Begin2.DVI

I7-41.
F=ω+λ 1 g+λ 2 h=x^2 +y^2 +z^2 +λ 1 g+λ 2 h
∂F
∂λ 1

=g=x+y+z−6 = 0

∂F

∂λ 2 =h= 3x+ 5y+ 7z−34 = 0

∂F

∂x

=2x+λ 1 + 3λ 2 = 0

∂F

∂y

=2y+λ 1 + 5λ 2 = 0

∂F

∂z

=2z+λ 1 + 7λ 2 = 0

Solve this system of equations and show

x= 1, y= 2, z= 3, λ 1 = 1, λ 2 =− 1

I7-42.

∇F=− 2 zˆe 2 + (3y−1)ˆe 3 and eˆn=xˆe 1 +yˆe 2 +zˆe 3 , withdS=

dxdy

|ˆen·ˆe 3 |=

dxdy

z

I=

∫ 1

1

∫+√ 1 −x 2

−√ 1 −x^2

(y−1)dydx=−π

Problem can also be solved using spherical coordinatesx= sinθcosφ, y= sinθsinφ, z=

cosθ

I7-44. (b)∂x∂~r=ˆe 1 +∂x∂yˆe 2 , ∂~r∂z=∂y∂zˆe 2 +ˆe 3 giving

E=

∂~r

∂x·

∂~r

∂x= 1 + (

∂y

∂x)

2

F=

∂~r

∂x·

∂~r

∂z=

∂y

∂x

∂y

∂z

G=∂~r∂z·∂~r∂z= (∂y∂z)^2 + 1

Show

dS=

√

EG−F^2 =

√

1 + (∂y

∂x

)^2 + (∂y

∂z

)^2

Normal to surface isN~ =

∂~r

∂z×

∂~r

∂x=−

∂y

∂x

ˆe 1 +ˆe 2 −∂y

∂z

ˆe 3 Note that if you use ∂~r∂x×∂~r∂z

you get−N~

The vector element of area isdS~=ˆendS= (−

∂y

∂x

ˆe 1 +eˆ 2 −∂y

∂z

ˆe 3 )dxdzTake the dot

product of both sides of this equation with ˆe 2 and show

|ˆen·ˆe 2 |dS=dxdz, or dS=

dxdz

|ˆen·ˆe 2 |

Here the absolute value sign is used to insure that the element of surface areadSis

positive.

Solutions Chapter 7