Question #138700
Evaluate both sides of the divergence theorem for the vector field A= 5xy ax + xy² ay + 4z az
defined in the region 1≤ x ≤ 3, −2 ≤ y ≤ 4 and −1≤ z ≤ 2 .
1
Expert's answer
2020-10-16T10:58:43-0400

Solution

Vector field

A=5xyax+xy2ay+4zazA=5xya_x+xy^2 a_y+4za_z

Del operator

=axddx+ayddy+azddz\nabla=a_x \frac{d}{dx}+a_y \frac{d}{dy}+a_z\frac{d}{dz}

Now divergence is written as

.A=d(5xy)dx+d(xy2)dy+d(4z)dz\nabla.A=\frac{d(5xy)}{dx}+\frac{d(xy^2)}{dy}+\frac{d(4z)}{dz}

.A=5y+2x+4\nabla.A=5y+2x+4

Now using divergence theorem

.AdV=.Adxdydz\intop \nabla . A dV=\iiint \nabla.A dx dy dz

=122413(5y+2x+4)dxdydz\int^2_{-1}\int^4_{-2} \intop^3_1(5y+2x+4) dx dy dz

So finally we get

.AdV=408\intop \nabla . A dV=408


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