Fundamental Theorem of Line integral

Use the Fundamental Theorem of Line Integrals to evaluate, where $C$: a smooth curve from $(0,0)$ to $(10,5)$.

$$\int_C(6y\,\mathbf{i} + 6x\,\mathbf{j})\cdot d\mathbf{r}$$

I am more familiar with integrating ds rather than dr. If anyone could help me get around this that would be great.

• They mean the same thing. It's like po-tay-to vs po-tah-to. – Muphrid Jul 4 '13 at 22:49
• Really? Po-tah-to? I've never lived in an English-speaking country, but this is the first time I've heard about anyone pronouncing (and not mispronouncing) it like that, dictionary.com doesn't register this pronunciation either. More to the point, I always get confused with notations for integrals. Even more so after passing a differential geometry course and integrating differential forms... – tomasz Jul 4 '13 at 22:58

First step, just parametrize your curve as

$$r = x\mathbf{i}+y\mathbf{j} = t\mathbf{i}+2t\mathbf{j} \implies d\mathbf{r} = (\mathbf{i}+2\mathbf{j})dt,\quad 0\leq t \leq 5.$$

Then the integral can be evaluated as

$$\int_C(6y\,\mathbf{i} + 6x\,\mathbf{j})\cdot d\mathbf{r}= \int_{0}^{5}(6(2t)\,\mathbf{i} + 6t\,\mathbf{j})\cdot (\mathbf{i}+2\mathbf{j})dt = \dots.$$

• @TonyP: You are welcome. – Mhenni Benghorbal Jul 5 '13 at 21:41
• what is the answer to this problem? – jain smit Jul 6 '13 at 17:57

Hint: The vector field $\mathbf{F}(x,y) = 6y\,\mathbf{i} + 6x\,\mathbf{j}$ is conservative. (Why?) And thus there exists a scalar potential $\phi$ such that $\nabla\phi = \mathbf{F}$. And $\int_{C} \mathbf{F}\cdot d\mathbf{r} = \phi(B)-\phi(A)$ where A and B are the start point and end point of $C$ respectively.

• what is the answer? – jain smit Jul 6 '13 at 18:21
• that is wrong... – jain smit Jul 7 '13 at 20:52