Question

Stokes theorem, line integral problem

Answer 1

well let's try to start with he line integral part first...

Answer 2

so my vertices would be (0,0,0), (0,3,3),(1,3,3), (1,0,0)

Answer 3

yep

Answer 4

so we need to write out those 4 lines in space parameterized
here's the first part\[C_1:x=0,y=t,z=t;~0\le t\le-3\]

Answer 5

+3

Answer 6

wht do my less-than or equal-to signs look like minus signs to me?
my browser must be messed up, yeah I am seeing what I wrote weird is all

Answer 7

ah, there we go I had to reload\[C_1:x=0,y=t,z=t;~0\le t\le3\]take your time man let me get this all straightened out on paper.

Answer 8

that way I'm less likely to make mistakes

Answer 9

\[C_1:x=0,y=t,z=t;~0\le t\le3\]\[C_2:x=t,y=3,z=3;~0\le t\le1\]\[C_3:x=1,y=3-t,z=3-t;~0\le t\le3\]\[C_4:x=1-t,y=0,z=0;~0\le t\le1\]

Answer 10

C2 is not same for C1 0<t<3

Answer 11

are you asking or telling?

Answer 12

C2 is 0<t<1 because the only parameter that varies is x, which varies from 0 to 1

Answer 13

your right

Answer 14

here are the integrals along each curve
please do double-check me on everything\[C_1:\int_0^3\langle0,0,0\rangle\cdot\langle0,1,1\rangle dt=0\]\[C_2:\int_0^1\langle t^2,324t,9t\rangle\cdot\langle1,0,0\rangle dt=\int_0^1t^2dt=\frac13\]\[C_3:\int_0^3\langle 1,4(3-t)^3,(3-t)^2\rangle\cdot\langle0,-1,-1\rangle dt\]\[=\int_0^3-4(3-t)^3-(3-t)^2dt=\left.(3-t)^4+\frac13(3-t)^3\right|_0^3=-90\]\[C_4:\int_0^1\langle(1-t)^2,0,0\rangle\cdot\langle-1,0,0\rangle dt=\int_0^1-(1-t)^2dt\]\[=\frac13(1-t)^3|_0^1=-\frac13\]adding the four integrals gives\[0-\frac13-90-\frac13=-90\]

Answer 15

ok now to use Stoke's theorem
let me do that on paper as well, one moment please

Answer 16

c2...108t but does not matter gets dotted with 0

Answer 17

\[4xy^3=4(3^3)t=108t\]yep you are right, but like you said it doesn't matter

Answer 18

looks good!

Answer 19

no worries

Answer 20

have to walk the dogs...be back in 10 minutes

Answer 21

\[\text{curl}\vec F=2yx\hat i-y^2\hat j+4y^3\hat k\]ok hopefully I'll have it all worked out by the time you are back

Answer 22

\[S:y=z\]\[g(x,y,z)=0=z-y\implies\nabla g=\langle0,-1,1\rangle\]\[\iint\limits_\sigma\text{curl}\vec F\cdot d\vec S=\iint\limits_D(\nabla\times\vec F)\cdot\nabla gdA\]let's sketch the region \(D\)...

Answer 23

so it is a simple regtangle
so we get...