Seminar assignments - Problems chapter 6 PDF

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Problems chapter 6...

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First, I made a mistake in class when talking about the vector field xi yj . F x, y x2 y 2 4 As pointed out by a couple of students, this field actually has negative divergence where it is defined (i.e., away from the origin). Algebraically, this is because ∇ F x, y

Fx Fy x y y 4 x2 y2 3 2y x 4 x2 y2 3 2x x2 y 2 4 x2 y 2 4 8 2 2 2 x y2 8 x y x2 y2 8y2 x2 y2 8x2 x2 y 2 5 x2 y 2 5 2 2 6x 6y 6 5 2 2 2 x y2 4 x y which is negative for all x, y 0, 0 . Geometrically, it is because the arrows pointing into any point (again, other than the origin) are bigger than the arrows pointing away from it. I had intended to show a field that had positive divergence at 0, 0 but smaller positive divergence away from it. A better example would have been xi yj G x, y . 1 x2 y 2 Here the calculation comes out as 2 ∇ G x, y 2 1 x y2 2 which is positive for all x, y

R2 , but greatest at 0, 0 .

[6.2] #8: Let F x, y

2x2 j.

3xyi

First, we evaluate C F ds directly. We need to parametrize C. Let L, B, R be the left, bottom and right line segments, and let T be the semicircle on the top. Note that T has radius 1 and center at 1, 0 , so it satisfies the equation x 1 2 y2 1. We can therefore parametrize the curves as Curve L B R T

x t 0, 2 t t, 2 2, t 1 cos t, sin t

Range for t 0 t 2 0 t 2 2 t 0 t π 0

1

x t 0, 1 1, 0 0, 1 sin t, cos t

Therefore F ds

F xt

x t dt

F xt

L

C

x t dt

Fx t

B

0, 0

0

6t, 2t2

0, 1 dt

6t, 8

1, 0 dt

0, 1 dt

2

0

0

Fx t T

2

2

x t dt

R

π

31

cos t sin t i

0 2

12 4

5 cos3 t

8 dt

0

0

cos t

3 dt

7 cos2 t

2

3 dt

by symmetry

0 7 cos t sin t π

t

16

7 cos2 t

π

0

8t

cos tj dt

0

2 2

sin ti

π

0

6t dt 3t2

cos t 2 j

2 1

2

0

π . 2

Meanwhile, let D be the area enclosed by the curve C. Green’s Theorem says that F ds C

F

∇

k dA

x

D

2x2

y

3xy

1

x 1

3x

dA

D

D 2

4x

dA

2

x dy dx 0 2

2

2

1

x

1

2

x dx

0

(calculation omitted) x 1 2x 3 2x x2 2 arcsin x 1 x2 2 6 0 π 4 . 2 Note: I used a computer algebra system to do this last integral. Nothing this complicated will appear on Friday’s test!

[6.2] #10: Call the ellipse C and call the region it encloses D. The work done by the field on the particle is F ds C

∇

F

k dA

x

4y

D

x D

3 dA D

3 area of D .

y

4y

3x

dA

x t dt

We have shown in class (and see the book, Example 3, p.430) that the area of an ellipse with horizontal and vertical radii a, b is πab. Therefore the area of D is 2π and the integral is 6π....