Advanced book on Mathematics Olympiad

Real Analysis 577

while the area element is

dσ=

1

cosα

dxdy,

αbeing the angle formed by the normal to the sphere with thexy-plane. It is easy to see

that cosα=za=

√

a^2 −x^2 −y^2

a. Hence the integral is equal to

− 2

∫∫

D

(

z

x

a

+x

y

a

+y

z

a

)a

z

dxdy=− 2

∫∫

D

(

x+y+

xy

√

a^2 −x^2 −y^2

)

dxdy,

the domain of integrationDbeing the diskx^2 +y^2 −ax≤0. Split the integral as

− 2

∫∫

D

(x+y)dxdy− 2

∫∫

D

xy

√

a^2 −x^2 −y^2

dxdy.

Because the domain of integration is symmetric with respect to they-axis, the second
double integral is zero. The first double integral can be computed using polar coordinates:
x=a 2 +rcosθ,y=rsinθ,0≤r≤a 2 ,0≤θ≤ 2 π. Its value is−πa
3
4 , which is the
answer to the problem.
(D. Flondor, N. Donciu,Algebra ̧ ̆si Analiz ̆a Matematica ̆(Algebra and Mathematical
Analysis), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1965) ̆

529.We will apply Stokes’ theorem. We begin with

∂φ

∂y

∂ψ

∂z

−

∂φ

∂z

∂ψ

∂y

=

∂φ

∂y

∂ψ

∂z

+φ

∂^2 ψ

∂y∂z

−

∂φ

∂z

∂ψ

∂y

−φ

∂^2 ψ

∂z∂y

=

∂

∂y

(

φ

∂ψ

∂z

)

−

∂

∂z

(

φ

∂ψ

∂y

)

,

which combined with the two other analogous computations gives

∇φ×∇ψ=curl(φ∇ψ).

By Stokes’ theorem, the integral of the curl of a vector field on a surface without boundary
is zero.
(Soviet University Student Mathematical Competition, 1976)

530.For the solution, recall the following identity.

Green’s first identity.Iffandgare twice-differentiable functions on the solid region
Rbounded by the closed surfaceS, then
∫∫∫

R

(f∇^2 g+∇f·∇g)dV=

∫∫

S

f

∂g

∂n

dS,

where∂g∂nis the derivative ofgin the direction of the normal to the surface.