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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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    @Chri$s$$t$ian That isn't at all constructi$v$e or helpful. If you $w$an$t$ to be an ass I am in chat, you can make fun of me all you $w$an$t$.2012-04-28

1 Answers 1

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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, 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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    One last thing I don't quit get is how you know which curve to subtract from which, to me it looks like the only option to get a valid answer is to add both.2012-04-28