My Math Forum cube flux

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 April 24th, 2015, 07:21 AM #1 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 cube flux calculating the flux of cube length's S, centered at the origin, with the field being: F=x^2(i)+y^2(j)+z^2(k) So the flux of the face of the cube in the (i) direction : F dot dA(i)= x^2*dA...but x at this surface is just S/2( cube's center is at origin) , so it (S^2)/4*dA, and dA is S^2...so its " (S^4)/4 now the negative (i) direction is -x^2*dA, but x is -S/2..So -(S^2)/4*dA... the addition of the flux of two surfaces is 0..if I do the same for the (k) and (j) components. I end of with a grand total of 0 flux.. is the correct?????????
 April 24th, 2015, 01:19 PM #2 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 seems the flux of a cube whos center is at origin is 0 for a lot of vector fields
 April 25th, 2015, 06:56 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 As long as there is no source or sink the flux through any region will be 0. Rather than integrating over each face, it is simpler to use the "Divergence Theorem": $\int\oint \vec{F}\cdot d\vec{S}= \int\int\int \nabla\cdot\vec{F} dV$ Essentially that says that the net flow through the region is equal to the change in the quantitiy in in the region- if more flows in than out, the amount increases, Here, $\vec{F}= x^2\vec{i}+ y^2\vec{j}+ z^2\vec{k}$ so $\nabla\cdot\vec{F}= 2x+ 2y+ 2z$. Find $\int_{-s/2}^{s/2}\int_{-s/2}^{s/2}\int_{-s/2}^{s/2} 2x+ 2y+ dz dzdydx$ Thanks from Kinroh
 April 26th, 2015, 09:41 AM #4 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 thanks. I am starrting to look at greens theorem. But Id like to practice the math of finding the flux without it Im looking at another flux now of a cylinder with its base on the x,y plane with F=*ln(x^2+y^2)...there is only F in radial direction so i dont have to consider the flux of the top and bottom of the cylinder The normal of a cylinder is the radial vector../a, where a is the radius length So F dot N, is (x^2+y^2)*ln(x^2+y^2)/a want to integrate using cylindrical coords, so I want x=acos& , y=asin& pluging this in to the dot product I end up with: a*ln(a^2) dS in cylindrical coords = a*d&*dz so double integral: a^2*ln(a^2)*d&*dz which is 2*Pi*a^2*ln(a^2)*height But according to the text I am looking at the answer is : 4*Pi*a^2*ln(a).... can someone point out my problem? Last edited by Kinroh; April 26th, 2015 at 09:51 AM.
 April 26th, 2015, 09:51 AM #5 Senior Member   Joined: May 2012 Posts: 205 Thanks: 5 Never mind the last post..... The text simplified the ln(a^2) to 2ln(a)... which would yield the same answer I got...

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