Respuesta :
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_{(5,2,4)}^{(7,9,-3)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz[/tex]
Parameterize the line segment (call it [tex]C[/tex]) by
[tex]\vec r(t)=(1-t)(5\,\vec\imath+2\,\vec\jmath+4\,\vec k)+t(7\,\vec\imath+9\,\vec\jmath-3\,\vec k)[/tex]
[tex]\vec r(t)=(2t+5)\,\vec\imath+(7t+2)\,\vec\jmath+(4-7t)\,\vec k[/tex]
with [tex]0\le t\le1[/tex]. Then
[tex]\vec r'(t)=2\,\vec\imath+7\,\vec\jmath-7\,\vec k[/tex]
and the line integral is
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1((7t+2)\,\vec\imath+(2t+5)\,\vec\jmath+3\,\vec k)\cdot\vec r'(t)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1((2t+5)+2(7t+2)-21)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(28t+18)\,\mathrm dt=\boxed{32}[/tex]
Alternatively, if we can show that [tex]\vec F[/tex] is conservative, then we can apply the fundamental theorem of calculus. We need to find [tex]f[/tex] such that [tex]\nabla f=\vec F[/tex], which requires
[tex]\dfrac{\partial f}{\partial x}=y[/tex]
[tex]\dfrac{\partial f}{\partial y}=x[/tex]
[tex]\dfrac{\partial f}{\partial z}=3[/tex]
Integrating both sides of the first equation with respect to [tex]x[/tex] gives
[tex]f(x,y,z)=xy+g(y,z)[/tex]
Differentiating both sides wrt [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}[/tex]
[tex]\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]
Differentiating wrt [tex]z[/tex] gives
[tex]\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}[/tex]
[tex]\implies h(z)=3z+C[/tex]
So we have
[tex]f(x,y,z)=xy+3z+C[/tex]
and [tex]\vec F[/tex] is conservative. By the FTC, we find
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=f(7,9,-3)-f(5,2,4)=\boxed{32}[/tex]