Let's circulate smartly
True or False?
There exists a smooth, simple, closed, positively oriented curve C in R 2 such that ∮ C ( 3 x 2 y + 4 y 3 ) d x + 1 2 x d y = 2 0 1 8
(from a recent test on Vector Calculus)
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1 solution
Yes, this looks very good; thank you! You have certainly done most of the work.
But, as you say, a bit more needs to be done. You are examining circles centered at the origin only, and it is conceivable (and it does in fact happen) that you get larger circulations for some other curves (but always less than 2 0 1 8 ). I trust that you or somebody else will be able to fill this gap.
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Well, Sir, even I reached the surface integral and the squared terms simply told me to first check for a circle..........Well, that worked just like Steven Sir..........but since this was an MCQ, I just instinctively went for the answer after this...........I think that I would have to learn more to complete the proof............(I have only done Stewart's Calculus, so I only have a VERY BASIC knowledge of multivariable calculus)
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Well, you guys were lucky that the circle worked. If I had picked a lower value for the integral, for example, 35 rather than 2 0 1 8 , your approach would have led you astray. Did you study the multivariable part of Stewart's Calculus? If so, you are ready to tackle this problem; all you need are double integrals to finish the job. The key question is: what is the largest possible value of ∫ ∫ D ( 1 2 − 3 x 2 − 1 2 y 2 ) d A , not just for disks centered at the origin but for any region D in the plane. Think about it!
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@Otto Bretscher – Hahaa yes.........I should have solved it more "Smartly"!!!!
@Otto Bretscher – I think I've got it now. I have updated the solution
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@Steven Chase – Yes exactly... very good! The exact value of the maximal circulation is 1 2 π , well below 2 0 1 8 . Thank you!
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@Otto Bretscher – Thank you as well. I'm glad I have the luxury of doing these interesting problems without a time limit :)
@Otto Bretscher – Ohhhh I see, yes.........there was a similar question to this in the Additional problems where we had to find the maximum value of a line integral instead of the double integral.................OHhhh.......Shucks......Why did it not strike me earlier..!!!! It is simply when the double integral is calculated over the ellipse, and the value comes out to be 12*pi........!! Thanks for reminding me, Sir........
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@Aaghaz Mahajan – Don't be too harsh on yourself, Comrade!
Recall Stoke's Theorem:
∫ C F ⋅ d ℓ = ∫ ∫ S ∇ × F ⋅ d S
Curl of the vector field:
F = ( 3 x 2 y + 4 y 3 , 1 2 x , 0 ) ∇ × F = ( 0 , 0 , 1 2 − 3 x 2 − 1 2 y 2 )
Surface integral:
∫ ∫ S ∇ × F ⋅ d S = ∫ ∫ S ∇ × F d x d y = ∫ ∫ S ( 1 2 − 3 x 2 − 1 2 y 2 ) d x d y
What is the maximum possible value of this integral? Obviously (or perhaps not so obviously), the integration region has to be limited to where the integrand is positive. By inspection, the integration region yielding the maximum integral is therefore the interior of the following ellipse:
4 x 2 + 1 y 2 = 1
The corresponding integral comes out to 1 2 π ≈ 3 7 . 6 9 9 < 2 0 1 8 . There is therefore no curve which satisfies the requirements of the problem.