Question

Consider the vector field f(x,y,z)=(2z+5y)i+(z+5x)j+(y+2x)kf(x,y,z)=(2z+5y)i+(z+5x)j+(y+2x)k.
a.find a function ff such that f=∇ff=∇f and f(0,0,0)=0f(0,0,0)=0. f(x,y,z)=f(x,y,z)=
b.suppose c is any curve from (0,0,0)(0,0,0) to (1,1,1).(1,1,1). use part
a.to compute the line integral ∫cf⋅dr∫cf⋅dr.

Answer 1

$\mathbf f(x,y,z)=(2z+5y)\,\mathbf i+(z+5x)\,\mathbf j+(y+2x)\,\mathbf k$

We want to find a scalar function $f(x,y,z)$ such that

$\mathbf f(x,y,z)=\nabla f(x,y,z)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j+\dfrac{\partial f}{\partial z}\,\mathbf k$

$\dfrac{\partial f}{\partial x}=2z+5y\implies f(x,y,z)=2xz+5xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=z+5x=5x+\dfrac{\partial g}{\partial y}$
$\implies\dfrac{\partial g}{\partial y}=z\implies g(y,z)=yz+h(z)$

$\dfrac{\partial f}{\partial z}=y+2x=2x+\dfrac{\partial g}{\partial z}$
$\implies\dfrac{\partial g}{\partial z}=y+\dfrac{\mathrm dh}{\mathrm dz}=y$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

where $C$ is an arbitrary constant.

So,

$f(x,y,z)=2xz+5xy+yz+C$

By the gradient theorem, the line integral along any path from (0, 0, 0) to (1, 1, 1) subject to the vector field $\mathbf f(x,y,z)$ is then

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)=2+5+1=8$

(where $\mathcal C$ is any path connecting the given points)

Answer 2

Following are the response to the given question:

Given:

$\bold{F(x, y, z) =(2z+5y)i +(3z +5x)j+(3y +2x)k}$

To find:

function=?

line integral=?

Solution:

For point a:

Consider the vector field:

$F(x, y, z) =(2z+5y)i +(3z +5x)j+(3y +2x)k:$

$f_x= 2z +5y \\\\f_y = 3z + 5x \\\\f_z = 3y + 2x$

Integrating $f_x = 2z +5y$ with respect to $x:$

$f(x,y,z) = \int (2z + 5y )dx = 2xz + 5xy + g( y,z) \\\\f_y = 3z + 5x+g'(y,z)$

So, $g(y,z) = 0$  

$f (x, y,z) = 2xz +5xy +h(z) \\\\ f_z = 3y = h'(z)$

So,   $h(z) = 3 yz$

Thus,  

$\boxed{\boxed{\boxed{ f(x, y, z) = 2xz + 5xy +3yz}}}$

For point b:

$\to \int_c F\ dr = f(1,1,1)- f(0,0,0)$

                $= [2(1)(1)+5(1)(1) +3(1)(1)] -[ 2(0)(0)+5(0)(0) +3(0) (0)]\\\\= (2+5+3)-0\\\\ = 10$

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