a. Parameterize [tex]C[/tex] by
[tex]\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath[/tex]
with [tex]0\le t\le1[/tex].
b/c. The line integral of [tex]\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath[/tex] over [tex]C[/tex] is
[tex]\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt[/tex]
[tex]=\displaystyle10\int_0^1\mathrm dt=\boxed{10}[/tex]
d. Notice that we can write the line integral as
[tex]\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)[/tex]
By Green's theorem, the line integral is equivalent to
[tex]\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy[/tex]
where [tex]D[/tex] is the triangle bounded by [tex]C[/tex], and this integral is simply twice the area of [tex]D[/tex]. [tex]D[/tex] is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.
