Chapter 16.3, Problem 17E

To determine

(a)

To find:

$f$ such that $F=\nabla f$

Solution:

$f\left(x,y, z\right)=y{e}^{xz}$

Explanation:

1) Concept:

If $F=\nabla f$ then $F\left(x,y\right)=f_{x}i+f_{y}j+f_{z}k$

2) Given:

$F=\nabla f,F\left(x,y\right)=\left(yz{e}^{xz}\right)i+e^{xz}j+xy{e}^{xz}k$

$C:r\left(t\right)=\left(t^2+1\right)i+\left(t^2-1\right)j+\left(t^2-2t\right)k 0\le t\le 2$

3) Calculation:

If $F=\nabla f$ then $f_x=yz{e}^{xz} \dots \left(1\right)$

$f_y=e^{xz} \dots \left(2\right)$

$f_z=xy{e}^{xz} \dots \left(3\right)$

Integrate $\left(1\right)$ with respect to $x$

$f\left(x,y, z\right)=\frac{yz{e}^{xz}}{z}+g\left(y, z\right)$

$f\left(x,y, z\right)=y{e}^{xz}+g\left(y, z\right) \dots \left(4\right)$

Notice that the constant of integration is a constant with respect to $x$, that is, a functionof $y$, so consider it as $g\left(y, z\right)$.

Now differentiate above equation with respect to $y$

$f_y\left(x,y, z\right)=e^{xz}{+g}^{\text{'}}\left(y,z\right)\dots \left(5\right)$

Comparing (2), and (4)

$g^{\text{'}}\left(y,z\right)=0$

Integrating with respect to $y$

$g_y\left(y,z\right)=h\left(z\right)$, where $h\left(z\right)$ is a constant.

Hence (4) becomes

$f\left(x, y, z\right)=y{e}^{xz}+h\left(z\right) \dots \left(6\right)$

Differentiate with respect to $z$,

$f_z\left(x,y, z\right)=yz{e}^{xz}+h^{\text{'}}\left(z\right)$

Comparing this equation with (1)

$h^{\text{'}}\left(z\right)=0$

Integrate with respect to $z$

$h\left(z\right)=K$

Therefore, (6) becomes,

$f\left(x, y, z\right)=y{e}^{xz}+K$

$f\left(x,y, z\right)=y{e}^{xz} taking K=0$

Conclusion:

$f\left(x,y, z\right)=y{e}^{xz}$ is a function such that $F=\nabla f$

(b)

To evaluate:

${\int }_C^{\square }F· dr$ along the given curve $C$

Solution:

$4$

Explanation:

1) Concept:

Fundamental theorem of line integral:

Let $C$ be a smooth curve given by the vector function $r\left(t\right), a\le t\le b$. Let $f$ be a differential function of two or three variables whose gradient vector $\nabla f$ is continuous on $C$. Then

${\int }_C^{\square }\nabla f ·dr=f\left(r\left(b\right)\right)-f\left(r\left(a\right)\right)$

2) Given:

$F=\nabla f,F\left(x,y\right)=\left(yz{e}^{xz}\right)i+e^{xz}j+xy{e}^{xz}k$

$C:r\left(t\right)=\left(t^2+1\right)i+\left(t^2-1\right)j+\left(t^2-2t\right)k 0\le t\le 2$

3) Calculation:

C is a smooth curve with initial point $r\left(0\right)$ and terminal point $\left(1\right)$

So, by using concept,

${\int }_C^{\square }\nabla f ·dr=f\left(r\left(2\right)\right)-f\left(r\left(0\right)\right)=f(5, 3, 0)-f(1, -1, 0)$

Since $r\left(t\right)=\left(t^2+1\right)i+\left(t^2-1\right)j+\left(t^2-2t\right)k$

$r\left(0\right)=\left(0+1\right)i+\left(0-1\right)j+\left(0-0\right)k=i-j$

Therefore,$r\left(0\right)=(1, -1, 0)$

$r\left(2\right)=\left(4+1\right)i+\left(4-1\right)j+\left(4-4\right)k=5i+3j$

Therefore,$r\left(0\right)=(5, 3, 0)$

From the answer of part (a),$f\left(x,y, z\right)=y{e}^{xz}$ so,

${\int }_C^{\square }\nabla f ·dr=f(5, 3, 0)-f(1, -1, 0)$

$=3e^0-\left(-1\right)e^0=3+1=4 since e^0=1$

Therefore,

${\int }_C^{\square }\nabla f ·dr=4$

Conclusion:

${\int }_C^{\square }\nabla f ·dr=4$