# Homework7_Solutions.pdf - ECE 333 Professor Amir Arbabi...

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Solutions Homework 7 ECE 333 Professor Amir Arbabi Problem 1 (a) G = ˆ x - yz 2 ˆ y + ( y + 3 z ) ˆ z R G · d l : There are three vectors we need to evaluate G along; h 0 , 0 , - 2 i , h 0 , 1 , 0 i , and h 0 , - 1 , 2 i Starting with h 0 , 0 , - 2 i : d l = - ˆ z ( - dz ) so our integral for the first vector becomes Z G · d l = Z 0 2 G · - ˆ z ( - dz ) (1) = Z 0 2 y + 3 zdz note that here: y = 0 (2) = Z 0 2 3 zdz (3) = 3 2 z 2 0 2 = - 6 (4) Next, h 0 , 1 , 0 i : d l = ˆ y ( dy ) Z G · d l = Z 1 0 G · ˆ y ( dy ) (5) = Z 1 0 - yz 2 dy note that here: z = 0 (6) = 0 (7) h 0 , - 1 , 2 i : Here we’ll want to re-parameterize in terms of some variable, t y = t dy = dt z = 2 - 2 t dz = - 2 dt 1
Solutions Homework 7 ECE 333 Professor Amir Arbabi now, G = ˆ x - t (2 - 2 t ) 2 ˆ y + ( t + 3(2 - 2 t )) ˆ z and d l = ˆ y ( dy ) + ˆ z ( dz ) = ˆ y ( dt ) + ˆ z ( - 2 dt ) so our integral for the final line segment becomes Z G · d l = Z 0 1 G · ( ˆ y dt - 2 ˆ z dt ) = - Z 0 1 t (2 - 2 t ) 2 dt - 2 Z 0 1 ( t + 3(2 - 2 t )) dt = - Z 0 1 4 t - 8 t 2 + 4 t 3 dt - 2 Z 0 1 - 5 t + 6 dt = - 2 t 2 - 8 3 t 3 + t 4 0 1 - 2 - 5 2 t 3 + 3 t 2 0 1 = [2 - 8 3 + 1] + 2[ - 5 2 + 3] = 1 3 + 7 = 22 3 And so our total integral along all three line segments is - 6 + 0 + 22 3 = 4 3 R ( ∇ × G ) · d s : Here, using right hand rule we can determine our vector normal to the surface is ˆ x which means our differential surface is d s = ˆ x dydz . Therefor, for ∇ × G we are only concerned with the ˆ x -component: ( ∇ × G ) x = ˆ x ∂y ( y + 3 z ) - ∂z ( yz 2 ) = ˆ x [1 + 2 yz ] 2
Solutions Homework 7 ECE 333 Professor Amir Arbabi Then our surface integral becomes Z ( ∇ × G ) · d s = Z 2 0 Z 1 - z 2 0 (1 + 2 yz ) dydz = Z 2 0 y + y 2 z 1 - z 2 0 dz = Z 2 0 1 - z 2 + (1 - z 2 ) 2 zdz = Z 2 0 1 - z 2 + z - z 2 + z 3 4 dz = z - z 2 4 + z 2 2 - z 3 3 + z 4 16 2 0 = 2 - 1 + 2 - 8 3 + 1 = 4 3 (b) A = A x ˆ x + A y ˆ y + A z ˆ z B = B x ˆ x + B y ˆ y + B z ˆ z i. ∇ · ( A × B ) = ( ∇ × A ) · B - A · ( ∇ × B ) ( ∇ × A ) · B = ∂A z ∂y - ∂A y ∂z ˆ x + ∂A x ∂z - ∂A z ∂x ˆ y + ∂A y ∂x - ∂A x ∂y ˆ z · B = B x ∂A z ∂y - ∂A y ∂z + B y ∂A x ∂z - ∂A z ∂x + B z ∂A y ∂x - ∂A x ∂y A · ( ∇ × B ) = A · ∂B z ∂y - ∂B y ∂z ˆ x + ∂B x ∂z - ∂B z ∂x ˆ y + ∂B y ∂x - ∂B x ∂y ˆ z = A x ∂B z ∂y - ∂B y ∂z + A y ∂B x ∂z - ∂B z ∂x + A z ∂B y ∂x - ∂B x ∂y 3
Solutions Homework 7 ECE 333 Professor Amir Arbabi ∇ · ( A × B ) = ∇ · (( A y B z - B y A z )

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