It's most easily explained if you look at this image (sorry for the bad quality!): hand-drawn diagram

I haven't drawn it too well, the vertex should be at the y-axis.

I'm not sure how to find the area of the shaded region. The height is variable, and I've got an equation for the parabola based on h. The width of each rectangle is 2.5. It's very similar to Riemann sums, but horizontal. I thought about using a series, but I'm not sure how.

Any ideas?


EDIT: I should have been more specific. I do know how to find the total area of the shaded region for a specific height, say h=36. However, I'm trying to generalize a function f(h) where I plug in h, and it spits out the total shaded area.

  • $\begingroup$ If you like interchange the roles of $x$ and $y$. Then recognizing the Riemann sum will be easier. Without interchange, the curve has equation of shape $y=b-ax^2$, so our rectangles have area width ($w$, $2.5$ in your case) times $\sqrt{(b-y_i)/a}$, where the $y_i$ are $w$, $2w$, and so on until we can't take the square root any more. If this is enough, good. Otherwise please type the exact description of the parabola, and I can give a full calculation. $\endgroup$ – André Nicolas Jan 28 '12 at 16:59
  • $\begingroup$ Oops! My comment above produces half the area. $\endgroup$ – André Nicolas Jan 28 '12 at 17:13
  • $\begingroup$ @AndréNicolas Thanks! It did help, but I'm trying to generalize a formula to work out the total shaded region for any value of h. Something like f(h) = total shaded area. $\endgroup$ – user1045696 Jan 29 '12 at 2:14
  • $\begingroup$ You will not find a pleasant closed form for this sum involving square roots. There are, however, very good approximations. $\endgroup$ – André Nicolas Jan 29 '12 at 3:41

Let the function $y(x)$ be defined: $y(x)=-\frac{h}{36^2}x^2+h$

This function is being restricted to the range of $y(x)\leq 0$. The important piece here is determining when $y(x)$ is zero: $$ \begin{align} y(x)&=0\\ \therefore x&=\pm 36 \end{align} $$ (This was included in your photo.) We can now make specifications on $y(x)$. That is, $y(x)$ is defined on the interval $[-36,36]$. The rectangles will be bounded by two points of the form: $$P_i=(x_i,y(x_i)):y(x_i)=2.5i$$ That is, they will be bounded by two points where the $y$ coordinate (of both points) is a multiple of $2.5$. We can thus see $x_i$ in terms of the $y$ coordinate is: $$\begin{align} y(x_i)&=2.5i\\ \therefore x_i&=\pm\sqrt{1296-\frac{3240i}{h}} \end{align}$$ The length of the $i$th rectangle will be the distance between $(x_i,y(x_i))$ and $(-x_i,y(x_i))$ Since the distance is a horizontal line, the length is simply $|2x_i|$. Thus, we come to the following statement for the total area of the rectangles: $$A(r)=5\sum_{i=1}^{r}|x_i| \quad r=\text{ the number of rectangles}$$ Determining $r$ is essentially the main crux here. I believe involving calculus is where this gets us to. Taking the definite integral with respect to $x$ would give us the total area of the parabola (dependent upon $h$): $$\int_{-36}^{36} y(x) dx=\left[\int y(x) dx\right]^{36}_{-36}=\left[\frac{-h}{3888}x^{3}+hx+C \right]^{36}_{-36}=48h$$ To determine an upperbound on $r$, one must establish the $r$ at which $$A(r)<48h$$ is no longer true. Manipulating this algebraically gives us: $$\frac{|x_1|+\dots+|x_r|}{h}=\frac{1}{h^2}\sum_{i=1}^{r}\sqrt{1296h^2-hi}<9.6$$ So, for a given $h$, you can determine $1$) the upperbound on the amount of rectangles and $2$) the total area of all the rectangles (presuming you use the upperbound as the amount of rectangles).

I can't see how to take this any further, so this is the best answer I think I can provide.


Let me call the height of each rectangle $2.5$. Note that the top of the bottommost rectangle intersects the $y$-axis at $y=2.5$. Note that the top right corner of this rectangle coincides with the point on the parabola with coordinates, say, $( x_1, 2.5)$. So, half the width of this rectangle is $x_1$ and can be obtained by finding the positive solution to: $$ 2.5= {-h\over 36^2}\cdot x_1^2+h $$

This gives $$\tag{1}x_1=\sqrt{36^2( h-2.5)\over h }.$$

So, the width of the bottommost rectangle is $2x_1$, and its area is $(2.5)\cdot 2x_1=5x_1$.

The top of the second rectangle from the bottom intersects the $y$-axis at $y=5$ and the top right corner of this rectangle coincides with the point on the parabola with coordinates, say, $( x_2, 5)$. You may argue as above to find its width (note that to find half the width, you may just take equation (1) and replace "2.5" by "5" and $x_1$ with $x_2$"), then its area.

Continue for each rectangle and then take the sum. Note that the number of rectangles is the integer part of $h/2.5$

  • $\begingroup$ The issue isn't finding the area of the rectangles, it's generalizing a formula where I plug in h, and it gives me the total area? $\endgroup$ – user1045696 Jan 29 '12 at 1:46
  • $\begingroup$ @user1045696 It would be $\sum\limits_{k=1}^{IP(h/2.5) } 5\cdot\sqrt{36^2(h-2.5*k)\over h }$. I haven't thought of whether this can be simplified... $\endgroup$ – David Mitra Jan 29 '12 at 1:51
  • $\begingroup$ What's IP mean? Also, at the top of the sigma, is there any way to indicate that h/2.5 should be rounded down? Cheers $\endgroup$ – user1045696 Jan 29 '12 at 2:08
  • $\begingroup$ @user1045696 The "IP" means "round 5/6 down" :) (IP=integer part; note this isn't standard notation...) $\endgroup$ – David Mitra Jan 29 '12 at 2:12
  • $\begingroup$ Oh I see. Thanks! Do you know if there is any standard notation? $\endgroup$ – user1045696 Jan 29 '12 at 2:15

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