Respuesta :
Parameterize each leg by
[tex]\vec r_1(t)=(1-t)(4\,\vec\imath)+t(-4\,\vec\imath)=(4-8t)\,\vec\imath[/tex]
[tex]\vec r_2(t)=(1-t)(-4\,\vec\imath)+t(4\,\vec\jmath)=(-4+4t)\,\vec\imath+4t\,\vec\jmath[/tex]
[tex]\vec r_3(t)=(1-t)(4\,\vec\jmath)+t(4\,\vec\imath)=4t\,\vec\imath+(4-4t)\,\vec\jmath[/tex]
each with [tex]0\le t\le1[/tex].
The line integrals along each leg (in the same order as above) are
[tex]\displaystyle\int_0^1((4-8t)\,\vec\jmath)\cdot(-8\,\vec\imath)\,\mathrm dt=0[/tex]
[tex]\displaystyle\int_0^1(16t(t-1)\,\vec\imath-4\,\vec\jmath)\cdot(4\,\vec\imath+4\,\vec\jmath)\,\mathrm dt=\int_0^1(64t(t-1)-16)\,\mathrm dt=-\frac{80}3[/tex]
[tex]\displaystyle\int_0^1(16t(1-t)\,\vec\imath+(8t-4)\,\vec\jmath)\cdot(4\,\vec\imath-4\,\vec\jmath)\,\mathrm dt=\int_0^1(64t(1-t)-4(8t-4))\,\mathrm dt=\frac{32}3[/tex]
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The total line integral then has a value of -16. We can confirm this by checking with Green's theorem. Notice that [tex]C[/tex] as given as clockwise orientation, while Green's theorem assumes counterclockwise. So we must multiply by -1:
[tex]\displaystyle\int_{-C}\vec f\cdot\mathrm d\vec r=-\iint_D\left(\frac{\partial(x-y)}{\partial x}-\frac{\partial(xy)}{\partial y}\right)\,\mathrm dA=-\int_0^4\int_{y-4}^{4-y}(1-x)\,\mathrm dx\,\mathrm dy=-16[/tex]
as required.
The line integral of f along each segment of the triangle is given below and this can be determined by integrating each segment of the triangle.
Given :
- f(x,y) = xy i + (x-y) j
- 'c' is the triangle from (4,0) to (-4,0) to (0,4) to (4,0)
First, parametrize each segment of the triangle.
[tex]\rm \bar{r_1}(t) = (1-t)4\hat{i}+t(-4)\hat{i}=(4-8t)\hat{i}[/tex]
[tex]\rm \bar{r_2}(t) = (1-t)(-4)\hat{i}+t(4)\hat{j}=(-4+4t)\hat{i}+4t\hat{j}[/tex]
[tex]\rm \bar{r_2}(t) = (1-t)(-4)\hat{j}+t(4)\hat{i}=(4t)\hat{i}+(4-4t))\hat{j}[/tex]
where [tex]\rm 0\leq t\leq1[/tex]
Now, the line integral of f along each segment of the triangle is given by:
[tex]\rm \int\limits^1_0 {((4-8t)\hat{j})(-8\hat{i})} \, dt = 0[/tex]
[tex]\rm \int\limits^1_0 {(16(t-1)\hat{i}-4\hat{j})(4\hat{i}+4\hat{j})} \, dt = -\dfrac{80}{3}[/tex]
[tex]\rm \int\limits^1_0 {(16(1-t)\hat{i}+(8t-4)\hat{j})(4\hat{i}-4\hat{j})} \, dt = -\dfrac{32}{3}[/tex]
The line integral of f along each segment of the triangle is given above and this can be determined by integrating each segment of the triangle.
For more information, refer to the link given below:
https://brainly.com/question/14502499
