- #1
I dun get it
- 17
- 0
Homework Statement
\int_{C}(x+yz)dx + 2xdy + xyzdz
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)
x=1+t, y=3t, z=1; 0\leq t \leq 1
x'(t)=1, y'(t)=3, z'(t)=0
\int^{1}_{0}(1+t+3t)*1dt + 2(1+t)*3dt + 0
=\int^{1}_{0}1+4t+6+6tdt
[7t+5t^{2}]^{1}_{0}=12
For C going from (2,3,1) to (2,5,2)
x=2, y=1+2t, z=t; 1\leq t \leq 2
x'(t)=0, y'(t)=2, z'(t)=1
\int^{2}_{1}0 + 2*2*2dt + 2(1+2t)tdt
=\int^{2}_{1}8+2t+4t^{2}dt
[8t+t^{2}+\frac{4}{3}t^{3}]^{2}_{1}=\frac{55}{3}
Total C is 12+\frac{55}{3} = \frac{91}{3}
However, the answer in the book says 97/3, so I'm not sure what I did wrong
