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# Vector field

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Consider the vector field
F(x,y)= (-yi+xj)/(x^2+y^2)

Question1)Show that F is the gradient of the polar angle function teta(x,y)=arctan(y/x) defined over the right half-plane x>0 .

Question2)Suppose that C is a smooth curve in the right half-plane x>0 joining two points :
A:(x1,y1) and B(x2,y2).Express "integral(F.dr)"on C, in terms of the polar coordinates (r1,teta1) and (r2,teta2) of A and B.

QUESTION3)Compute directly from the definition of the line integrals:
"integral(F.dr)" on C1 where C1 is the upper half of the unit circle running from (1,0) to (-1,0);
and "integral(F.dr)" on C2 where C2 is the lower half of the same unit circle.

QUESTION4)Since F=Grad(teta) at any point of the plane where vector F is defined (not just in the right half plane x>0), th vector field F ought to be conservative (path-independant).
THIS IS TRUE IN SOME REGIONS, but not in others.
a) Give an example of a region in which vector F is conservative, and justify your answer using the fundamental theorem of calculus for line integrals.

b) Give another example of a region in which F is not conservative, and explain why this does not contradict the fundamental theorem.

##### Solution Summary

This solution provides a vector field and shows that it is the gradient of a given polar angle function, then gives characteristics of a smooth curve and shows how to express the integral in polar coordinates.
It also shows how to compute integrals using the definition of line integrals, and how to give regions where a vector is conservative and not conservative.

Solution provided by:
###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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