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Flux integral over a parabolic cylinder
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[QUOTE="Forcefedglas, post: 5483597, member: 586793"] [B]Homework Statement [/B] Evaluate ##\int\int_S \textbf{F}\cdot\textbf{n} dS ## where ##\textbf{F}=(z^2-x)\textbf{i}-xy\textbf{j}+3z\textbf{k}## and S is the surface region bounded by ##z = 4-y^2, x=0, x=3## and the x-y plane with ##\textbf{n}## directed outward to S. [B]The attempt at a solution[/B] I've worked out the correct answer but can't seem to fully understand why that is. I tried splitting up the flux integral into 3 separate surfaces: 1 for the parabola at x=3, another for the parabola at x=0, and lastly a parametric surface between them. At each parabola I just evaluated the flux integral in cartesian coordinates, which were ##\int_{-2}^2 \int_0^{4-y^2}(2y\textbf{j}+\textbf{k})\cdot((z^2-x)\textbf{i}-xy\textbf{j}+3z\textbf{k})##, which worked out to be 256/5 and 0, at x=0 and x=3 respectively. I parameterized the parabolic cylinder as ##\mu\textbf{i}+\lambda\textbf{j}+(4-\lambda^2)\textbf{k}##, so the flux integral for this was ## \int_0^2 \int_0^3(2\lambda\textbf{j}+\textbf{k})\cdot((z^2-x)\textbf{i}-\mu\lambda\textbf{j}+(12-3\lambda^2)\textbf{k})d\mu d\lambda## which works out to be 48, which is the correct answer. This might seem like a dumb question but I've been staring at it for hours and can't understand why the value of the flux integral at the x=0 parabola is ignored. I considered the possibility that it excludes the surfaces at x=0 and x=3 but similarly worded questions did not do this. Any help/tips will be appreciated, thanks! [/QUOTE]
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Flux integral over a parabolic cylinder
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