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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $x$ or $y$. Then find the area of the region bounded by $x+y^2= 6$ and $x+y=0$

this is what i got but it i wrong $6\cdot 3 - 3^3/3 - 3^2/2 - 6\cdot (-2) - (-2)^3/3 - (-2)^2/2$

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    Walk me through the steps you took to get to ((6*3)-((3)^3/3)-((3^2)/2))-(6*-2)-((-2)^3/3)-(-2^2)/22011-07-30
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    @marc: I assumed that your (-2^2)/2 in the last term was supposed to be ((-2)^2)/2, i.e., $(-2)^2/2$; let me know if I was wrong.2011-07-30
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    Also, I think the points of intersection are (0,0) and (-1/36, 1/6)... I may be wrong tho2011-07-30
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    +1 to Silver's first comment. (To borrow from the younger ones' language,) Walkthroughs are just as important in figuring out where a calculation went wrong as in figuring out a game you're playing.2011-07-30
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    @Silver: No, they’re $(-3,3)$ and $(2,-2)$.2011-07-30
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    Can someone explain how you got those points of intersection? Sry for being noob... But isn't it just a downward sloping line with a right opening parabola with a vertex at the origin...? (I just jumped on this question cuz I was like, "Finally! An easy question I can actually answer!")2011-07-30
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    @silver: you read, as I did, "x+y^2= 6 x+y=0" as $x+y^2=0$ and $6 x+y=0$, while Brian spotted it was meant to be $x+y^2=6$ and $x+y=0$.2011-07-30
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    ooooooooooooooooooooooooooooooooh, gotcha, thanks a lot Henry!2011-07-30

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