Question

consider the conservative vector field ( mathbf{f}(x, y, z)=(z + y) mathbf{i}+(3 z + x) mathbf{j}+(3 y + x) mathbf{k} ).
a) find a function ( phi ) such that ( mathbf{f}=
abla phi ) and ( phi(0,0,0)=0 ).
( phi(x, y, z)= )
b) suppose ( c ) is any curve from ( (0,0,0) ) to ( (1,1,1) ). use part (a) to compute the work done along ( c ).
(hint: because ( mathbf{f} ) is conservative, its integrals are \path independent. \)
( int_{(0,0,0)}^{(1,1,1)} mathbf{f} cdot d mathbf{r}= )

Explanation:

Part (a)

Step1: Integrate $F_x$ w.r.t $x$

$\phi = \int (z + y)dx = xz + xy + g(y,z)$

Step2: Differentiate $\phi$ w.r.t $y$

$\frac{\partial \phi}{\partial y} = x + g_y(y,z) = 3z + x \implies g_y(y,z)=3z$

Step3: Integrate $g_y$ w.r.t $y$

$g(y,z) = 3yz + h(z)$

Step4: Differentiate $\phi$ w.r.t $z$

$\frac{\partial \phi}{\partial z} = x + 3y + h'(z) = 3y + x \implies h'(z)=0 \implies h(z)=C$

Step5: Use $\phi(0,0,0)=0$ to find $C$

$\phi(0,0,0)=0 + 0 + 0 + C = 0 \implies C=0$

Step1: Apply Fundamental Theorem for line integrals

$\int_{C} \mathbf{F} \cdot d\mathbf{r} = \phi(1,1,1) - \phi(0,0,0)$

Step2: Compute $\phi(1,1,1)$

$\phi(1,1,1)=(1)(1)+(1)(1)+3(1)(1)=1+1+3=5$

Step3: Compute the integral

$\int_{C} \mathbf{F} \cdot d\mathbf{r}=5 - 0=5$

Answer:

$\phi(x,y,z)=xy + xz + 3yz$

Part (b)