# VC5.21

Let the given surface S is cube (OCBA,GDEF) which is bounded by planes x=0,y=0,z=0,x=2,y=2,z=2. The given cube meets the xy-plane in the square of OABC where the coordinates of A,B,C are (2,0),(2,2) and (0,2) respectively. Thus the curve C bounding S is square OABC.

Hence,by Stokes'theorm we have $\iint _{S}(\nabla \times F)\cdot ndS=\int _{C}F\cdot dr\,$

=$\int _{C}[(y-z+2)i+(yz+4)-xzk]\cdot (dxi+dyj+dzk)\,$

=$\int _{C}[(y-z+2)dx+(yz+4)dy-xzdz]\,$

=$\int _{C}[(y+2)dx+4dy]\,$ as on C,z=0 and dz=0.

=$\int _{{OA}}[(y+2)dx+4dy+4dy]+\int _{{AB}}[(y+2)dx+4dy]+\int _{{BC}}[(y+2)dx+4dy]+\int _{{CO}}[(y+2)dx+4dy]\,$ --(1)

Along OA,y=0,dy=0 and x varies from 0 to 2. Along AB,x=2,dx=0 and y varies from 0 to 2. Along BC,y=2,dy=0 and x varies from 2 to 0. and along CO,x=0,dx=0 and y varies from 2 to 0.

Therefore,from(1),$\iint _{S}(\nabla \times F)\cdot ndS\,$

=$\int _{{x=0}}^{{2}}2dx+\int _{{y=0}}^{{2}}dy+\int _{{x=2}}^{{0}}4dx+\int _{{y=2}}^{{0}}4dy\,$

=$2[x]_{0}^{2}+4[y]_{0}^{2}+4[x]_{2}^{0}+4[y]_{2}^{0}=4+8-8-8=-4\,$