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I am trying to find the area of a shaded area for the equations $x = y^2 - 2$ and $x = e^y$

The range in from y = -1 to y = 1

This tells me that I need to find

$$\int_{-1}^{1}y^2 - 2$$ and $$\int_{-1}^{1}e^y$$

$$\int_{-1}^{1}y^2 - 2$$ is $y^3/3 - 2y$ = $-10/3 $ Which is wrong so I am guessing I am suppose to do absolute value so I magic it to $10/3$

$$\int_{-1}^{1}e^y$$ is $e^y$ = $2.71... - .3678...$

Which is obviously wrong and so is the answer I get.

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What do you want to compute? Two integrals? Or the area between the curves $y = 1$, $y=-1$, $x = y^2 - 2$ and $x = e^y$? –  Patrick Da Silva Apr 28 '12 at 16:26
    
I have the picture already, it is provided by the horrible person named Stewart in his book. He cuts it off from y = -1 to y =1 –  user138246 Apr 28 '12 at 16:29
    
Oops! My apologies: I misread one of the curves. Ignore my previous comment. You will indeed be integrating from $y=-1$ to $y=1$. –  Brian M. Scott Apr 28 '12 at 16:31
    
I have never voted anything down and won't start to do so, but in my view such a lousy question is an insult to the community. –  Christian Blatter Apr 28 '12 at 17:14
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@Christian That isn't at all constructive or helpful. If you want to be an ass I am in chat, you can make fun of me all you want. –  user138246 Apr 28 '12 at 17:20
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1 Answer

Divide the region into horizontal slices. At each value of $y$ between $-1$ and $1$, the slice at $y$ extends from $y^2-2$ on the left to $e^y$ on the right, so its length is $$\text{right}-\text{left}=e^y-(y^2-2)=e^y-y^2+2\;,$$ and its area is $$dA=(e^y-y^2+2)\,dy\;,$$ its length times its (vertical) width $dy$. Thus, you should be calculating $$\int_{-1}^1(e^y-y^2+2)\,dy=\left[e^y-\frac{y^3}3+2y\right]_{-1}^1\;.$$ Can you finish it from there?

Added: You asked how I knew to subtract the parabola from the exponential. Think about calculating the area of a horizontal strip of width $dy$ taken at some particular value of $y$: I need to multiply the width $dy$ by the length of the strip. The strip runs between two $x$-coordinates, $y^2-2$ and $e^y$, so its length is the larger of these numbers minus the smaller.

Digression: How long is a line segment that runs from $x=-1$ to $x=3$? The segment runs from $-1$ up to $0$, for a length of $1$, and then from $0$ to $3$, for an additional length of $3$, making a total length of $4$. It’s not coincidence that $3-(-1)=4$: the length is always the bigger $x$-coordinate minus the smaller one. In this example you can see that subtracting $-1$ has the effect of adding to $3$ the length of the part of the segment that’s to the left of $0$, which is exactly what’s needed to make it come out right.

Back to the problem: I know that $y^2-2<e^y$, because between $y=-1$ and $y=1$ the parabola lies entirely to the left of the exponential, so it always gives the smaller $x$-coordinate. (For that matter, I also know it because $y^2-2<0$ when $-1\le y\le 1$, and $e^y$ is always positive.) Thus, I know that the length of a skinny rectangle at height $y$ must be $e^y-(y^2-2)$, and its area must be $\big(e^y-(y^2-2)\big)dy$.

Note that if the integration had been with respect to $x$, so that I had vertical strips of width $dx$, I’d subtract the bottom $y$-coordinate from the top one. It’s exactly the same idea: I’m getting the length of a strip by subtracting the small coordinate from the large one.

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Actually I am confused, I have no idea how you got that formula. To be it looks like I want to add both the areas since they are bounded by the axis and y= -1 and y= 1. –  user138246 Apr 28 '12 at 16:48
    
@Jordan: It’s much simpler in the long run to treat it as a single area, bounded on the left by $x=y^2-2$ and on the right by $x=e^y$. When you divide that region into very skinny horizontal rectangles, the rectangle at height $y$ runs from a lefthand endpoint of $y^2-2$ to a righthand endpoint of $e^y$. To get its length, you just subtract the lefthand $x$-coordinate from the righthand one, as I did in the first displayed line. It’s just like getting the length of a vertical strip by subtracting the bottom $y$-coordinate from the top one. –  Brian M. Scott Apr 28 '12 at 16:53
    
I don't quite understand this method, and to further complicate things I don't know what dA is. –  user138246 Apr 28 '12 at 16:55
    
@Jordan: $dA$ is the infinitesimal area of one slice, its length times its width. You can get a quick overview of the relevant formulas here, though unfortunately without an explanation of where they came from. –  Brian M. Scott Apr 28 '12 at 17:01
    
@Jordan: The best pictures that I can quickly find explaining the idea are here. –  Brian M. Scott Apr 28 '12 at 17:08
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