question 7 math 1

1
Q7 - div. V= 2x + 1 + 2z 00 0 2 π 22rcosθ 2 rcosθ +1+2 zdzrdr dθ 00 2 π 2 2 r 2 cosθz +rz +rz 2 2rcosθ 0 dr dθ 00 2 π 2 2 r 2 cosθ ( 2rcosθ)+r ( 2rcosθ ) +r ( 2rcosθ ) 2 dr dθ 00 2 π 2 4 r 2 cosθ 2 r 3 co s 2 θ+2 rr 2 cosθ+ 4 r4 r 2 cosθ + r 2 co s 2 θ dr dθ 00 2 π 2 6 r2 r 3 cos 2 θr 2 cosθ +r 2 co s 2 θdrdθ 0 2 π 3 r 2 r 4 2 co s 2 θr 3 3 cosθ+ r 3 3 cos 2 θ 2 0 0 2 π 12 8 cos 2 θ8 3 cosθ+ 8 3 cos 2 θdθ 0 2 π 12 4( 1 +cos2 θ )− 8 3 cosθ+ 4 3 ( 1 +cos2 θ ) ( 12 8 3 θ4 3 sin 2 θ8 3 sinθ ) 2 π 0 = ( 24 π16 3 π00 ) ¿ 56 π 3

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Math 1

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Page 1: Question 7 Math 1

Q7-

div. V= 2x + 1 + 2z

∭00 0

2π 22−rcosθ

2rcosθ+1+2 z dz rdr dθ

∬0 0

2π 2

2 r2 cosθz+rz+r z2 2−rcosθ0

dr dθ

∬0 0

2π 2

2 r2 cosθ(2−rcosθ)+r (2−rcosθ )+r (2−rcosθ )2dr dθ

∬0 0

2π 2

4 r2 cosθ−2r3 co s2θ+2 r−r2cosθ+4 r−4 r2 cosθ+r2co s2θdr dθ

∬0 0

2π 2

6 r−2r3 co s2θ−r2 cosθ+r2 co s2θdr dθ

∫0

3 r2− r4

2cos2θ− r

3

3cosθ+ r

3

3co s2θ2

0dθ

∫0

12−8co s2θ−83cosθ+ 8

3co s2θdθ

∫0

12−4(1+cos2θ)−83cosθ+ 4

3(1+cos2θ )dθ

(12−83θ−4

3sin 2θ−8

3sinθ)2π

0=(24 π−16

3π−0−0)

¿ 56π3