Begin2.DVI

I7-38. (continued)

Method II Use spherical coordinatesx= 3 sinθcosφ, y= 3 sinθsinφ, z= 3 cosθ

dS=3dθ3 sinθdφ, F~·eˆn=

x^2 +y^2

3

I=

∫∫

S

(3)(x^2 +y^2 ) sinθdθdφ=

∫π/ 2

θ=0

∫ 2 π

φ=0

(3)(9 sin^2 θcos^2 φ+ 9 sin^2 θsin^2 φ) sinθ dθdφ

I=27

∫π/ 2

0

∫ 2 π

0

sin^3 θ dθdφ= 36π

I7-40. G=x+y−2 = 0andD^2 =F(x,y) =x^2 +y^2 one finds

gradF=∇F= 2xˆe 1 + 2yeˆ 2 andgradG=∇G=ˆe 1 +ˆe 2 Let

~r=xˆe 1 +yˆe 2 denote position vector to point on circle,

then d~rdx = ˆe 1 +dxdyˆe 2 is tangent to circle. Differentiate

equation of circle to show 2 x+2ydydx= 0ordydx=−xy, then

one can writed~rdx=ˆe 1 −xyˆe 2 is tangent to circle. Observe thatd~rdx·∇F= 2x+ 2y−yx= 0

which shows∇F is perpendicular tod~rdxor∇Fis perpendicular to line. The slope of

the line is− 1 and the vectors~r 1 = 2ˆe 1 , ~r 2 = 2ˆe 2 point to points on the line so that

~α=~r 1 −~r 2 is a direction vector of the line. Note that~α·∇G= (2ˆe 1 − 2 ˆe 2 )·(ˆe 1 +ˆe 2 ) = 0

so that∇Gis perpendicular to line.

The quantityD^2 is a minimum when the circle just touches the line and at this

point of contact the normal to the circle is∇Gand this vector is also perpendicular

to the line and has the same direction as∇F. Consequently, there exists a scalarλ

such that∇F+λ∇G= 0at this point of contact.

HereH=F+λG=x^2 +y^2 +λ(x+y−2)with

∂H

∂λ

=x+y−2 = 0 constraint equation

∂H

∂x

=2x+λ= 0

∂H

∂y

=2y+λ= 0

where the last two equations are a necessary condition for H to have an extreme

value. Solve the above system of equations and showx= 1, y= 1, λ=− 2 so that the

minimum distance from the origin to the line is

√

2.

Solutions Chapter 7