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I have to find the region $E$ where

$$ \iiint_E (1-x^2 - 2y^2 - 3z^2) dV $$

has maximum value, but I'm not sure how to start.

I was thinking of getting the derivative of the integral and then finding the extrema with the usual $f_x = 0, f_y = 0$ equation system, but even if I get the second derivative of it, I'm stuck with a single integral and I'm not sure if that's right.

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What is the interval over which $\int (1-x^2) dx$ is a maximum? What is the surface over which $\int (1-x^2-2y^2) dx dy$ is a maximum? Can you now see what the region $E$ should be? –  user17762 May 14 '12 at 22:57
    
Um, so the best region for the first integral would be from 0 to 1, the best region for the second integral would be from 0 to 1 and 0 to $\frac{1}{\sqrt{2}}$, so I suppose the last integral should be maximized with the region $((x,y,z) : 0 \leq x \leq 1, 0 \leq y \leq \frac{1}{\sqrt{2}}, 0 \leq \ z \leq \frac{1}{\sqrt{3}})$. Is that right? –  user1002327 May 14 '12 at 23:06
    
Not quite. For the first, the interval is $[-1,1]$ i.e. $(1-x^2) \geq 0$. For the second, the surface is $x^2 + 2y^2 \leq 1$ i.e. $1 - x^2 - 2y^2 \geq 0$. Can you see why? –  user17762 May 14 '12 at 23:07
    
Oh, you're right, I was thinking of integrating the integral and not the function. Yes, the best interval for the first one should be [-1, 1]. And for the second one, yes, it's a paraboloid, so that region is the ellipse which encloses the positive part of the surface. –  user1002327 May 14 '12 at 23:11
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Exactly. I guess now it should be clear what $E$ should be? And feel free to provide your own answer and accept it. You even get a badge for it. :) –  user17762 May 14 '12 at 23:13
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1 Answer 1

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Maximizing the integral means integrating over positive integrands. This means the region that maximizes the integral is

$$ E = \left\{ (x,y,z) : 1 -x^2 - 2y^2 -3y^2 \geq 0\right\} \\ = \left\{ (x,y,z) : x^2 + 2y^2 + 3y^2 \leq 1 \right\} $$

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