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Compute the flux integral for the field $\mathbf{F} = x\mathbf{i} + y\mathbf{j} + 5\mathbf{k}$, where $S$ is the boundary of the cylinder enclosed by $x^2+z^2=1$ and the planes $y=0$ and $x+y=2$.

How many surfaces are there? $S_1:$ base ($y=0$), $S_2:$ body of the cylinder, $S_3:$ top surface, plane $x+y = 2$.

The first two surfaces are ok, but I am stuck with $S_3$. I'm quite sure this integral is incorrect:

Evaluating the integral for $S_3$: plane $x+y=2$.

$\begin{array}{rcl} \displaystyle\iint_S \mathbf{F} \cdot \hat{\mathbf{n}}dS &=& \displaystyle\int_{-1}^1 \displaystyle\int_0^{2-x} \left < 2-y,y,5\right>\cdot \dfrac{\left < 1,1,0 \right >}{\sqrt{2}} dy dx \\ &=&\dfrac{1}{\sqrt{2}}\displaystyle\int_{-1}^1 \displaystyle\int_{0}^{2-x} 2 dydx \\ &=&\dfrac{1}{\sqrt{2}}\displaystyle\int_{-1}^1 2(2-x)dx = \dfrac{1}{\sqrt{2}}\displaystyle\int_{-1}^1 (4-2x) dx \\ &=&\dfrac{1}{\sqrt{2}} (4x-x^2) \bigg\vert_{-1}^1 \\ &=& \dfrac{1}{\sqrt{2}} (4-1 +4 +1) \\ &=& \dfrac{8}{\sqrt{2}}\end{array}$

I am quite sure something is wrong with the bounds of the integral. What should it be?

cgo
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1 Answers1

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Here is the idea: first parametrize the base disc (at $y= 0$) of the cylinder. This is $(u,v)\mapsto (u \cos v, 0, u\sin v)$, with $u \in [0,1]$, $v\in [0,2\pi]$. The top surface $S_3$ is a piece of plane $x+y=2$ cut out by the cylinder. You can look at this piece as the graph of a function over the $xz$-plane, with $y$ as function values. So, $y= 2-x$ becomes $y = 2-u\cos v$, and we get $$ \Sigma: [0,1] \times [0,2\pi] : (u,v)\mapsto (u \cos v, 2-u\cos v, u\sin v). $$

The surface integral becomes $$ \begin{align*} \iint_{S_3} \mathbf{F}\cdot\mathbf{\hat n}\, dS &= \int_0^1 \int_0^{2\pi} (u\cos v, 2-u\cos v, 5)\cdot (u,u,0)\, dv du\\ &= \int_0^1 \int_0^{2\pi} 2u \, dv du\\ &= 2\pi. \end{align*} $$

Ernie060
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