Finding Curl from a vector field picture

sjrrkb
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Homework Statement


I need to analyze these pictures for my homework and find out the curl of the vector field at the point (red) on the picture.


Homework Equations



http://i1242.photobucket.com/albums/gg525/sjrrkb/ScreenShot2012-11-26at61615PM.png

The Attempt at a Solution


basically I can calculate curl and divergence and evaluate them mathematically...but when it comes to analyzing a picture and determining the curl at a given point on a vector field I am totally lost. Any assistance...in the form of answers, places I can go to get help, or just suggestions would be appreciated.
 
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For the k component of curl, imagine a loop in the xy plane around the point in question, now imagine a line integral around the loop, what is the net result? Consider this for each axis and you should get close. You'll need to be careful with convention, to decide on +-.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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