I am trying to compute a triple integral to find the volume of the region bounded by the following curves,

$$ x^2 + 2y^2=2 $$ $$ x+y+2z=2 $$ $$ z=0 $$

How will I go about finding the integration limits for my triple integral of this bounded region?

My thinking for this problem was the following. Since I know that $z=0$, then from region $2$, I put $z=0$, and then I get that $x+y=2$ , and then solving for one of the variables, I get that $x = 2 - y$ and $y = 2 - x$. Then I can substitute this into the first equation, but then I get confused in which way I need to order the integration limits, and how to approach this problem. What are the limits for $z$ , $y$ , and $x$. Also based on the integration limits, which way will I integrate, $dxdydz$ , $dxdzdy$ , $dydxdz$ , $dydzdx$, $dzdxdy$, $dzdydx$ ?


Since $x+y=2$ does not intersect the elipse $x^2+2y^2=2 $ in $Oxy$, the bounds are

$-1 \le y \le 1$

$-\sqrt{2-2y^2} \le x \le \sqrt{2-2y^2}$

$0 \le z \le (2-x-y)/2$

The order of integration is $dzdxdy$

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  • $\begingroup$ When computing the integrals, I get a negative value, and volume can't be negative. $\endgroup$ – Viktor Raspberry Nov 7 '17 at 4:16

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