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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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    ooooooooooooooooooooooooooooooooh, gotcha, thanks a lot Henry!2011-07-30

2 Answers 2

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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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    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?2011-07-30
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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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    @André: [Alternatively...](http://fooplot.com/index.php?&type0=0&type1=2&y0=-x&px1=6-s%5E2&py1=s&smin1=-7&smax1=7&xmin=-12&xmax=12&ymin=-7&ymax=7)2011-07-30