Chapter 16.3, Problem 16E

To determine

(a)

To find:

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

Solution:

$f\left(x,y, z\right)=x{y}^{2}z+x^2{z}^2$

Explanation:

1) Concept:

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

2) Given:

$F=\nabla f,F\left(x,y\right)=(y^{2}z+2x{z}^2)i+2xyzj+\left(x{y}^2+2x^{2}z\right)k$

$C:x=\sqrt{t}, y=t+1, z=t^2, 0\le t\le 1$

3) Calculation:

If $F=\nabla f$ then $f_x=y^{2}z+2x{z}^2 \dots \left(1\right)$

$f_y=2xyz \dots \left(2\right)$

$f_z=x{y}^2+2x^{2}z \dots \left(3\right)$

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

$f\left(x,y, z\right)=x{y}^{2}z+x^2{z}^2+g\left(y, z\right) \dots \left(4\right)$

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 $y$

$f_y\left(x,y, z\right)=2xyz{+g}^{\text{'}}\left(y,z\right)\dots \left(5\right)$

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

$f\left(x,y, z\right)=x{y}^{2}z+x^2{z}^2+h\left( z\right) \dots \left(6\right)$

Differentiate with respect to $z$,

$f_z\left(x,y, z\right)=x{y}^2+2x^{2}z+h^{\text{'}}\left(z\right)$

Comparing this equation with (3)

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

Integrate with respect to $z$

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

Therefore, (6) becomes,

$f\left(x,y, z\right)=x{y}^{2}z+x^2{z}^2+K$

$f\left(x,y, z\right)=x{y}^{2}z+x^2{z}^2 taking K=0$

Conclusion:

$f\left(x,y, z\right)=x{y}^{2}z+x^2{z}^2$ is a function such that $F=\nabla f$

(b)

To evaluate:

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

Solution:

$5$

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:

$F=\nabla f,F\left(x,y\right)=(y^{2}z+2x{z}^2)i+2xyzj+\left(x{y}^2+2x^{2}z\right)k$

$C:x=\sqrt{t}, y=t+1, z=t^2, 0\le t\le 1$

3) Calculation:

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

So, by using concept,

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

Since $x=\sqrt{t}, y=t+1, z=t^2 ,$

$r\left(t\right)=\left(\sqrt{t},t+1,t^2\right)$

$r\left(0\right)=(0, 1, 0) and r\left(1\right)=(1, 2, 1)$

From the answer of part (a),$f\left(x,y\right)=x{y}^{2}z+x^2{z}^2{\int }_C^{\square }\nabla f ·dr$ becomes

${\int }_C^{\square }\nabla f ·dr=\left(\left(1\right)\left(4\right)\left(1\right)+1^2{1}^2\right)-0\mathrm{ }=5-0=5$

Therefore,

${\int }_C^{\square }\nabla f ·dr=5$

Conclusion:

${\int }_C^{\square }\nabla f ·dr=5$