Recent content by danni7070

Ok. The teacher told me to use Stoke's Theorem which I did but the answer was 0 and there is an example in the book with simular results which says that the theorem "fails" in that particular example. benorin: I'm not sure what div(curlF) = 0 tells me in this example other than I just know...

the b) is still unsure because what I get from the curl part is -ze^z(2+z) and that confuses me. There is an example in the book but as almost always, they are pretty "easy" :P

Ahh, very nice! thanks. Now I get it. \int\int\int_D 3dV = 3\cdot volume of cone = 3\cdot \frac{1}{3} \cdot \pi \cdot 1^2 \cdot 1 = \pi

Thanks, this is solved. The teacher did not set this problem correctly up and it made me confused. In fact this was easy :P But here is another problem! Let S be 0 <= z = 1- \sqrt{x^2+y^2} and F : R^3\rightarrow R^3, (x,y,z) \rightarrow (x,z^2e^z+y,z) be vector field. a)...

Homework Statement Let D be an area in R^3 and S be its surface. D fulfills the Divergence theorem. Let N be the unit normal on S and let the volume, V, be known. Let (\overline{x},\overline{y}, \overline{z}) coordinates of the centre of mass of D be known (and the density delta is...

\frac{n^7}{(n+1)^7} behaves like \frac{n^7}{n^7} for large n so in your example it's just 1 if the limit goes to infinity. Can you see it?

\frac{n}{n+1}*\frac{2^n}{2^{n+1}}*\frac{(x+2)^{n+1}}{(x+2)^n} I get this if I just skip the (-1)^n part cos you just know it is 1, -1, 1, -1 for different values of n which tells you that the sum is alternating.

If I have (a_n + b_n)^n = c_n where a_n is convergent and b_n divergent. Is c_n then divergent? And what if a_n and b_n were divergent, would c_n be divergent also? but what if they were both convergent then surely c_n is convergent right? I can't see a rule or a theorem that tells me...

sorry I meant e^t * e^-t = 1 because e^t * 1/e^t = 1...

(e^t+e^{-t})^2 = e^{2t}+2+e^{-2t} Not 100% sure but is e^t+e^-t = 1 ?

\sqrt{e^{2t}+2+e^{-2t}} = e^t+e^{-t} I just can't see it.

Yeah, thanks dynamicsolo that's exactly what I was doing. I always forget what's the denominator and numerator cause I've learned it my own language and it's like there isn't any room for more names :) I'm on a pause in mathematics. The test is over and it was ok. But for now I have to...

I was just wondering if this was the right way to solve this limit problem. \lim_{x\rightarrow\infty} (\sqrt{x+1} - \sqrt{x})^\frac{1}{ln(x)} Multiply both sides... (\frac{1}{\sqrt{x+1}+\sqrt{x}})^\frac{1}{ln(x)} = 0^0 Wich is undefined. Any suggestions?

Ok thanks a bunch guys. The rest is simple \lim_{x\rightarrow\pi/2} \frac{sin(x)+x*cos(x)}{-sin(x)} = \frac{\pi/2}{-\pi/2} = -1 Edit: It is not pi/2 there it is supposed to be 1/-1

Ok I think I got it! The above equals to \frac{x*sin(x)-\pi/2}{cos(x)} : [\frac{0}{0}]