Question

Check my answers?
Also, can anyone work with me to find the line integral for question 6 c)

Answer 1

so
\(ϕ=2x^3 y-2y^2 z^2+z^3 \)

how do i start finding line integral from (2,1,1),(2,0,1)

Answer 2

i think you cam simply take the difference of potential

Answer 3

\(\phi\) is the potential function right ?

Answer 4

\(\Large \phi _1 -\phi_2\)
?

Answer 5

Work done = \(\phi(2,0,1) - \phi(2,1,1) \)

Answer 6

since the Force is conservative

Answer 7

but won't taking difference , equivalent to differentiating ?

Answer 8

yes \[\phi \] is potential function
work done
\[\int\limits_{2,1,1}^{2,0,1}F.dr=\int\limits_{2,1,1}^{2,0,1}d(\phi)\]

Answer 9

because it is conservative force field

Answer 10

the line integral is asked for what ?

F , F.dr or [hi or...

Answer 11

*phi

Answer 12

[hi lol

Answer 13

work done = line integral (F.dr)

Answer 14

yes @ganeshie8

Answer 15

please also check the question 2 a)

Answer 16

yes 2 a) looks good wonderfully done :)

Answer 17

let me quickly calculate the line integral....by that time, verify whether my '[hi' is correct or not :P

Answer 18

partials of [hi should equal the components of F

Answer 19

-14 is correct ?

Answer 20

x partial is 6x^2y but 'i' component is 3x^2y
so my [pi is incorrect ?

Answer 21

looks so... checking...

Answer 22

i think phi is correct

Answer 23

phi is ok

Answer 24

-14 is correct too ?

Answer 25

are you guys sure they are asking line integral of F.dr only ?

Answer 26

that part is sure, phi doesn't look right to me

Answer 27

how would i get correct phi....

Answer 28

http://www.wolframalpha.com/input/?i=grad%282x%5E3y-2y%5E2z%5E2%2Bz%5E3%29

Answer 29

you have started with : \(\large F = \nabla \phi\)
but the \(\phi \) function you got is not satisfying above,right ?

Answer 30

right,
what other method than integrating individual components can be used ??
because my integration part is correct...

Answer 31

but
\[d(\phi)=F.dr\]
\[d(\phi)=[3x^2yi+(x^3-2yz^2)j+(3z^2-2y^2z)(dxi+dyj+dyk)\]

Answer 32

\[d(\phi)=[3x^2y+(x^3−2yz^2)+(3z^2−2y^2z)]\]
now integrating
\[\phi=2x^3y-2y^2z^2+z^3\]

Answer 33

that is what @hartnn also got

Answer 34

and what about
\(\large F = \nabla \phi\)
not getting satisfied?

Answer 35

ohh i have'nt checked that

Answer 36

"x partial is 6x^2y but 'i' component is 3x^2y"

Answer 37

what am i missing ?

Answer 38

okay let me grab pen and paper

Answer 39

yes ur right @hartnn

Answer 40

\(ϕ=2x^3 y-2y^2 z^2+z^3+c \)

\(∫_{2,1,1}^{2,0,1}F.dr = ∫_{2,1,1}^{2,0,1}(dϕ)\)
\(=ϕ(2,0,1)-ϕ(2,1,1) \\
=2(8)(0)-2(0)(1)+1+c-2(8)(1)+2(1)(1)-1-c \\
= -16+2=-14
\)

any mistake anyone ?

Answer 41

no i dont see any

Answer 42

but
\[F \neq (grad)\phi\]

Answer 43

^
thats what i am still confused about

Answer 44

\(\nabla\)
`\(\nabla\)`

Answer 45

ok oh thanks :)
and that book oppenheim is awesome

Answer 46

ikr!
that book actually sparked my interest in signals and systems :)

Answer 47

yeah their approach is awesome :)

Answer 48

so what is the conclusion of this problem :(

Answer 49

*waiting for the genius to reply*

Answer 50

yeah ^_^

Answer 51

try this
\(\large \phi = x^3y - y^2z^2 + z^3\)

Answer 52

that satisfies, but how would i get it ?

Answer 53

Since we know that the force is conservative from part A, we can use a simple line integral to find the potential function

Answer 54

\(\large \phi(x_1,y_1,z_1) - \phi(0,0,0) = \int_C F.dr \)

Answer 55

\(\large \phi(x_1,y_1,z_1) = \int_C F.dr + \phi(0,0,0) \)

Answer 56

take any simply path and evaluate the line integral