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OpenStudy (anonymous):

Calculate the line integral of the vector field F = (xy)i + (x − y)j along C, the triangle composed of three segments. C1 is the line segment from (3, 0) to (-3, 0). C2 is the line segment from (-3, 0) to (0, 3). C3 is the line segment from (0, 3) to (3, 0).

OpenStudy (anonymous):

Do you know how to parametrize a line segment?

OpenStudy (anonymous):

\[ \mathbf{r}(t) = P_1+(P_2-P_1)t,\,t\in[0,1] \]

OpenStudy (anonymous):

That parametrization goes from \(P_1\) to \(P_2\) when \(t\) goes from \(0\) to \(1\).

OpenStudy (anonymous):

so, for each segemtn i parameterize?

OpenStudy (anonymous):

Each line segment needs its own parametrization and thus its own integral.

OpenStudy (anonymous):

so for c2, (-3,0) to (0,3) the parameterization would be r(t) = -3 + t(6) ??

OpenStudy (anonymous):

It's a vector.

OpenStudy (anonymous):

yes so then i take the derivative and then dot it with the F(r(t)) right/

OpenStudy (anonymous):

I was saying you had it wrong because you didn't even have a vector, for one.

OpenStudy (anonymous):

But yes, in general you got the right idea.

OpenStudy (anonymous):

hmm... im stumped on getting the vector, then...

OpenStudy (anonymous):

wait would the vector be <3,3>

OpenStudy (anonymous):

I'll do the first one, but after that it's on you.

OpenStudy (anonymous):

I just need guidance is all, If i can see how one is done i can do the others...

OpenStudy (anonymous):

"the line segment from (3, 0) to (-3, 0)"\[ \mathbf r(t)=\mathbf p_1 +(\mathbf p_2-\mathbf p_1)t = \begin{bmatrix} 3 \\ 0 \end{bmatrix} + \left( \begin{bmatrix} 3 \\ 0 \end{bmatrix} - \begin{bmatrix} -3 \\ 0 \end{bmatrix} \right)t = \begin{bmatrix} 3 \\ 0 \end{bmatrix} + \begin{bmatrix} 6 \\ 0 \end{bmatrix}t = \begin{bmatrix} 3+6t \\ 0 \end{bmatrix} \]

OpenStudy (anonymous):

\[ r'(t) = \begin{bmatrix} 6 \\ 0 \end{bmatrix} \]\[ \mathbf F\circ \mathbf{r} (t) = \begin{bmatrix} xy \\ x-y \end{bmatrix} \circ \begin{bmatrix} 3+6t \\ 0 \end{bmatrix} = \begin{bmatrix} (3+6t)(0) \\ (3+6t)-(0) \end{bmatrix} = \begin{bmatrix} 0 \\ 3+6t \end{bmatrix} \]\[ \int \mathbf F\cdot d\mathbf s= \int_{t_0}^{t_1}[\mathbf F\circ \mathbf r(t)]\cdot \mathbf r'(t) \;dt = \int_0^1 \begin{bmatrix} 0 \\ 3+6t \end{bmatrix}\cdot \begin{bmatrix} 6 \\ 0 \end{bmatrix}dt =\int_0^10\;dt = 0 \]

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