Author Topic: Evaluating a line integral (Read 1096 times) Tweet Share

/0 · « **on:** May 14, 2010, 04:14:44 pm »

Let $\mathbf{\mu}$ and $\mathbf{\lambda}$ be constant vectors, and let $\mathbf{r} = x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$

Defining $\mathbf{F} = (\mathbf{\mu} \cdot \mathbf{r})\mathbf{\lambda}+(\mathbf{\lambda} \cdot \mathbf{r})\mathbf{\mu}$

Show that $\int_L \mathbf{F} \cdot d\mathbf{r}=(\mathbf{\lambda} \cdot \mathbf{r})(\mathbf{\mu} \cdot \mathbf{r})$ where L is the straight line from $(0,0,0)$ to the point $P(\mathbf{r})$

From my midsemester -.-

Mao · « **Reply #1 on:** May 17, 2010, 11:43:56 am »

With absolutely no regards about being rigorous, we apply Definition 13 (section 16.2, Stewart) over the path. Firstly, define some definitions:
$\mathbf{r}(t) = \mathbf{0} + t (\mathbf{r}-\mathbf{0})=t\mathbf{r}$ [confusing notation, yeah, the r on the RHS denote some point in R3 space], thus, $\mathbf{r}'(t) = \mathbf{r}$
$\mathbf{F}(\mathbf{r}(t)) = t \mathbf{F}(\mathbf{r})$ [again, the r on the RHS denote the point in R3 space]

$\int_L \mathbf{F}\cdot d\mathbf{r} = \int_0^1 \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\; dt = \int_0^1 (\mathbf{F}(\mathbf{r}) \cdot \mathbf{r}) t \; dt$

$I = \left((\mu \cdot \mathbf{r})(\lambda \cdot \mathbf{r}) + (\lambda \cdot \mathbf{r})(\mu \cdot \mathbf{r})\right) \int_0^1 t\; dt = (\mu \cdot \mathbf{r})(\lambda \cdot \mathbf{r})$

Yeah, that should do it, there are probably easier ways, but I've very rusty in this area.

/0 · « **Reply #2 on:** May 19, 2010, 02:28:24 am »

Thanks Mao, after a while of staring at your solution I think i've got it

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Author Topic: Evaluating a line integral (Read 1096 times) Tweet Share

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Evaluating a line integral

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