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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)/2 – user13327 Jul 30 '11 at 2:27
@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. – Brian M. Scott Jul 30 '11 at 2:41
+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. – J. M. Jul 30 '11 at 2:52
@Silver: No, they’re $(-3,3)$ and $(2,-2)$. – Brian M. Scott Jul 30 '11 at 2:52
@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 $x+y=0$. – Henry Jul 30 '11 at 3:02

You did essentially everything right.

The parabola and line meet at $y=-2$ and $y=3$. So our area is $$\int_{-2}^3 (6-y^2 -(-y))\,dy$$ It is clear from your answer that you indeed ended up calculating $$\int_{-2}^3 (6-y^2 +y)\,dy$$

An antiderivative is given by $$6y-\frac{y^3}{3}+\frac{y^2}{2}$$

So the answer is $$\left(6(3)-\frac{3^3}{3}+\frac{3^2}{2}\right)-\left(6(-2)-\frac{(-2)^3}{3}+\frac{(-2)^2}{2}\right)$$

Your next to last term has a sign error. You have $-\frac{(-2)^3}{3}$ but when you "open" my brackets above, it should turn to $+$. (Then, when you expand $(-2)^3$, you get a $-$ again! Minus signs are evil.)

Minor Comment: You did not remove all parentheses before integrating. That left another layer of them, and may have contributed to the minus sign slip.

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3 is also incorrect, i tried everything, but thanks for your help everyone – Marc Aoun Jul 30 '11 at 5:01
If you mean that the intersection points are not $-2$ and $3$, either your solution book is mistaken, or the equations have not been given correctly. The picture by @Henry shows the $y$-coordinates of the intersection points clearly, and I calculated the intersection points algebraically, we get a simple quadratic equation in $y$. What ultimate answer does the book give? – André Nicolas Jul 30 '11 at 5:24

Here is the sketch.

2 curves

You can count the squares and part squares to see the answer is a little over 20.

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Nice! Wish I knew how to do that. – André Nicolas Jul 30 '11 at 3:21
@André: I used GeoGebra – Henry Jul 30 '11 at 3:23
@André: Alternatively... – J. M. Jul 30 '11 at 3:38

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