- #1

- 239

- 17

- Homework Statement
- let ##f : R^3 → R## the function ##f(x,y,z)=(\frac {x^3} {3} +y^2 z)##

let ##\gamma## :[0,## \pi ##] ##\rightarrow## ##R^3## the curve ##\gamma (t)##=(cos t, t cos t, t + sin t) oriented in the direction of increasing t.

The work along ##\gamma## of the vector field F=##\nabla f## is

- Relevant Equations
- vector field integral

let ##f : R^3 → R## the function ##f(x,y,z)=(\frac {x^3} {3} +y^2 z)##

let ##\gamma## :[0,## \pi ##] ##\rightarrow## ##R^3## the curve ##\gamma (t)##(cos t, t cos t, t + sin t) oriented in the direction of increasing t.

The work along ##\gamma## of the vector field F=##\nabla f## is:

what i did is

##F = (x^2, 2yz, y^2)##

then i compute the integral

##\int_0^\pi F(\gamma (t)) (\gamma'(t) )dt##

but doing so becomes a very long integral. is there any short cut theorem, that shortens such evaluation? The curl of F is zero, could it be useful?

let ##\gamma## :[0,## \pi ##] ##\rightarrow## ##R^3## the curve ##\gamma (t)##(cos t, t cos t, t + sin t) oriented in the direction of increasing t.

The work along ##\gamma## of the vector field F=##\nabla f## is:

what i did is

##F = (x^2, 2yz, y^2)##

then i compute the integral

##\int_0^\pi F(\gamma (t)) (\gamma'(t) )dt##

but doing so becomes a very long integral. is there any short cut theorem, that shortens such evaluation? The curl of F is zero, could it be useful?