3.3 Multivariable Differential and Integral Calculus 181
∮Cydx+zdy+xdz=−A√
3 ,
whereAis the area of the disk bounded byC. Observe thatCis the circumcircle of the
triangle with vertices( 1 , 0 , 0 ),( 0 , 1 , 0 ), and( 0 , 0 , 1 ). The circumradius of this triangle
is
√ 6
3 ,soA=2
3 π. The answer to the problem is therefore−2 π√ 3
3. Example.Orthogonal to each face of a polyhedron construct an outward vector with
length numerically equal to the area of the face. Prove that the sum of all these vectors
is equal to zero.
Solution.We exhibit first an elementary solution based on vector operations. Consider
the particular case of a tetrahedronABCD. The four vectors are^12
−→
BC×
−→
BA,^12
−→
BA×
−→
BD,
1
2−→
BD×
−→
BC, and^12−→
DA×
−→
DC. Indeed, the lengths of these vectors are numerically equal to
the areas of the corresponding faces, and the cross-product of two vectors is perpendicular
to the plane determined by the vectors, and it points outward because of the right-hand
rule. We have
−→
BC×−→
BA+
−→
BA×
−→
BD+
−→
BD×
−→
BC+
−→
DA×
−→
DC
=
−→
BC×
−→
BA+
−→
BA×
−→
BD+
−→
BD×
−→
BC+(
−→
BA−
−→
BD)×(
−→
BC−
−→
BD)
=
−→
BC×
−→
BA+
−→
BA×
−→
BD+
−→
BD×
−→
BC+
−→
BA×
−→
BC−
−→
BA×
−→
BD
−
−→
BD×
−→
BC+
−→
0 =
−→
0.
Thisproves that the four vectors add up to zero.
In the general case, dissect the polyhedron into tetrahedra cutting the faces into
triangles by diagonals and then joining the centroid of the polyhedron with the vertices.
Sum up all vectors perpendicular to the faces of these tetrahedra, and note that the vectors
corresponding to internal walls cancel out.
The elegant solution uses integrals. LetSbe the polyhedron and assume that its
interiorVis filled with gas at a (not necessarily constant) pressurep. The force that the
gas exerts onSis
∫∫
Sp−→ndA, where−→n is the outward normal vector to the surface ofthe polyhedron anddAis the area element. The divergence theorem implies that
∫∫Sp−→ndA=∫∫∫
V∇pdV.Here∇pdenotes the gradient ofp. If the pressurepis constant, then the right-hand side
is equal to zero. This is the case with our polyhedron, wherep=1. The double integral
is exactly the sum of the vectors under discussion, these vectors being the forces exerted
by pressure on the faces.