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I have the double integral ∫∫x dA bounded by the curves y=1, y=-x, and y = √x

I drew out the graph, but I'm having trouble determining what the bounds for the integrals are. Is x from -y to y^2, and y from 0 to 1, or is x from 0 to 1 and y from -x to √x (or vice versa)?

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    The square root of x is not x^2! Try again, and I'm sure you'll get it...2011-12-14
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    Whoops, I entered it wrong on wolfram alpha to generate that graph image. On my own paper, I have the right graph, but the question still remains. Removed the graph image to avoid confusion.2011-12-14
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    Once that is sorted out, split the integral into two integrals, at $x =0$. The upper function is "1" throughout, while the lower function is $-x$ on [-1,0] and $\sqrt x$ on (0,1]2011-12-14
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    You could also integrate $dy$, using only one integral...2011-12-14
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    Also, once you figure it out, you are encouraged to *answer your own question*. Obviously this isn't possible in many situations, but I suspect you can take the hints and roll with them.2011-12-14
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    The problem is that I don't believe the top bound for both integrals is 1 as you said, so I'm confused if you're right with your hints or not2011-12-14

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$\int_0^1 \int_{-y}^{y^2} dx \ dy$ is what you would get with horizontal "representative rectangles". So in this case, the inner integration gives:

$\int_0^1 (y^2 - (-y)) dy$, which is the integral I hinted at.

If you want to integrate dx,

$A = \int_{-1}^0 (1 - (-x)) dx + \int_0^1 (1 - \sqrt x) dx$

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    Regarding you doubts that y = 1 is the "upper" function in the *dx* approach, I'd suggest that you redraw the region and see that no other region is bounded by **all three equations.**2011-12-14
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    Thanks, I really only needed the bounds to practice setting up integrals with curves given, but you confirmed that my initial guess was correct2011-12-14
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    I was the victim of serial downvoting on 5 January 2012. Ask me anything.2012-02-14
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    @TheChaz What happened!?!??!2012-02-21
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    @Peter: Hah! Can't quite remember... somebody called me a pinhead or something, then a bunch of my answers were downvoted.2012-02-21
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    @TheChaz Didn't you report it or something? That is so non-scientific to do, seems kind of inquisitive.2012-02-21
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    @Peter, apparently there are measures in place to detect and reverse serial downvoting (you can search meta for the details), but I don't really care. I get on here to help people, crack jokes, and find collaborators for when I become a legitimate mathematician (not necessarily in that order).2012-02-21
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    @TheChaz =) That was my reaction to that. I discovered that many people tend to get a little worked up when discussing maths, kind of looking for the other's mistake but not providing their insight, correction or proof. I guess it's an aspect to work on.2012-02-21