Sarahisme

yep that helps a lot thanks! i was told this aswell, that [math]dS=\left| \frac{\partial \vec{r}}{\partial u}\times\frac{\partial \vec{r}}{\partial v}\right|\,du\,dv[/math] so we want to evaluate this integral... and that gives [math] \frac{\pi}{4} [/math]

how do i work out what dS is? that is i get the same normal as you except for the du^dv parts. what does the ^ stand for? or does du^dv = [math]\frac{du}{dv}[/math] ? how does this look for the integral we need to evaluate? = i used the unit normal here, instead of the normal that you used and so i get a different answer...(as you can see above) is it possible to do this without wedge product stuff? i.e. just by using cross and dot products

i thought it was a cylinder... i guess not, lol this is what we want to do yeah? [math] \int \int_{S} \nabla \times \vec{F} \dot \vec{\hat{n}} dS [/math] and you can work out that [math] \nabla \times \vec{F} =[/math] i + j + k is this ok so far? but then i don't see how to get the n part, or the dS part.... :S or well i don't know how to use the n part once i get it....

i don't understand wedge stuff. but i got the unit normal field as (sin(v), -cos(v), u) by taking the cross product of the paramaterization of r. hmm but once we have this unit normal field, how do use it in the formula? that is how can we integrate with u's and v's in there?

Hello all, I am completely new to this stokes theorem bussiness..what i have got so far is the nabla x F part, but i am unsure of how to find N (the unit normal field i think its called). any suggestions people? i get that nabla x F or curl(F) = i + j + k cheers -sarah

yeah i can see it could prove quite useful in the near future for me, i will keep it in mind. Thanks for all the help matt & severian!

[math] I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{\frac{1}{cos(\theta)}} f®drd\theta + \int_{\frac{\pi}{4}}^{\frac{-\pi}{4}} \int_{\frac{-1}{cos(\theta)}}^{0} fdrd\theta [/math]

i sort of understand the summation notation, but i had never seen it before you showed me so its still a little difficult to grasp. Thank you very much for showing me it though.

lol, ok i see. i like your method of thinking... its quite unique

what do you think of this as an answer for part (b)...? [math]\nabla \times (f®\vec{v}) = \frac{f'®}{\sqrt{x^2+y^2+z^2}} [(yc - zb)i - (xc - za)j + (xb - ya)k] [/math]

lol, you've lost me...

ok hold on. i don't understand how you did this step... [math] \nabla \sqrt{\vec{a} \cdot \vec{a}} = \frac{1}{|\vec{a}|} \vec{a} \cdot \nabla \vec{a} [/math] also, once there i can't quite see where to carry on too..... i think i am missing a key thing about this question (although i don't know what it is!, lol ) p.s. my reasoning behind the answer to (a) being 0 is this.... if [math] |\vec{v} \times \vec{r}|[/math] is a scalar quantity (just a number), so [math] \nabla |\vec{v} \times \vec{r}|[/math] is [math] {\mathbf i} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+ {\mathbf j} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}+{\mathbf k} \frac{\partial |\vec{v} \times \vec{r}|}{\partial x}[/math] then isnt the derivative of a scalar 0?

ok things are getting confusing.... this is what i have done for part (b) now: when put into polar coordinates: [math] \int \int f(\sqrt{x^2+y^2})dxdy = \int \int f®rdrd\theta [/math] And then using the condtions that [math] |y| \leq |x| [/math] So then [math] |rsin\theta| \leq |rcos\theta| [/math] [math] \Rightarrow |tan\theta| \leq 1 [/math] So [math] -\pi/4 \leq \theta \leq \pi/4 [/math] So there are the limits for theta. Then using the other condition that [math] |x| \leq 1 [/math] [math] |rcos\theta| \leq 1 [/math] [math] \frac{-1}{cos\theta} \leq r \leq \frac{1}{cos\theta} [/math] So then the integral becomes… [math] I = \int^{\pi/4}_{-pi/4} \int^{1/cos\theta}_{-1/cos\theta} f®rdrd\theta [/math]

hmm thats kind of what i was thinking but then isnt the answer for part (a) just 0? also i am unsurewhat it means when it says "f® is a function of r = |r|"?

Hello, I am confused by what this question is asking, and i was wondering if anyone could suggest the best way to approach it? Thanks sarah

right yep, i should have seen that. ok so how does this look....?? (a) [math] I = \int_0^{2\pi}\int_{0}^1f®rdrd\theta [/math] (b) [math] I = 2\int_{-\pi/4}^{\pi/4}\int_0^{\frac{1}{cos\theta}} f®rdrd \theta [/math] © [math] I = \int_{0}^{2\pi}\int_{0}^{cos\theta}f(tan\theta)rdrd\theta [/math]

i thought i had it going from 0 to another value...? which part of the question are you referring too btw?

Hi all, i am little puzzled by these questions, not sure if i am doing it correctly or just blundering along! what do you guys think? cheers Sarah for (a) i get: [math] I = \int_0^{2\pi}\int_{-1}^1f®rdrd\theta [/math] for (b) i get: [math] I = \int_{-\pi/4}^{\pi/4}\int_{\frac{-1}{cos\theta}}^{\frac{1}{cos\theta}}f®rdrd\theta [/math] for © i get: [math] I = \int_{0}^{2\pi}\int_{0}^{cos\theta}f(tan\theta)rdrd\theta [/math] if people want the working just shout, i just don't have quite enough time to put it up at the moment. -sarah

hmmm ok, so here is what i think is going on.... i think it is a 'trick' kind of question. where the linear manner part just trying to throw you off. the first thing it states is that the larvae have a constant mortality rate which implies that the survival of the larvae is not density dependent. hmm....

and so , do you understand it to be density dependent or independent?

i would love to be, but thats the question from the book sorry ;p

Hi all, Experimental observations in the lab revealed that the larvae of an insect species experiences a constant rate of mortality regardless of its age. That is qx is constant. However, the time from egg hatching to pupation increase in a linear manner with the density at which the larvae are raised. In this example is larval survival (number pupating/number hatched) density dependent? i was just wondering you thoughts on this question... i think that larval survival is density dependent, but the answers say otherwise, would anyone be able to help explain this too me? thanks Sarah

don't worry i've figured it out! thanks for the assistance Yggdrasil