Not sure if we can chalk it up textbook error, or just something we're not seeing...
Ah nevermind, feeling rather silly...
Yep, I just got that. Makes sense now. Thank you so much!
Although, can I ask, how would one know (without knowing the answer) that the period is in fact 2l and not l, like I wrongly assumed it was, just from reading the problem/looking at the image. Ah man, feeling rather stupid.
http://en.wikipedia.org/wiki/Fourier_series
Everything I've looked up suggests the 2 should be there. I'm also pretty sure I've been using the 2 in all my fourier coefficients calculations up until now. :( I thought the two arises from the fact that ω=2pi/T.
I'm sorry. I guess I'll run...
I'm almost certain the 2pi factor belongs there. At least it certainly does in the equations for the coefficients of a fourier series, where the 2pi emerges from ω. Unless for some reason those equations are modified in this case? I know the book tends to leave off the 2pi, but that's when...
So actually, my An includes the prefactor of 2h/\pi^{2}. There's on the other hand, is independent of the constants out front. That's why I was thinking that maybe the two answers are equivalent, in that they factored those pre-factors out, but then the fact that our sin terms are different is...
I used y= 4hx/l for 0<x<l/4
2h-4hx/l for l/4<x<l/2
0 for l/2<x<l
So this basically gave me two equations for An, where the first is equal to
An = 2/l (from 0 to l/4)∫(4hx/l)*sin(2\pinx/l)dx
and the second, for l/4<x<l/2
An = 2/l ∫(2h-4hx/l)*sin(2\pinx/l)dx
Do those look ok...
Homework Statement
A string of length l has a zero initial velocity and a displacement y_{0}(x) as shown. (This initial displacement might be caused by stopping the string at the center and plucking half of it). Find the displacement as a function of x and t.
See the following link for...
Hm, ok, I'm not sure I follow here. If I computed the dot product of two cross products, isn't that wrong, in that I should first compute the cross product of (BxC) x (CxA), before taking the dot product that I did in the step before?
Thanks for the response!
Ok, well, I know that (AxB)_{i} = ε_{ijk}A_{j}B_{k}. And I can say that, (BxC)_{i} = ε_{ilm}B_{l}C_{m}.
If I write that all as one term,
ε_{ijk}A_{j}B_{k}ε_{ilm}B_{l}C_{m}
then that equals,
(δ_{jl}δ_{km} - δ_{jm}δ_{kl})A_{j}B_{k}B_{l}C_{m}
and I know...
Homework Statement
Hi all,
Here's the problem:
Prove, in tensor notation, that the triple scalar product of (A x B), (B x C), and (C x A), is equal to the square of the triple scalar product of A, B, and C.
Homework Equations
The Attempt at a Solution
I started by looking at the triple...
Homework Statement
Find the center of mass of a solid of density \delta = 1 enclosed by the spherical coordinate surface \rho = 1-cos\phi.
Homework Equations
The Attempt at a Solution
I'm a bit confused about how to start here, mainly because the surface is defined by spherical...
Ah ok. How's this?
$$\int_{0}^{\pi/2}\int_{0}^{1}\int_{0}^{r} r(6+4rsinθ) dzdrdθ$$
Thanks again for all the help by the way. Definitely starting to feel a bit better about these conversions.
Yeah, you're absolutely right. I realized that right after I submitted that last post.
Here are my new integrals.
Cartesian Coordinates:
$$\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{0}^{\sqrt{x^{2}+y^{2}}} 6+4y dzdxdy$$
Cylindrical Coordinates:
I'm not sure about this one, particularly...
Hm, I am thoroughly confused now, haha. I thought the region was the inside of the cone?
What I'm picturing: If I drew a cylinder and placed a cone directly inside it, so that the top of the cone meets the top of the cylinder, the region I'm looking at is inside the cone.
Should your theta limits be from 0 to $$\theta/2$$ since it's only the first octant?
So this is what my integral in cartesian coordinates looks like:
$$\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{1} 6+4y dzdxdy$$
I guess I don't quite understand ##\rho##, in that case. I...
Hehe fair enough. I get 2pi^2.
Looks like a nice and neat enough answer -- hopefully that's what you got!
I went through the problem a second time, though, and now I'm finding myself confused about one of the trig substitutions I made. Once I've expanded (1-cos(2ϕ)/2)^2, I'm left with a few...
Homework Statement
Set up triple integrals for the integral of f(x,y,z)=6+4y over the region in the first octant that is bounded by the cone z=(x^2+y^2), the cylinder x^2+y^2=1 and the coordinate planes in rectangular, cylindrical, and spherical coordinates.
Homework Equations...
Ok, so when I expand (1-cos(2\phi)/2)^2, I get 1/4 - (cos(2\phi))/2 + (cos^2(2\phi))/4
Replacing the last term with a half-angle formula, I have
1/4 - (cos(2\phi))/2 +1/8 + (cos(2\phi))/8
At that point, I take the integral of that with respect to \phi.
I wind up with 1/4*\phi -...
Ok, so as far as my limits of integration go, those are ok?
I also get (1/3)\rho^3, and evaluating at 2sin\phi I get 8/3∫∫sin^4(\phi) d\phid\theta. I'm not sure where I ought to be getting 16 from?
As for ∫∫sin^4(\phi) d\phid\theta, I think I'm simply screwing up some basic calculus. I...
Homework Statement
Find the volume enclosed by the spherical coordinate surface ρ = 2sin∅
Homework Equations
dV = ∫∫∫(ρ^2)sin∅dρd∅dθ
The Attempt at a Solution
(Sorry about my notation!)
Alright, here's what I've done so far...
Since the region is a torus, centered...
Homework Statement
Find the volume of the region bounded by the cylinder x^2 + y^2 =4 and the planes z=0, and x+z=3.
Homework Equations
V = ∫∫∫dzdxdy
V=∫∫∫rdrdθ
The Attempt at a Solution
Alright, so I feel as though I'm missing a step somewhere along the way, but here's what I've gotten...
Proof -- Express in Clyndrical Coordinates
Homework Statement
Show that when you express ds^2 = dx^2 + dy^2 +dz^2 in cylindrical coordinates, you get ds^2 = dr^2 + r^2d^2 + dz^2.
Homework Equations
x=rcosθ
y=rsinθ
z=z
The Attempt at a Solution
EDIT// I was really over thinking...
A, B, and P are co-linear, right? And I'm saying that based on the fact that AP and BP are anti-parallel vectors.
And since they all lie along the same line, I wouldn't need three points, and so a parametrization using AP is just as valid as one using BP, even though they appear different?
I'm...
Homework Statement
Find a parametrization of the equation of the line formed by the points A, B, and P.
A(2,-1,3) B(4,3,1) P(3,1,2)
Homework Equations
x=x_0+v_1*t
y=y_0+v_2*t
z=z_0+v_3*t
The Attempt at a Solution
Alright, so, I've already determined that P is equidistant from the points...
Thanks, I think I got it. I just get totally thrown off when the index isn't in the correct form, or when the series is blatantly obvious, haha.
Thank you, though!