Quick help with line integrals

  • #1

Homework Statement


[tex]\int_{C}(x+yz)dx + 2xdy + xyzdz[/tex]

C goes from (1,0,1) to (2,3,1) and (2,3,1) to (2,5,2)

The Attempt at a Solution



For C going from (1,0,1) to (2,3,1)
[tex] x=1+t, y=3t, z=1; 0\leq t \leq 1[/tex]
[tex]x'(t)=1, y'(t)=3, z'(t)=0[/tex]

[tex]\int^{1}_{0}(1+t+3t)*1dt + 2(1+t)*3dt + 0[/tex]
[tex]=\int^{1}_{0}1+4t+6+6tdt[/tex]
[tex][7t+5t^{2}]^{1}_{0}=12[/tex]



For C going from (2,3,1) to (2,5,2)
[tex] x=2, y=1+2t, z=t; 1\leq t \leq 2[/tex]
[tex]x'(t)=0, y'(t)=2, z'(t)=1[/tex]

[tex]\int^{2}_{1}0 + 2*2*2dt + 2(1+2t)tdt[/tex]
[tex]=\int^{2}_{1}8+2t+4t^{2}dt[/tex]
[tex][8t+t^{2}+\frac{4}{3}t^{3}]^{2}_{1}=\frac{55}{3}[/tex]

Total C is [tex]12+\frac{55}{3} = \frac{91}{3}[/tex]
However, the answer in the book says 97/3, so I'm not sure what I did wrong
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
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Hi I dun get it! :smile:

Your 55/3 should be 8*1 + 1*3 + (4/3)*7 = 8 + 3 + 28/3 = 61/3 :wink:
 

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