Question

\[F=(ye^{xy}\ln(1+z^2))i + (xe^{xy}\ln(1+z^2)+1)j + (\frac{2ze^{xy}}{1+z^2})k\]

The potential function of F, I worked out to be:
\[f(x,y,z)=e^{xy}\ln(z^2+1)+y+c\]

\[G = (ye^{xy}\ln(1+z^2)+x)i + (xe^{xy}\ln(1+z^2)+1)j + (\frac{2ze^{xy}}{1+z^2})k\]

Let C be the curve x=t, y=0, z=t(1-t) with \[0\le t \le 1\] which connects \[x_{0}=(0,0,0)\] with \[x_{1}=(1,0,0)\]

Find \[\int\limits_C G.dr\]

Answer 1

@sazap10 , Would you mind typing only the question separately because I can't distinguish between what you have done and what is given in the question

Answer 2

Let \[G = (ye^{xy}\ln(1+z^2)+x)i + (xe^{xy}\ln(1+z^2)+1)j + (\frac{2ze^{xy}}{1+z^2})k\] Let C be a curve of the form x=t, y=0, z=t(1-t) with \[0\le t \le 1\] which connects \[x_{0}=(0,0,0)\] with \[x_{1}=(1,0,0)\] Find \[\int\limits_C G.dr\]

Answer 3

@experimentX will help you in this :)

Answer 4

what is G??

Answer 5

If I am not wrong , G is flux ?? Am I right??

Answer 6

a vector field i think

Answer 7

Change all x, y, and z to t's
take a dot product .. and integrate.

Answer 8

dr = idx + jdy + kdz

Answer 9

i think u are supposed to use the potential function f

Answer 10

but i idk how since G is different from F

Answer 11

F and G are same ... expect for the fact that there is extra x on i component of G

Answer 12

then could i use the potential function f to find the circulation? i did that and got the answer as 0. Which doesnt seem right to me

Answer 13

it's only changing around x axis ... i don't think these two points are equipotential surface.

Answer 14

Change all x,y,z to t's and integrate
\[ \int_{0}^{1}(ye^{xy}\ln(1+z^2)+x)dx\]

Answer 15

G is a conservative field, then the work done from point A to point B is

fB) - f(A) where f is the potential of the conservative Field G

Answer 16

The answer is
f(1,0,0)-f(0,0,0) =ln(1)