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When doing horizontal integration... why do we write the function as a function of y with the x by itself? What is the high level point of that?

So here is my work. Is this right? I am asked to find the area enclosed between these equations:

$$ 4x + y^2 = 12, x = y$$

First thing's first is to graph it and find the points of intersection:

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Point of intersection:

$$ \frac{12 - y^2}{4} = y$$

$$ 12 - y^2 = 4y$$

$$y^2 + 4y -12 = 0$$ $$(y - 2)(y+6) = 0$$ $$y = {2,-6}$$

So the points are $(2,2) , (-6, -6)$.

So the integral...why do we write it as a function of y? Can't we write it as a function of x still?:

$$ \int_{-6}^2 - y \, dy$$ $$= \int_{-6}^2 3 \cdot \frac{-1}{4} y^2 \, dy$$ $$= [ 3y - \frac{1}{4} \cdot \frac{y^3}{3} - \frac{y^2}{2} ]_{-6}^2$$

$$ 6 - \frac{2}{3} - 2 - (-18 + 18 -18) = 21 \frac{1}{3}$$

Does this look right? More importantly, why do we use the equation where y is the variable?

Jwan622
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1 Answers1

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You could write it as a function of whatever you want; as long as you do the integral correctly, either will give you the area between the two curves. However, choosing one variable (in this case, $y$) to set up the integral for is sometimes easier than the other.

Here, for instance, writing this as a function $f(y)$ and integrating with respect to $y$ lets us write this area as a single integral $\int_{-6}^2 f(y)\, dy$. Here your $f(y)$ should be $\frac{12 - y^2}{4} - y$.

However, if we tried to integrate $dx$, we would run into two inconveniences. First, setting it up as a function $g(x)$ of $x$, we would have $y = \pm\sqrt{\text{something}}$, and we would have to make a choice of sign here. Secondly, when trying to write this integral, we have to integrate the top function minus the bottom function right? But these change; from $x = -6$ to $x = 2$, the top function is $y = x$, and the bottom is the $4x + y^2 = 12$ function. However, at $x = 2$, we switch; the area is now enveloped on top by the positive square root $y = \sqrt{12 - 4x}$, and on the bottom by the negative square root.

Thus, in trying to do the $dx$ integral, we would a) have to work with more annoying functions, and b) have to split our integral up into two integrals. It's easier to integrate $dy$!

Chris
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