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I have trouble in this problem:

Let $S$ be the surface obtained by intersecting $x^2+ y^2 \leq 1$ with $x+2y+z = 5$, and let $F = 2x\mathbf i - z\mathbf j + x\mathbf k$. Suppose that $S$ is given upwards pointing orientation. Then compute the flux of $F$ upwards across $S$.

First, I find that: $x^2+y^2 \leq 1$ is a solid cylinder in $\mathbb R^3$. It is a cylinder with its centre running along the $z$-axis and with radius $1$. Then, $x+2y+z = 5$ is a plane in $\mathbb R^3$, so it is slicing this cylinder with a plane and computing the flux across this slice. But then when computing the flux I got $-8\pi$, which is wrong.

kipf
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Shiloh Dynastie
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1 Answers1

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The plane $x+2y+z=5$ is parametrized by $$ (x,y)\mapsto\pmatrix{x\\y\\5-x-2y}\,,\quad x^2+y^2\le 1\,. $$ As explained here, $\mathbf{n}\,dS$ is $$ \mathbf{n}\,dS=\pmatrix{1\\0\\-1}\times\pmatrix{0\\1\\-2}\,dx\,dy=\pmatrix{1\\2\\1}\,dx\,dy\,. $$ Then the integral is $$ \int_{plane} F\cdot\mathbf{n}\,dS=\int\limits_{x^2+y^2\le 1}(2x-2z+x)\,dx\,dy=\int\limits_{x^2+y^2\le 1}x-2(5-x-2y)\,dx\,dy=-10\,\pi\, $$ because the terms containing $x$ or $y$ in the integrand are zero.

Kurt G.
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