Question

Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line segment from ​(1​,2​,3​) to ​(5​,7​,- 2​) and evaluating the line integral of of F = yi + x j+ 3k along the segment. Since F is conservative, the integral is independent of the path.

Answer 1

$\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k$

is conservative if there is a scalar function $f(x,y,z)$ such that $\nabla f=\vec F$. This would require

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

$\dfrac{\partial f}{\partial z}=3$

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

$f(x,y,z)=xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$f(x,y,z)=xy+h(z)$

$\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C$

$f(x,y,z)=xy+3z+C$

so $\vec F$ is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

$\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r$

$=f(5,7,-2)-f(1,2,3)=\boxed{18}$