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$D=${$(x,y)\in R^2 : x^2+9y^2 \le 1$, $3x+9y^2 \le 1$}

1 ) find the boundary $\vartheta D$ of $D$ , and sketch D. What shoud I do in order to find the boundary of a given domain as :

$D=${$(x,y)\in R^2 : A \le n , B \le n' $} (with A,B inequalities)

2 ) And how do I calculate the flux of the vector field $F(x,y,z) = z^2+y^2+x^2$ coming out of $\vartheta D$ ?

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    $3x+9y^2=1$ is a parabola (sideways) and $x^2+9y^2=1$ is an ellipse. Where do they intersect? – 2017-02-18

1 Answers 1

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$D$ looks like this:

enter image description here

It is the region bounded by the blue curve (on le the left) and the pink one (on the right). Both curves intersect in $(0,-1/3)$ and $(0,1/3)$. Therefore, you can write $D$ as follows: $$ D=\{(x,y)\;|\; -\frac{1}{3} \le y \le \frac{1}{3}, -\sqrt{1-9y^2} \le x \le \frac{1-9y^2}{3} \} $$

and for the boundary $\partial D$ $$ \partial D = \left\{(x,y)\;|\; x=-\sqrt{1-9y^2}, x\in [-1,0] \right\}\cup \left\{(x,y)\;|\; \frac{1-9y^2}{3}, x\in [0,\frac{1}{3}] \right\} $$

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    Can you also help me calculate the Flux coming out of D (the second part of the question) – 2017-02-19
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    $F$ is not a vector field, as you have defined it. I would suggest posting this as a separate question. – 2017-02-20