My Math Forum Flux of the vector field through the surface

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 August 5th, 2017, 12:44 AM #1 Newbie   Joined: Aug 2017 From: Germany Posts: 1 Thanks: 0 Flux of the vector field through the surface Hello! I need help with determining the flux of the vector field. I'm enclosing link to my math.stackexchange question. https://math.stackexchange.com/quest...gh-the-surface
 August 5th, 2017, 03:23 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 To begin with, you are mistaken about the surface. It is not any part of a sphere. Looking at just the first octant, the surface consists of straight lines connecting points on the two given semi-circles. We can take, as parametric equations for the surface, y= z= s, for $0\le s\le 1$, and, for each s, x lies on the line between 0 and $\sqrt{1- s^2}$: $x= t\sqrt{1- s^2}$ for $0\le t\le 1$. We can write that as the vector function $\vec{V}= t\sqrt{1- s^2}\vec{i}+ s\vec{j}+ s\vec{k}$. Differentiating with respect to s, $\vec{V_s}= -\frac{2st}{\sqrt{1- s^2}\vec{i}+ \vec{j}+ \vec{k}$. Differentiating with respect to t, $\vec{V_t}= \sqrt{1- s^2}\vec{i}$. The "vector differential of surface area" is given by the cross product of those vectors, $\sqrt{1- s^2}$(\vec{j}- \vec{k}\right)dsdt\$. To find the flux of the given vector function through that surface, take the dot product of the vector function, written in terms of s and t, and the vector differential of surface area and integrate. Last edited by Country Boy; August 5th, 2017 at 03:34 AM.

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