Question

(1 point) Let F (72+72) i + (2y +62 + 6 sin(y*)) 3+ (2x + 6y + 2e=") R. (a) Find curl F. curl F = <0,0,0> (b) What does your answer to part (a) tell you about Sc F. dr where is the circle (2 – 30)2 + (y - 35)2 = 1 in the ty-plane, oriented clockwise? SF. dr = 0 (c) If C is any closed curve, what can you say about SF. dr? SoF dr = 0 (d) Now let C be the half circle (1 – 30)2 + (y – 35)2 = 1 in the ry-plane with y > 35, traversed from (31,35) to (29, 35). Find Sc. dr by using your result from (C) and considering C plus the line segment connecting the endpoints of C. ScF. dr = 481509.54 Note: You can earn partial credit on this problem. Activate Windows

Answer 1

Let

FL, y, z) = (72 +7.7)i + (2y +62 + 6 sin(y)j + (7.x + 6y + 2e)k

(a)
2 k curl F=TXF dr 72 +7.73 2y +62 +6 sin(y) 70 + 6y +2e=*

= (5, - (7х + бу + 2e). (2) + 6 + 6 sin(s) i+ -(72 + 7х3) д -(7х + бу + 2e**) дх ду дz

+ д - (2y + 62 + де sin(у?)) — д -(72 +7х3) ) ду

= ( 6-6)i + (7 – 7)j + (0 - 0)k = 0

ANSWER:
curl F = 0

(b) Let be the circle
(2 – 30)2 + (y - 35)2 = 1
in
fic
-plane, oriented clockwise. So, is a closed path in
fic
-plane.

By Stoke's theorem,
F.dr = = S curl F.25 = 0 JS
since
curl F = 0
where is the surface on
fic
-plane bounded by the circle .

(c) Let be any closed curve in
fic
-plane. Again by Stoke's theorem
F.dr = = S curl F.25 = 0 JS
since
curl F = 0
where is the surface bounded by the closed curve .

(d) Let be the half-circle
(2 – 30)2 + (y - 35)2 = 1
in
fic
-plane with
y > 35
, transversed from
(31.35)
to
(29.35)
. Let's first draw and show it's direction/orientation.

c B = (29, 35) A = (31, 35)

Now, let
C1
denotes the line joining the endpoints of i.e.
(31.35)
and
(31.35)
, traversed from
(29.35)
to
(31.35)
.

In other words
C1 = {(1 – t)(29, 35,0) + (31,35,00<t<1}

= {(29 + 2t, 35,00 <t<1}

Let's plot draw in on the graph.

C B = (29, 35) A = (31, 35) Ci

Now, observe and
C1
together constitute a closed curve on
fic
-plane traversing counter clockwise. We call
C=C UC1
.

From the part (c), we know that
1. F.di = 0
as is a closed curve.

Now,

F.dr F.di = F.dr + Son F.dr CUC

$\implies 0=\int_{C}\vec{F}.d\vec{r}+\int_{C_1}\vec{F}.d\vec{r}$

de F.dr F.dr с

(1

So if we can compute
F.de Ci
, we can get
F.dr c
. So, let's find
F.de Ci
.

By definition,
Jan F.dr = F(C(t)).C|(t) dt 0
where
FL, y, z) = (72 +7.7)i + (2y +62 + 6 sin(y)j + (7.x + 6y + 2e)k
and $C_1(t)=(29t+2)\hat i+35\hat j+0\hat k$

So, ${C_1}'(t)=29\hat i+0\hat j+0\hat k$ and $\vec{F}(C_1(t))=(7(29t+2)^3)\hat i+(70+6\sin(35^3))\hat j+(7(29t+2)+210+2)\hat k$ .

Thus
FC (t)).C (t) = 29 x 7 x (29+ + 2)3

= 29 x 7 x (293+ + 3 x 292 x 2 +3 x 29t x 22 +23)
$=7\times29^4t^3+42\times 29^3t^2+84\times 29^2t+29\times 56$

Hence, $\int_{C_1}\vec{F}.d\vec{r}=\int_0^1(7\times29^4t^3+42\times 29^3t^2+84\times 29^2t+29\times 56) \ dt$ $=\frac{7\times 29^4}{4}+\frac{42\times 29^3}{3}+\frac{84\times 29^2}{2}+29\times 56$

So, from
(1)
we get,
F.dr =- -Son F.dr = -1616133.75

ANSWER: