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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.

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  • $\begingroup$ They mean the same thing. It's like po-tay-to vs po-tah-to. $\endgroup$ – Muphrid Jul 4 '13 at 22:49
  • $\begingroup$ 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... $\endgroup$ – tomasz Jul 4 '13 at 22:58
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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

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

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

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

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