SigepBrandon
New member
- Joined
- Feb 17, 2011
- Messages
- 39
The question reads let \(\displaystyle \overrightarrow{F}(X,Y,Z)=\frac{Z^{2}}{X}\overrightarrow{i}+\frac{Z^{2}}{Y}\overrightarrow{j}+2Z*LN(XY)\overrightarrow{k}\)
evaluate
\(\displaystyle \int_{c}\overrightarrow{F}\cdot d\overrightarrow{s}\)
where C is the path of straight line segments from P=(1,2,1) to Q=(4,1,7) to R=(5,11,7) and then back to P.
my approach was to show that F is conservative and closed, therefore, the result is zero. So I took the grad of F which came out to be -
\(\displaystyle < \frac{z(2x-z)}{x^{2}}, \frac{z(2x-z)}{y^{2}}, 2(\frac{z}{x}+\frac{z}{y}+ln(xy))>\)
The curl does turn out to be zero, but my question is:
The grad is defined for the path chosen, but undefined for x,y less than or equal to 0, and since a conservative function is path independent, I'm not sure If i can make this argument.. any help or ideas would be greatly appreciated.
thanks,
bs.
evaluate
\(\displaystyle \int_{c}\overrightarrow{F}\cdot d\overrightarrow{s}\)
where C is the path of straight line segments from P=(1,2,1) to Q=(4,1,7) to R=(5,11,7) and then back to P.
my approach was to show that F is conservative and closed, therefore, the result is zero. So I took the grad of F which came out to be -
\(\displaystyle < \frac{z(2x-z)}{x^{2}}, \frac{z(2x-z)}{y^{2}}, 2(\frac{z}{x}+\frac{z}{y}+ln(xy))>\)
The curl does turn out to be zero, but my question is:
The grad is defined for the path chosen, but undefined for x,y less than or equal to 0, and since a conservative function is path independent, I'm not sure If i can make this argument.. any help or ideas would be greatly appreciated.
thanks,
bs.