Finding Area between Parabolas

**NewUser01** · July 4th 2012, 07:06 AM

The question is:

A shape has been formed by three intersecting parabolas, the equations of these parabolas are:

1. $y=-x^2-6x-9$
2. $y=x^2-4.5$
3. $y=-x^2+6x-9$

Using calculus, find the area of the shape formed by the intersecting parabolas.

My attempt:

Well, firstly the graph of them:

I'm guessing that the question means to find that area in the middle underneath the red curve. Well, that's the assumption I've made in answering the question. Anyways, firstly is some info that isn't present on the graph that isn't hard to work out. The x-intercepts of the lower parabolas are -3 and 3. The point of intersection that the lower parabolas have with the middle one is at x=-1.5 and x=1.5. The x-intercepts of the red parabola are $\pm\sqrt{4.5}$ .

I've worked out the area of that shape to be 3.022 sq. units. Firstly I found the area above the lower parabolas for $x=-3$ to $x=3$ . I did this by finding the area from one of the x-intercepts to zero and then just doubling it. This yielded in 18.

Then I found the area bounded by the x-axis for the red parabola (the middle one). The area was 12.728 sq units.
Then I found the area from one of the x-intercepts to the point that the two parabolas touch ( $x=\pm1.5$ ). This gave a value of 1.125 sq units. So then I worked out the area of the shape as:

$Area=18-12.728-2(1.125)$
$Area=3.022$

Is this approach of finding the area correct? Thanks in advanced.

**Reckoner** · July 4th 2012, 07:38 AM

Originally Posted by NewUser01

Is this approach of finding the area correct? Thanks in advanced.

It doesn't have to be that complicated. You can find the area using two integrals. We know the lower parabolas intersect the upper parabola at $\textstyle x=\pm\frac32,$ and the lower parabolas intersect each other at $x=0.$ Thus, the area of the middle region is simply

$A = \int_{-3/2}^0\left[\left(x^2 - \frac92\right) - \left(-x^2-6x-9\right)\right]\,dx + \int_0^{3/2}\left[\left(x^2 - \frac92\right) - \left(-x^2+6x-9\right)\right]\,dx$

And due to symmetry, we have

$A = 2\int_0^{3/2}\left[\left(x^2 - \frac92\right) - \left(-x^2+6x-9\right)\right]\,dx$

$= 2\int_0^{3/2}\left(2x^2 - 6x + \frac92\right)\,dx$

$= 2\left[\frac23x^3 - 3x^2 + \frac92x\right]_0^{3/2}$

$= 2\left(\frac94 - \frac{27}4 + \frac{27}4 - 0\right)$

$= \frac92.$

Your approach should work, but I'm not sure how you determined the area "from one of the x-intercepts to the point that the two parabolas touch" to be 1.125.

**skeeter** · July 4th 2012, 07:49 AM

using symmetry ...

$A = 2\int_0^{1.5} (x^2-4.5) - (-x^2+6x-9) \, dx = 4.5$

**Plato** · July 4th 2012, 07:55 AM

This $2\int_0^{1.5} {[{x^2} - 4.5 + {{\left( {x - 3} \right)}^2]}dx}$ will also give the answer.

**NewUser01** · July 4th 2012, 09:16 PM

Wow, that approach is much easier. What was I thinking? Thanks for the help everyone.

Question solved.

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