Advanced book on Mathematics Olympiad

576 Real Analysis

Using the identity

∑

n≥ 1

1

n^2 =

π^2

6 , we obtain

∫∞

0

F(t)dt=

π^3

12

.

(Gh. Sire ̧tchi,Calcul Diferen ̧tial ̧si Integral(Differential and Integral Calculus),

Editura ̧Stiin ̧tifica ̧ ̆si Enciclopedic ̆a, Bucharest, 1985)

526.The integral from the statement can be written as

∮

∂D

xdy−ydx.

Applying Green’s theorem forP(x, y)=−yandQ(x, y)=x, we obtain
∮

∂D

xdy−ydx=

∫∫

D

( 1 + 1 )dxdy,

which is twice the area ofD. The conclusion follows.

527.It can be checked that div

−→

F =0 (in fact,

−→

F is the curl of the vector fieldeyz

−→

i +

ezx

−→

j +exy

−→

k). IfSbe the union of the upper hemisphere and the unit disk in the

xy-plane, then by the divergence theorem

∫∫

S

−→

F ·−→ndS=0. And on the unit disk

−→

F ·−→n =0, which means that the flux across the unit disk is zero. It follows that the

flux across the upper hemisphere is zero as well.

528.We simplify the computation using Stokes’ theorem:

∮

C

y^2 dx+z^2 dy+x^2 dz=− 2

∫∫

S

ydxdy+zdydz+xdzdx,

whereSis the portion of the sphere bounded by the Viviani curve. We have

− 2

∫∫

S

ydxdy+zdydz+xdzdx=− 2

∫∫

S

(z,x,y)·−→ndσ,

where(z,x,y)denotes the three-dimensional vector with coordinatesz, x,andy, while

−→n denotes the unit vector normal to the sphere at the point of coordinates(x,y,z).We

parametrize the portion of the sphere in question by the coordinates(x, y), which range

inside the circlex^2 +y^2 −ax=0. This circle is the projection of the Viviani curve onto

thexy-plane.

The unit vector normal to the sphere is

−→n =

(x

a

,

y

a

,

z

a

)

=

(

x

a

,

y

a

,

√

a^2 −x^2 −y^2

a

)

,