Question

6. [-/7 Points) DETAILS MY NOTES ASK YOUR TEACHER This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A vector field F is shown. Use the interpretation of divergence derived in this section to determine whether div F is positive or negative at P1 and at P2. Pi P2.

please i need answer for both for a thumb up

Answer 1

Pic-1) At the point $P_1$ the field lines are towards origin,

So divergence is negative.

At the point  $P_2$ the field lines are moving outwards from origin,

So divergence is positive.

Pic-2)

The curl of the given field is

$\text{curl }{\bf F}=\begin{vmatrix} \hat i &\hat j &\hat k\\ {\partial \over \partial x} &{\partial \over \partial y} & {\partial \over \partial z}\\ x+y^2 & y+z^2 & z+x^2 \end{vmatrix} \\ \\ ~~~~~~~~~~~=\hat i[0-2z]-\hat j[2x-0]+\hat k[0-2y] \\ ~~~~~~~~~~~=-2[z\hat i+x\hat j+y\hat k]$

Also the surface enclosed by the curve C is
T +y +z = 7
that can be written as
2= 7 - 7 - y

Let $f(x,y,z)=x+y+z-7=0$

Now the unit normal vector is

$\hat{n}={ \bigtriangledown f\over\| \bigtriangledown f\| }={\left \langle 1, 1, 1 \right \rangle \over \sqrt{3}}$

So the Stokes theorem we get

$\int_{C}{\bf F} \cdot d{\bf r} = \iint_D \text{Curl }{\bf F} \cdot d{\bf S}$ --------------------(1)

Now, take D is the region in the xy-plane, by taking z=0, we get $0\leq x\leq 7, 0\leq y \leq 7-x$

Thus, by (1) we get using  $z=7-x-y$

$\int_{C}{\bf F} \cdot d{\bf r} =-{2\over \sqrt{3}} \int_0^7\int_{0}^{7-x} ([7-x-y]+x+y)dydx \\ \\ ~~~~~~~~~~~~~~~={-14\over \sqrt{3}}\int_0^7(7-x) dx \\ ~~~~~~~~~~~~~~~={-14\over \sqrt{3}}\times {49\over 2}\\ ~~~~~~~~~~~~~~~=-{343\over \sqrt{3}}$