Recent content by mcfc

If instead of \oint cos^2 \theta d\theta we had \oint {1 \over 1 +cos^2 \theta } d\theta how would that be solved?

I'm unsure how to do this problem: (a + 2b)\nabla(\nabla \cdot \vec u) - b \nabla \times \nabla \times \vec u - (3a + 2b)c\nabla T(r)= \vec 0 \hat u = U_r \hat r + u_\theta \hat \theta +u_z \hat z a,b,c constants how would I solve this for u?

Thanks. I've been confused because I've seen two forms of the polar laplacian: 1)\frac {\partial}{\partial r}( {1 \over r}\frac{\partial T r}{\partial r}) and 2) {1 \over r} \frac{\partial}{\partial r}(r \frac{\partial T}{\partial r}) how are these equivalent?

\nabla ^2 T = \frac {\partial}{\partial r}( {1 \over r}\frac{\partial rT}{\partial r})=0 is the polar form used, which implies {1 \over r}\frac{\partial rT}{\partial r}= A and integrate to get my (incorrect) result above... What am I missing!?

My result is : T= \frac{1}{2} Ar + \frac{B}{ r} A, B constants. Also in this example, I'm not sure how to apply the boundary conditions... But the result I saw(without proof) involved logarithms...?

Hi, Yes, sorry for the confusion

I need to solve: \nabla ^2 T = 0 with T=T(r) and r=a/T=T1 and r=b/T=T2 Can anyone offer advice as to the solution?

sorry...I don't follow(again)

HI That does makes sense, but I can't see how to define an explicit equivalence relation...?

If I have a subset, how do I define an equivalence relation. I understand it has to satisfy three properties:transitive, symmetric and reflexive, but I'm not sure how to give an explicit definition of the equivalence relation, for example on I where I=\{(x,y) : 0 \le x\le 1 \ \& \ 0 \le y \le 1\}

Consider the heat equation in a radially symmetric sphere of radius unity: u_t = u_{rr}+{2 \over r}u_r \ for \ (r,t) \in (0,1) x (0,\infty) with boundary conditions \lim_{r \rightarrow 0}u(r,t) < \infty ; \ u(1,t)=0\ for \ t >0 Now, using separation of variables u=R(r)T(t) leads to the...

I have an integral \int \int_S x^2 + yz \ dS and wish to transform to spherical polar coordinates. How does dS become dS = r^2 \sin \theta d\theta d\phi ?? Where surface S is x^2 + y^2 + z^2 = 1

Thanks, Q1-how would I evaluate: G(y) = 1-F(2-y)\ \mbox{to give the cdf? Do I just substitute 2-y into }{1 \over 4} x^2 \mbox{ to give } G(Y) = 1 - {1 \over 4} (2-y)^2 = 4y - {1 \over 4}y^2\ for \ 0 \leq y \leq 2 \mbox{for the cdf of Y? and so for the pdf: } 4-{1 \over 2}y Q2-Why are...

To show it's odd: look at values in the intervals? -\sin(-\pi) - \sin({\pi }) = 0 \sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0 -\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0 do I need to show anything else?

Hello I'm not too sure if this is the correct location for my post, but it's the best fit I can see! The cdf of the continuous random variable X is F(x)=\left\{\begin{array}{cc}0&\mbox{ if }x< 0\\ {1\over 4} x^2 & \mbox{ if } 0 \leq x \leq 2\\ 1 &\mbox{ if } x >2\end{array}\right...