Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to evaluate line integral of $$(2xy-z) dx + (yz) dy + (x) dz$$ over any path from $(1,0,0)$ to $(2,1,4)$.

I thought of integrating once from $(1,0,0)$ to $(2,0,0)$ then to $(2,1,0)$ and lastly to $(2,1,4)$.

...but that wouldn't necessarily be valid for any path. I think I need to convert the expression in the form $d(f(x, y, z))$ but I'm being unable to do so.

share|cite|improve this question
That is not a conservative field. Are you sure the problem is asking for a path-independent solution, or could it be asking you to integrate over any path you so choose? – apnorton Jan 31 '13 at 14:42
@anorton yes. F = (2xy−z)i+(yz)j+(x)k Evaluate ∫F.dr along any path from (1,0,0) to (2,1,4) – Shashwat Black Jan 31 '13 at 15:09

Let $C$ be the path of integration, $\mathbf{F}(x,y,z)=(2xy−z,yz,x)$ the vector field, and $d\mathbf{r}=(dx,dy,dz)$ an infinitesimal displacement along $C$.

Choose $C$ to be the line segment from $(1,0,0)$ to $(2,1,4)$. A parametrization for the line is given by $\mathbf{r}(t)=(1,0,0)+t(1,1,4)=(t+1,t,4t)$ for $t \in [0,1]$.

Note that $\mathbf{F}(\mathbf{r}(t))=(2t^2-2t,4t^2,t+1)$ and $\mathbf{r}'(t)=(1,1,4)$. Now we can evaluate the line integral:

$$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_{t=0}^{1} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \,dt = \int_{t=0}^{1} 6t^2+2t+4 \,dt = \left[2t^3+t^2+4t\right]_{t=0}^{1} = 7$$

However, it is to be noted that $\boldsymbol{\nabla}\times\mathbf{F}=(-y,-2,-2x)\ne\mathbf{0}$, so the integral depends on $C$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.