Why Is My Line Integral Calculation Incorrect?

AI Thread Summary
The discussion centers on the calculation of a line integral for the vector function v=(y², 2x(y+1), 0) from point a=(1,0,0) to point b=(2,2,0), with the correct answer being 11. The original poster attempted to parameterize the path using x=1+t and y=1+t, leading to an incorrect integral result of 6. Participants pointed out that the parameterization was flawed, as it did not correctly represent the starting point a. Alternative parameterizations were suggested, including x=t and y=2t-2, to ensure the correct path is followed. The discussion highlights the importance of accurate parameterization in line integral calculations.
asdf1
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Calculate the line integral of the function v=(y的平方, 2x(y+1), 0) from the
point a=(1,0,0) to the point b=(2,2,0)
The correct Ans: 11

However, when I was calculuating the problem,
I supposed(is that the right word?) to make the parameter equations,

x=1+t
y=1+t
z=0
where 0<t<1
but when I integrated the function, the answer is 6.
 
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i don't know why, but i see your function as some chinese characters, can you retype it in some other way:).
 
Are you sure about that parametrization of your path?

I'd say

\left\{ \begin{array}{l}<br /> x = t \\ <br /> y = 2t - 2 \\ <br /> z = 0 \\ <br /> \end{array} \right

or

\left\{ \begin{array}{l}<br /> x = t + 1 \\ <br /> y = 2t \\ <br /> z = 0 \\ <br /> \end{array} \right
 
asdf1 said:
Calculate the line integral of the function v=(y的平方, 2x(y+1), 0) from the
point a=(1,0,0) to the point b=(2,2,0)
The correct Ans: 11

However, when I was calculuating the problem,
I supposed(is that the right word?) to make the parameter equations,

x=1+t
y=1+t
z=0
where 0<t<1
but when I integrated the function, the answer is 6.

when t= 0, your parametrization gives (1, 1, 0), there is no value of t that gives (1, 0, 0).
 
sorry, i mistyped~
a=(1,0,0)
 
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