2 dz dy 0 d 1 2 dz a x dza z 2 a x a z dz 106 for

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2 dz ,dy = 0 d 1 =− 2 dz a X + dza z ¿ (− 2 a X + a z ) dz
106. For each of the ff pairs of points obtain the equation for the straightline passing through the points:a.) (1,2,0) and (3,4,0), b) (0,0,0) and (2,2,-1) 0 0
C 1 = 1, C 2 = 0 Equation is y = x + 1, z = 0. (b) dx 2 0 = dy 2 0 = dz 1 0 dx 2 = dy 2 =− dz 1 2 x = 1 2 y + C 1 =− z + C 2 0 = 0 + C 1 = 0 + C 2 C 1 = C 2 = 0 Equation is x = y =− 2 z . (c) dx 3 1 = dy 2 1 = dz 4 1 ¿ 2 = Φ 3 = dz 3 1 2 x = 1 3 y + C 1 = 1 3 z + C 2 1 2 = 1 3 + C 1 = 1 3 + C 2 C 1 = 5 6 ,C 2 = 1 6 1 2 x = 1 3 y + 5 6 = 1 3 z + 1 6 Equation is 3 x + 2 y = 5, 3 x 2 z = 1. 107. Convert into Cartesian coordinates each of the ff. a.) (2, 5 π 6 , 3 ¿ in cylindrical coordinates; b.) (4, 4 π 3 , 1 ¿ ) ) ¿ )
108. Convert into cylindrical coordinates the ff points specified in Cartesian coordinates: (a.) (-2,0,1);b.)(1,- 3 , 1 ¿
Solution: 2 2 2
110. Convert into spherical coordinates the ff points specified in Cartesian coordinates (a.) (-2,0,1), b.) (-3, 3 , 2 ¿ Solution: 2 4 2
1 + sinπt x ( ¿ ) ¿ ¿ 2 ¿ 1 cos ( πt ) ¿ ¿ ¿ 0 ¿ T ( x , y , z,t ) = T ¿ +4 z 2 where To is constant. Find the shapes of the constant – temperature surfaces for each of the ff values: a.) t = 0, b.) t = 0.5s and c)t = 1s Solution : (a) T ( x, y ,z , 0 )= T 0 ( x 2 + 4 z 2 ) Constant temperature surfaces are given by ( x 2 + 4 z 2 )= ¿ constant, which are elliptic cylinders. 112. Using the divergence theorem, find the surface integral of the vector field (xax +yay + zaz)over each of the following closed surface (a) the surface of a cube of sides 1m; (b) the surface of a cylinder of radius 1/√π m and length 2 m, and (c) the surface of a sphere of radius 1/(π) 1/3 m. S
Reference: Elements of Engineering Electromgnetics by Nannapaneni Rarayana Rao
113. T ( x, y ,z , 1 )= T 0 ( x 2 + 16 y 2 + 4 z 2 ) Constant temperature surfaces are given by ( x 2 + 16 y 2 + 4 z 2 )= ¿ constant, which are ellipsoids.

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