Question

Consider the specific vector field: [ mathbf{V}(x, y, z)=aleft(x^{2}+y^{2} ight){-y mathbf{i}+x mathbf{j}}+b z mathbf{k} . ] Consider a closed curve ( C ) bounding a 2 -d disk ( R ) in the ( x-z ) plane: [ egin{aligned} C: & mathbf{r}(s)=c cos (t) mathbf{i}+0 mathbf{j}+c sin (t) mathbf{k}=langle c cos (t), 0, c sin (t) angle \ R: & mathbf{r}(u, v)=x(u, v) mathbf{i}+0 mathbf{j}+z(u, v) mathbf{k}=langle x(u, v), 0, z(u, v) angle . end{aligned} ] Without using Stokes' theorem, formally evaluate [ oint_{C} mathbf{V}(mathbf{r}(s)) cdot mathrm{d} mathbf{r}(s)=? ? ? ] (You should get a very simple result.) Now using Stokes' theorem, formally evaluate [ oint_{C} mathbf{V}(mathbf{r}(s)) cdot mathrm{d} mathbf{r}(s)=? ? ? ] (You should get a very simple result.) Hint: Explicitly calculate ( abla imes mathbf{V} ) Hint: Try to think what the result means geometrically

          Consider the specific vector field:
[
mathbf{V}(x, y, z)=aleft(x^{2}+y^{2}
ight){-y mathbf{i}+x mathbf{j}}+b z mathbf{k} .
]
Consider a closed curve ( C ) bounding a 2 -d disk ( R ) in the ( x-z ) plane:
[
egin{aligned}
C: & mathbf{r}(s)=c cos (t) mathbf{i}+0 mathbf{j}+c sin (t) mathbf{k}=langle c cos (t), 0, c sin (t)
angle \
R: & mathbf{r}(u, v)=x(u, v) mathbf{i}+0 mathbf{j}+z(u, v) mathbf{k}=langle x(u, v), 0, z(u, v)
angle .
end{aligned}
]
Without using Stokes' theorem, formally evaluate
[
oint_{C} mathbf{V}(mathbf{r}(s)) cdot mathrm{d} mathbf{r}(s)=? ? ?
]
(You should get a very simple result.)
Now using Stokes' theorem, formally evaluate
[
oint_{C} mathbf{V}(mathbf{r}(s)) cdot mathrm{d} mathbf{r}(s)=? ? ?
]
(You should get a very simple result.)
Hint: Explicitly calculate ( 
abla 	imes mathbf{V} )
Hint: Try to think what the result means geometrically

Show more…

Consider the specific vector field:
[
mathbfV(x, y, z)=aleft(x^2+y^2
ight)-y mathbfi+x mathbfj+b z mathbfk .
]
Consider a closed curve ( C ) bounding a 2 -d disk ( R ) in the ( x-z ) plane:
[
eginaligned
C: mathbfr(s)=c cos (t) mathbfi+0 mathbfj+c sin (t) mathbfk=langle c cos (t), 0, c sin (t)
angle R: mathbfr(u, v)=x(u, v) mathbfi+0 mathbfj+z(u, v) mathbfk=langle x(u, v), 0, z(u, v)
angle .
endaligned
]
Without using Stokes' theorem, formally evaluate
[
ointC mathbfV(mathbfr(s)) cdot mathrmd mathbfr(s)=? ? ?
]
(You should get a very simple result.)
Now using Stokes' theorem, formally evaluate
[
ointC mathbfV(mathbfr(s)) cdot mathrmd mathbfr(s)=? ? ?
]
(You should get a very simple result.)
Hint: Explicitly calculate (
abla imes mathbfV )
Hint: Try to think what the result means geometrically

Added by Tee S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert William Mead

10/10/2023

Step 1

The curve C is a circle of radius c in the x-z plane, so y = 0 along C. Show more…

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Consider the specific vector field: V(x, y, z) = a(x^2 + y^2){-y i + x j} + b z k. Consider a closed curve C bounding a 2-d disk R in the x–z plane: C : r(s) = c cos(t)i + 0j + c sin(t) k = <c cos(t), 0, c sin(t)>; R : r(u, v) = x(u, v)i + 0j + z(u, v) k = <x(u, v), 0, z(u, v)>. Without using Stokes’ theorem, formally evaluate ∮_C V(r(s)) · dr(s) = ??? (You should get a very simple result.) Now using Stokes’ theorem, formally evaluate ∮_C V(r(s)) · dr(s) = ??? (You should get a very simple result.) Hint: Explicitly calculate ∇ × V Hint: Try to think what the result means geometrically

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Transcript

00:01 The other vector field here, v of x, y, z is a times x squared plus y squared, negative y times i plus x times j, plus bz times k.

00:14 So if we write this out, that'll be i is going to be a times x squared plus y squared, minus...

00:23 Oh, no, excuse me.

00:25 So our components here, a times x squared plus y squared times negative y times i, plus x times a times x squared plus y squared times j, plus b times z times k.

00:48 Which is great, that is our...

00:50 This is our curve in a much less awkward way.

00:56 Our vector field, excuse me.

00:57 We have a closed curve, let's see.

01:02 C times cosine ti, c times sine tk, which is just...

01:06 It's a circle in the xz plane.

01:12 Going over this region here, which is just the disk in the xz plane.

01:19 And we'd like to first, without stokes ' theorem, evaluate this.

01:24 That's going to be the integral as t goes from 0 to 2 pi.

01:31 Let's see, our vector field dotted with r of s, so let's see, r prime of s.

01:39 Of course it's going to be c times, what do we got this? cosine t, division is 2 minus sine t, so we got 0, and then cosine t on the c.

01:51 So we can pull out our constant c, and we end up with, let's see.

02:01 Negative sine t times all this, so we have...

02:07 So negative y is 0, that's just all going to go to 0.

02:12 Plus 0 times all of this is 0, plus cosine t times b times z, which is going to be b times cosine t times z is sine t.

02:32 And i suppose this should be c squared, because this c turns into the r prime.

02:39 Dt, which is c squared times the integral as t goes from 0 to 2 pi.

02:44 We can pull out a b as well...

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