### Select your language

Suggested languages for you:

Americas

Europe

Q13E

Expert-verified
Found in: Page 781

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

### Answers without the blur.

Just sign up for free and you're in.

# (a) Find a function $$f$$ such that $${\bf{F}} = \nabla f$$ and (b) use part (a) to evaluate $$\int_C {\bf{F}} \cdot d{\bf{r}}$$ along the given curve $$C$$.$${\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}$$, $$C$$ is the line segment from $$(1,0, - 2)$$ to $$(4,6,3)$$

a) The potential function $$f$$ of vector field $${\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}$$ is$$f(x,y,z) = xyz + {z^2}$$.

b) The value of $$\int_C \nabla f \cdot d{\bf{r}}$$ along the curve $$C$$ is 77.

See the step by step solution

## Step 1: Given data

Vector field is $${\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}$$ and line segment from $$(1,0, - 2)$$ to $$(4,6,3)$$.

## Step 2: Relation between the potential function $$f$$ and vector field $${\bf{F}}$$

Relation between the potential function $$f$$ and vector field $${\bf{F}}$$.

$$\nabla f = {\bf{F}}$$

Consider vector function $${\bf{r}}(t),a \le t \le b$$ with a smooth curve $$C$$. Consider $$f$$ is a differentiable function two or three variables of gradient function $$\nabla f$$ and is continuous on curve $$C$$. Then,

$$\int_C \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))$$

## Step 3: Find the potential function $$f$$ such that $$\nabla f = {\bf{F}}$$

a)

Vector field is $${\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}$$.

Consider $$\nabla f = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}$$.

Write the relation between the potential function $$f$$ and vector field $${\bf{F}}$$.

Substitute $${f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}$$ for $$\nabla f$$,

$${\bf{F}} = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}$$

Compare the equation $${\bf{F}} = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}$$ with $${\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}$$.

$$\begin{array}{l}{f_x}(x,y,z) = yz\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\{f_y}(x,y,z) = xz\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\{f_z}(x,y,z) = xy + 2z{\rm{ }}\,\,\,\,\,\,{\rm{(3) }}\end{array}$$

Integrate equation (1) with respect to $$x$$.

\begin{align} f(x,y,z) & =\int{y}zdx \\ & =yz\int{d}x \\ & =yz(x)+g(y,z)\left\{ \because \int{d}t=t \right\} \\ & f(x,y,z)=xyz+g(y,z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4) \end{align}

Apply partial differentiation with respect to $$y$$ on both sides of equation (4).

\begin{align} & {{f}_{y}}(x,y,z)=\frac{\partial }{\partial y}(xyz+g(y,z)) \\ & =\frac{\partial }{\partial y}(xyz)+\frac{\partial }{\partial y}(g(y,z)) \\ & =xz\frac{\partial }{\partial y}(y)+{{g}^{\prime }}(y,z) \\ & =xz(1)+{{g}^{\prime }}(y,z)\left\{ \because \frac{\partial }{\partial t}(t)=1 \right\} \end{align}

$${f_y}(x,y,z) = xz + {g^\prime }(y,z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$

Compare the equations $$(2)$$ and (5).

$${g^\prime }(y,z) = 0$$

Apply integration on both sides of equation.

$$\begin{array}{c}\int {{g^\prime }} (y,z)dy = \int 0 dy\\g(y,z) = h(z)\end{array}$$

Here,

$$h(z)$$ is function of $$z$$.

Substitute $$h(z)$$ for $$g(y,z)$$ in equation (4),

$$f(x,y,z) = xyz + h(z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(6)$$

Apply partial differentiation with respect to $$z$$ on both sides of equation (6).

\begin{align} {{f}_{z}}(x,y,z) & =\frac{\partial }{\partial z}(xyz)+\frac{\partial }{\partial z}(h(z)) \\ & =xy\frac{\partial }{\partial z}(z)+{{h}^{\prime }}(z) \\ & =xy(1)+{{h}^{\prime }}(z)\left\{ \because \frac{\partial }{\partial t}(t)=1 \right\} \\ & {{f}_{z}}(x,y,z)=xy+{{h}^{\prime }}(z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7) \end{align}

Compare the equation (3) and ( 7 ).

$${h^\prime }(z) = 2z$$

Integrate the equation with respect to $$z$$.

\begin{align} h(z) & =\int{(2z)}dz \\ & =2\left( \frac{{{z}^{2}}}{2} \right)+K\left\{ \because \int{(t)}dt=\frac{{{t}^{2}}}{2} \right\} \\ & ={{z}^{2}}+K \end{align}

Here,

$$K$$ is constant.

Consider the value of $$K$$ as 0 .

Substitute $${z^2} + K$$ for $$h(z)$$ in equation $$(6)$$,

$$f(x,y,z) = xyz + {z^2} + K$$

Substitute 0 for $$K$$,

$$\begin{array}{c}f(x,y,z) = xyz + {z^2} + 0\\ = xyz + {z^2}\end{array}$$

Thus, The potential function $$f$$ of vector field $${\bf{F}}(x,y) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}$$ is $$f(x,y,z) = xyz + {z^2}$$.

## Step 4: Determine the value of $$\int_C \nabla f \cdot d{\bf{r}}$$ along the curve $$C$$

b)

Consider vector function $${\bf{r}}(t),a \le t \le b$$ with a smooth curve $$C$$. Consider $$f$$ is a differentiable function two or three variables of gradient function $$\nabla f$$ and is continuous on curve $$C$$. Then,

$$\int_C \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(8)$$

The initial point of curve is $$(1,0, - 2)$$ so it is lower limit $$a$$ and terminal point is $$(4,6,3)$$ so it is upper limit $$b$$.

Find the common value $$\int_C \nabla f \cdot d{\bf{r}}$$ by the use of equation (8).

\begin{align} & \int_{C}{\nabla }f\cdot d\mathbf{r}=f(4,6,3)-f(1,0,-2) \\ & =\left[ (4)(6)(3)+{{3}^{2}} \right]-\left[ (1)(0)(-2)+{{(-2)}^{2}} \right]\quad \left\{ \because f(x,y,z)=xyz+{{z}^{2}} \right\} \\ & =72+9-0-4 \\ & =77 \end{align}

Thus, the value of $$\int_C \nabla f \cdot d{\bf{r}}$$ along the curve $$C$$ is 77.

### Want to see more solutions like these?

Sign up for free to discover our expert answers

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.