Recent content by Archimedess

This is not right. The lower limit for ##r## is not ##0##. Should it be 1? Even if i use 1 the solution is not right

Nope.. ##1 \le y \le 2## is correct

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..

Yes, sorry edited my post The problem is now how do i write that half semicircle

Yes, i notice that ##1\leq y \leq 2## and ##0\leq x \leq ?## the half semicircle

I don't know another way to solve it.. i mean is not required but is the only way i know this time

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...

Ok, I got it thank you so much everyone

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...

##1/e##... yeah, change of variables definitely helps.. thanks