Question

Test the divergence theorem for the function v = (xy) X+ (2yz) y+ (3zx) Z. Take as your volume the cube shown in Fig. 1.30, with sides of length 2.

Answer 1

$$V=\int_0^2\int_0^2\int_0^2(y+2z+3x)dxdydz\\V=2\int_0^2\int_0^2(y+2+3x)dxdy \\V=2^2\int_0^2(3+3x)dx\\V=48\\ V_1=\int_0^2\int_0^2\int_0^2(y)dydzdx=8\\ V_2=\int_0^2\int_0^2\int_0^2(3x)dxdydz=24\\V_3=\int_0^2\int_0^2\int_0^2(2z)dxdydz=16\\8+24+16=48$$