Answer to Question #140207 in Physics for Jayshree

Question

test the divergence theorem for the function v= xyî+2yzj+3zxk Take as your volume the cube with side of length 1

Answer 1

$$div\ \vec{v}=y+2z+3x$$

The volume integral of it over the volume of the cube is equal to

$$V=\smallint^1_0 dx\smallint^1_0 dy \smallint^1_0 dz (y+2z+3x)\\=\smallint^1_0 dx\smallint^1_0 dy (y+1+3x)\\=\smallint^1_0 dx(1.5+3x)=3$$

$$V_1=\smallint^1_0 dx\smallint^1_0 dy (3x)=1.5\\V_2=\smallint^1_0 dx\smallint^1_0 dz (2z)=1\\\\V_3=\smallint^1_0 dy\smallint^1_0 dz (y)=0.5$$

$$V_1+V_2+V_3=1.5+1+0.5=3=V$$

The surface integral across the surface of the cube is equal to the sum of the surface integrals across each of the 6 faces of the cube, and is thus equal to the volume integral of it over the volume of the cube.