Evaluate the integral Integral from (5 comma 2 comma 4 )to (7 comma 9 comma negative 3 )y dx plus x dy plus 3 dz by finding parametric equations for the line segment from ​(5 ​,2​,4​) to ​(7 ​,9​,negative 3​) and evaluating the line integral of Fequals yiplusxjplus3k along the segment. Since F is​ conservative, the integral is independent of the path.

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]

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