Question

F(x, y, z) = yzexzi + exzj + xyexzk,

Answer 1

$\nabla f(x,y,z)=\mathbf F(x,y,z)\iff\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j+\dfrac{\partial f}{\partial z}\,\mathbf k=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k$

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

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

$\dfrac{\partial f}{\partial z}=xye^{xz}=xye^{xz}+\dfrac{\mathrm dh}{\mathrm dz}$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

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

$\mathbf r(t)=(t^2+5)\,\mathbf i+(t^2-1)\,\mathbf j+(t^2-5)\,\mathbf k$
$\implies\mathbf r(0)=5\,\mathbf i-\mathbf j-5\,\mathbf k$
$\implies\mathbf r(5)=30\,\mathbf i+24\,\mathbf j+20\,\mathbf k$

By the gradient theorem, we have for any path $\mathcal C$ originating at the point (5, -1, -5) and terminating at the point (30, 24, 20),

$\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=f(\mathbf r(5))-f(\mathbf r(0))=f(30,24,20)-f(5,-1,-5)$
$\implies\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=24e^{600}+e^{-25}$