VC4.27

From Example Problems
Jump to: navigation, search

Let OX,OY,OZ be the coordinate axes and i,j,k be the unit vectors along these axes.

Let S1,S2,S3,S4,S5,S6 be the faces of the given rectangular parallelopiped.

Therefore,\iint _{S}F\cdot ndS=\iint _{{S_{1}}}F\cdot ndS+\iint _{{S_{2}}}F\cdot ndS+\iint _{{S_{3}}}F\cdot ndS+\iint _{{S_{4}}}F\cdot ndS+\iint _{{S_{5}}}F\cdot ndS+\iint _{{S_{6}}}F\cdot ndS\, --(1)

For the face S1,x=a,n=i \iint _{{S_{1}}}F\cdot ndS=\int _{{y=0}}^{{a}}\int _{{z=0}}^{{a}}(2yxi-yzj+x^{2}k)\cdot idydz\, where x=a.

=\int _{{y=0}}^{{a}}\int _{{z=0}}^{{a}}(3ya)dydz=2a[{\frac  {y^{2}}{2}}]_{0}^{a}\cdot [z]_{0}^{a}=a^{4}\, --(2)

For the face S2,x=0,n=-i. \iint _{{S_{2}}}F\cdot ndS=\int _{{y=0}}^{{a}}\int _{{z=0}}^{{a}}(2yxi-yzj+x^{2}k)\cdot (-i)dydz=0\, where x=0 --(3)

For the face S3,y=a,n=j \iint _{{S_{3}}}F\cdot ndS=\int _{{x=0}}^{{a}}\int _{{z=0}}^{{a}}(2yxi-yzj+x^{2}k)\cdot jdxdz\, where y=a =\int _{{x=0}}^{{a}}\int _{{z=0}}^{{a}}(-az)dxdz=-a[{\frac  {z^{2}}{2}}]_{0}^{a}\cdot [x]_{0}^{a}=-{\frac  {a^{4}}{2}}\, --(4)

For the face S4,y=0,n=-j. \iint _{{S_{4}}}F\cdot ndS=\int _{{x=0}}^{{a}}\int _{{z=0}}^{{a}}(2yxi-yzj+x^{2}k)\cdot (-j)dxdz=0\, where y=0 --(5)

For the face S5,z=a,n=k. \iint _{{S_{5}}}F\cdot ndS=\int _{{y=0}}^{{a}}\int _{{x=0}}^{{a}}(2yxi-yzj+x^{2}k)\cdot kdxdy\, where x=a \int _{{y=0}}^{{a}}\int _{{x=0}}^{{a}}x^{2}dxdy=[{\frac  {x^{3}}{3}}]_{a}^{a}\cdot [y]_{0}^{a}={\frac  {a^{4}}{3}}\, --(6)

For the face S6,z=0,n=-k. \iint _{{S_{6}}}F\cdot ndS=\int _{{y=0}}^{{a}}\int _{{x=0}}^{{a}}(2yxi-yzj+x^{2}k)\cdot (-k)dxdy\, where z=0 -\int _{{y=0}}^{{a}}\int _{{x=0}}^{{a}}x^{2}dxdy=-[{\frac  {x^{3}}{3}}]_{0}^{a}\cdot [y]_{0}^{a}=-{\frac  {a^{3}}{3}}\, --(7)

Using(2),(3),(4),(5),(6)and(7),(1) reduces to

\iint _{S}F\cdot ndS=a^{4}-{\frac  {1}{2}}a^{4}+{\frac  {1}{3}}a^{3}-{\frac  {1}{3}}a^{3}={\frac  {1}{2}}a^{4}\,

Main Page