VC4.27

From Exampleproblems

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, $\iint_S F\cdot n dS=\iint_{S_1}F\cdot n dS+\iint_{S_2}F\cdot n dS+\iint_{S_3}F\cdot n dS+\iint_{S_4}F\cdot n dS+\iint_{S_5}F\cdot n dS+\iint_{S_6}F\cdot n dS\,$ --(1)

For the face S1,x=a,n=i $\iint_{S_1}F\cdot n dS=\int_{y=0}^{a}\int_{z=0}^{a}(2yxi-yzj+x^2k)\cdot i dy dz\,$ where x=a.

= $\int_{y=0}^{a}\int_{z=0}^{a}(3ya)dy dz=2a[\frac{y^2}{2}]_0^a \cdot[z]_0^a=a^4\,$ --(2)

For the face S2,x=0,n=-i. $\iint_{S_2}F\cdot n dS=\int_{y=0}^{a}\int_{z=0}^{a}(2yxi-yzj+x^2k)\cdot(-i)dy dz=0\,$ where x=0 --(3)

For the face S3,y=a,n=j $\iint_{S_3}F\cdot n dS=\int_{x=0}^{a}\int_{z=0}^{a}(2yxi-yzj+x^2k)\cdot j dx dz\,$ where y=a = $\int_{x=0}^{a}\int_{z=0}^{a}(-az)dx dz=-a[\frac{z^2}{2}]_0^a\cdot[x]_0^a=-\frac{a^4}{2}\,$ --(4)

For the face S4,y=0,n=-j. $\iint_{S_4}F\cdot n dS=\int_{x=0}^{a}\int_{z=0}^{a}(2yxi-yzj+x^2k)\cdot(-j)dx dz=0\,$ where y=0 --(5)

For the face S5,z=a,n=k. $\iint_{S_5}F\cdot n dS=\int_{y=0}^{a}\int_{x=0}^{a}(2yxi-yzj+x^2k)\cdot k dx dy\,$ where x=a $\int_{y=0}^{a}\int_{x=0}^{a} x^2 dx dy=[\frac{x^3}{3}]_a^a\cdot[y]_0^a=\frac{a^4}{3}\,$ --(6)

For the face S6,z=0,n=-k. $\iint_{S_6}F\cdot n dS=\int_{y=0}^{a}\int_{x=0}^{a}(2yxi-yzj+x^2k)\cdot(-k) dx dy\,$ where z=0 $-\int_{y=0}^{a}\int_{x=0}^{a} x^2 dx dy=-[\frac{x^3}{3}]_0^a\cdot[y]_0^a=-\frac{a^3}{3}\,$ --(7)

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

$\iint_S F\cdot n dS=a^4-\frac{1}{2}a^4+\frac{1}{3}a^3-\frac{1}{3}a^3=\frac{1}{2}a^4\,$

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