Question

Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,

Answer 1

$\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k$

We're looking for a scalar function $f$ such that $\nabla f=\vec f$. This means

$\dfrac{\partial f}{\partial x}=yze^{xz}$

$\dfrac{\partial f}{\partial y}=e^{xz}$

$\dfrac{\partial f}{\partial z}=xye^{xz}$

In the second PDE, integrate both side with respect to $y$:

$\dfrac{\partial f}{\partial y}=e^{xz}\implies f(x,y,z)=ye^{xz}+g(x,z)$

Differentiate both sides of this with respect to $x$:

$yze^{xz}=yze^{xz}+\dfrac{\partial g}{\partial x}$

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

$\implies f(x,y,z)=ye^{xz}+h(z)$

Differentiate this with respect to $z$:

$xye^{xz}=xye^{xz}+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=K$

for some constant $K$, so that

$f(x,y,z)=ye^{xz}+K$

By the fundamental theorem of calculus, this line integral has a value of

$\displaystyle\int_C\nabla f\cdot\mathrm d\vec r=f(\vec r(5))-f(\vec r(0))=f(30,22,20)-f(5,-3,0)=\boxed{22e^{600}+3}$

since

$\vec r(5)=30\,\vec\imath+22\,\vec\jmath+20\,\vec k$

$\vec r(0)=5\,\vec\imath-3\,\vec\jmath$