how to set up a? D7.4. (a) Evaluate the closed line integral of H about the rectangular path P1(2, 3, 4) to P2(4, 3, 4) to P3(4, 3, 1) to P4(2, 3, 1) to P1, given H = 3zax − 2x3az A/m. (b) Determine the quotient of the closed line integral and the area enclosed by the path as an approximation to (∇×H)y. (c) Determine (∇×H)y at the center of the area. Ans. 354 A; 59 A/m2; 57 A/m2
do it as a sum of four line integrals. from P1 to P2. P2 to P3. P3 to P4 and P4 to P1. Notice that each path depends on only one coordinate. On P1 to P2 for example dy=0 and dz=0. For this path, we have:\[\int\limits_{2}^{4}12axdx=6a (4^2-2^2)=48a\]Since z=4, constant. This should be enough to figure out the rest , but not sure how you end up with an answer of 354 since the constant a is not given?
I apologize for the confusion, ax and az are unit vectors
parametrize the line ... line can be written in parametric form as r(t) = (initial point)+t(final point), t goes from 0 to 1 eg P1(2, 3, 4) to P2(4, 3, 4) x = 2 + (4-2)t => dx = 2dt y = 3 + 0 t => dy = 0 z = 4 + 0t => dz = 0 dr = i dx + j dy + k dz change everything to t's .. take a dot product .. and evaluate it.
Oh, that makes much more sense. Yeah, assuming you are doing the work/tangential line integral and not the normal/flux one, that should work.
you shouldn't need to parameterize the curve on this one. For instance, the work integral from P1 to P2 is:\[\int\limits_{2}^{4}<3z,0,-2x^3><dx,dy,dz>\]Since z=4 (constant) and dz=0 on this path, we get and integral in x alone:\[\int\limits_{2}^{4}12dx=12(4-2)=24\]
I'm assuming that \[ax=\hat{i}\] and,\[az=\hat{k}\]in the above.
The next integral would be:\[\int\limits_{4}^{1}-2x^3dz=128\int\limits_{1}^{4}dz=384\]
From P3 to P4, we have:\[\int\limits_{4}^{2}3dx=-6\]
And from P4 back to P1 we get:\[\int\limits_{1}^{4}-2(2)^3dz=-16(4-1)=-48\]
Summing the individual integrals gives us the total line integral along this rectangle. 24+384-6-48=354
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