Chapter 16.3, Problem 18E

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:

$F=\nabla f,F\left(x,y\right)=\sin yi+\left(x\cos y+\cos z\right)j-y\sin zk$

$C:r\left(t\right)=\sin ti+tj+2tk 0\le t\le \pi /2$

3) Calculation:

If $F=\nabla f$ then $f_x=\sin y \dots \left(1\right)$

$f_y=x\cos y+\cos z \dots \left(2\right)$

$f_z=-y\sin z \dots \left(3\right)$

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

$f\left(x,y, z\right)=x\sin y+g\left(y, z\right)\dots \left(4\right)$

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

Now differentiate above equation with respect to $y$

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

Comparing (2), and (4)

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

Integrating with respect to $y$

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

Hence (4) becomes

$f\left(x,y, z\right)=x\sin y+y\cos z+\mathrm{h}\left(\mathrm{z}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\dots \left(6\right)$

Differentiate with respect to $z$,

$f_z\left(x,y, z\right)=-y sin z+h\text{'}\left(z\right)$

Comparing this equation with (3)

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

Integrate with respect to $z$

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

Therefore, (6) becomes,

$f\left(x,y, z\right)=x\sin y+y\cos z+\mathrm{K}$

$f\left(x,y, z\right)=x\sin y+y\cos z taking K=0$

Conclusion:

$f\left(x,y, z\right)=x\sin y+y\cos z$ is a function such that $F=\nabla f$

(b)

To evaluate:

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

Solution:

$1-\frac{\pi }{2}$

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:

$F=\nabla f,F\left(x,y\right)=\sin yi+\left(x\cos y+\cos z\right)j-y\sin zk$

$C:r\left(t\right)=\sin ti+tj+2tk 0\le t\le \pi /2$

3) Calculation:

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

So, by using concept,

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

Since $r\left(t\right)=\sin ti+tj+2tk$

$r\left(\pi /2\right)=\sin \frac{\pi }{2}i+\frac{\pi }{2}j+2\left(\frac{\pi }{2}\right)k=i+\frac{\pi }{2}j+\pi$

Therefore,$r\left(0\right)=\left(1, \frac{\pi }{2}, \pi \right)$

$r\left(0\right)=\sin \left(0\right)i+0j+2\left(0\right)k =0$

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

From the answer of part (a),$f\left(x,y, z\right)=x\sin y+y\cos z$ so,

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

$=\left(1·\sin \frac{\pi }{2}+\frac{\pi }{2}·\cos \pi \right)-\left(0 \right)$

$=1-\frac{\pi }{2}-0, since \cos \pi =-1, \sin \frac{\pi }{2}=1$

$=1-\frac{\pi }{2}$

Therefore,

${\int }_C^{\square }\nabla f ·dr=1-\frac{\pi }{2}$

Conclusion:

${\int }_C^{\square }\nabla f ·dr=1-\frac{\pi }{2}$