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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... – The Chaz 2.0 Dec 14 '11 at 16:02
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. – StudentProgramee Dec 14 '11 at 16:04
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] – The Chaz 2.0 Dec 14 '11 at 16:05
You could also integrate $dy$, using only one integral... – The Chaz 2.0 Dec 14 '11 at 16:06
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. – The Chaz 2.0 Dec 14 '11 at 16:15
up vote 0 down vote accepted

$\int_0^1 \int_{-y}^{y^2} dx \ dy$ 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. – The Chaz 2.0 Dec 14 '11 at 16:27
Thanks, I really only needed the bounds to practice setting up integrals with curves given, but you confirmed that my initial guess was correct – StudentProgramee Dec 14 '11 at 16:35
I was the victim of serial downvoting on 5 January 2012. Ask me anything. – The Chaz 2.0 Feb 14 '12 at 0:17
@TheChaz What happened!?!??! – Pedro Tamaroff Feb 21 '12 at 2:01
@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. – Pedro Tamaroff Feb 21 '12 at 3:41

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