Sure, that's what i did in my post i think..
I was trying something different like: ##\int_{1}^{2}\int_{0}^{\sqrt{1-(y-1)^2}}\frac{y}{x^2+y^2}dxdy##
But i think is way more easy with polar coordinates once you find the right values..
Hello there,
I'm struggling in this problem because i think i can't find the right ##\theta## or ##r##
Here's my work:
##\pi/4\leq\theta\leq\pi/2##
and
##0\leq r\leq 2\sin\theta##
So the integral would be: ##\int_{\pi/4}^{\pi/2}\int_{0}^{2\sin\theta}\sin\theta dr d\theta##
Which is equal to...
This is from a past exam of my course (we can't use any table during exam), a friend of mine says that you don't need it and you can find it anyway.. I doubt it.
What i did is: ##R=\rho\frac{l}{\Sigma}=\rho\frac{l}{d_2^2\pi-d_1^2\pi}##
The problem is that I don't have ##\rho##. Is there a way to find ##R## without knowing it?
Many thanks.
Since ##d=q+p \implies q=d-p##
\begin{cases}
\frac{1}{I_1}+\frac{1}{d-I_1}=\frac{1}{f}\\
\frac{1}{I_2}+\frac{1}{d-I_2}=\frac{1}{f}
\end{cases}
Is this correct?
I'm always struggling understand how to determine if a two variable function has global minima, I know that if I find a local minima and the function is convex than the local minima is also a global minima, in this case is really difficult to determine if the function is convex.
Sorry if I...
The thing is..
if i substitute in ##-u\ln u-\lambda(u^2+v^2-2)=0## the results are correct
else if i substitute in ##u\ln u+\lambda(u^2+v^2-2)=0## the results are not correct..
I don't understand what I'm doing wrong
Ok, here's what I got
##u\ln u+\lambda(u^2+v^2-2)=0##
##\begin{cases}
\ln u+1+2\lambda u=0\\
2\lambda v=0\\
u^2+v^2-2=0
\end{cases}##
The solutions are: ##(1/e,\pm\sqrt{2-1/e^2},0)## and ##(\pm\sqrt{2},0,\frac{\ln\pm\sqrt{2}+1}{\pm2\sqrt{2}})##
By substitution I get ##-1/e## and...