# Area betwixt two graphs

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• Apr 10th 2010, 10:47 PM
Archduke01
Area betwixt two graphs
Consider the area between the graphs http://euler.vaniercollege.qc.ca/web...168a3bf431.png and http://euler.vaniercollege.qc.ca/web...6e724df531.png. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals
where http://euler.vaniercollege.qc.ca/web...ea9131f391.png , http://euler.vaniercollege.qc.ca/web...7973f0e8c1.png, http://euler.vaniercollege.qc.ca/web...61d577bdc1.png and
http://euler.vaniercollege.qc.ca/web...8785a207f1.png
http://euler.vaniercollege.qc.ca/web...77bbb02e81.png

***

First off, to find a, b and c I believe we have to find the intercepts, and to do that we need to make the equations equal to each other and solve for x. But I'm having problems because of the y values when trying to isolate them.

$\displaystyle x + 3y = 36$
$\displaystyle y = - \frac {x}{3}+12$

$\displaystyle x+4=y^2$
$\displaystyle y = \sqrt{x+4}$
I'm stuck there. The square root poses a problem for me.

$\displaystyle - \frac{x}{3} + 12 = \sqrt{x+4}$

How can I isolate and solve for x?
• Apr 10th 2010, 10:59 PM
dwsmith
Square:

$\displaystyle \big(\sqrt{x+4}\big)^2=\big(12-\frac{x}{3}\big)^2$
• Apr 10th 2010, 11:32 PM
Archduke01
Quote:

Originally Posted by dwsmith
Square:

$\displaystyle \big(\sqrt{x+4}\big)^2=\big(12-\frac{x}{3}\big)^2$

$\displaystyle x+4 = - \frac {x^2}{9} + 144$
$\displaystyle \frac {x^2}{9} + x - 140 = 0$

Do I use the quadratic formula to solve for x? I have a feeling this is wrong because it only gives me 2 values for x and I'm looking for 3.
• Apr 10th 2010, 11:59 PM
mr fantastic
Quote:

Originally Posted by Archduke01
$\displaystyle x+4 = - \frac {x^2}{9} + 144$ Mr F says: The right hand side here is totally wrong.

Most problems in calculus can be traced back to poor skills in basic algebra. You are advised to take much greater care with your work and to thoroughly review all the pre-calculus material you're expected to know at the calculus level.

$\displaystyle \frac {x^2}{9} + x - 140 = 0$

Do I use the quadratic formula to solve for x? I have a feeling this is wrong because it only gives me 2 values for x and I'm looking for 3.

A simple sketch of the two graphs (with the required area shaded - something you should have done as the very first step) shows that two solutions are expected. Why on Earth do you expect three?
• Apr 11th 2010, 04:46 AM
Archduke01
Quote:

Originally Posted by mr fantastic
A simple sketch of the two graphs (with the required area shaded - something you should have done as the very first step) shows that two solutions are expected. Why on Earth do you expect three?

Because there are 3 unknown values; a, b, and c...
• Apr 11th 2010, 05:20 AM
skeeter
as Mr. F stated, a sketch of the graph will reveal what is required ...

you may be better served by integrating w/r to y to find this area.

$\displaystyle A = \int_c^d f(y) - g(y) \, dy$
• Apr 11th 2010, 08:06 AM
Archduke01
Quote:

Originally Posted by Archduke01
Because there are 3 unknown values; a, b, and c...

Finding the 2 solutions gives me answers for 2 values ... so how do I find the third one?
• Apr 11th 2010, 09:14 AM
dwsmith
Quote:

Originally Posted by Archduke01
Finding the 2 solutions gives me answers for 2 values ... so how do I find the third one?

By third one, do you mean the vertex of of the parabola?
• Apr 11th 2010, 09:23 AM
Archduke01
Quote:

Originally Posted by dwsmith
By third one, do you mean the vertex of of the parabola?

Ah, so that's what it is! Thank you. Up until now, I've never had to find the vertex to calculate the area ... is it because one of the graphs are not a function?
• Apr 11th 2010, 09:29 AM
dwsmith
By your integral setups, you are integrating over x; therefore, you need the x coordinates. Since you need xs, you need to know the the vertex.

Or you could integrate over like skeeter suggested over y. Then you will only need the y coordinates.

Did you ever get the foiling process right?

• Apr 11th 2010, 09:33 AM
Archduke01
Quote:

Originally Posted by dwsmith
By your integral setups, you are integrating over x; therefore, you need the x coordinates. Since you need xs, you need to know the the vertex.

Or you could integrate over like skeeter suggested over y. Then you will only need the y coordinates.

Did you ever get the foiling process right?

Oh indeed I corrected that, I got 60 and 21 as 2 of my values, which are correct.

To find the vertex it's -b/2a. Our function is $\displaystyle y^2 = (x+4)$ ... How would I proceed to find the vertex? Is there a different formula?

And yes I know I could integrate w/r to y, but the first part of the question asks me to do it the way portrayed in the topic title, and the 2nd part asks me to find the area using integration w/r to y.
• Apr 11th 2010, 09:36 AM
dwsmith
Since this parabola is situated on the x axis, you can set y=0 and obtain the vertex.
• Apr 11th 2010, 09:49 AM
Archduke01
$\displaystyle y^2 = (x+4)$

$\displaystyle -b/2a = -4/2 = -2$

Which is wrong... the answer is supposed to be -4, but how? This is a straightforward procedure.

EDIT; I see what you mean now, with setting y = 0, but why is it that the formula failed? Shouldn't they both have worked?
• Apr 11th 2010, 09:51 AM
dwsmith
This is a horizontal parabola. You need to solve this equation.

$\displaystyle x=y^2-4$

Just pretend x is y and solve that.

Hint: there is no b.
• Apr 11th 2010, 09:54 AM
AllanCuz
Quote:

Originally Posted by Archduke01
Consider the area between the graphs http://euler.vaniercollege.qc.ca/web...168a3bf431.png and http://euler.vaniercollege.qc.ca/web...6e724df531.png. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals
where http://euler.vaniercollege.qc.ca/web...ea9131f391.png , http://euler.vaniercollege.qc.ca/web...7973f0e8c1.png, http://euler.vaniercollege.qc.ca/web...61d577bdc1.png and
http://euler.vaniercollege.qc.ca/web...8785a207f1.png
http://euler.vaniercollege.qc.ca/web...77bbb02e81.png

***

First off, to find a, b and c I believe we have to find the intercepts, and to do that we need to make the equations equal to each other and solve for x. But I'm having problems because of the y values when trying to isolate them.

$\displaystyle x + 3y = 36$
$\displaystyle y = - \frac {x}{3}+12$

$\displaystyle x+4=y^2$
$\displaystyle y = \sqrt{x+4}$
I'm stuck there. The square root poses a problem for me.

$\displaystyle - \frac{x}{3} + 12 = \sqrt{x+4}$

How can I isolate and solve for x?

I'm not sure that what you have above is correct. In fact the area between 2 functions is the larger function - the smaller function over some bound (either x or y)

So I would write

$\displaystyle A=\int_{b}^a g(y) - h(y)dy$

Let us re-arrange your equations such that

$\displaystyle x= 36 - 3y$

$\displaystyle x= y^2 - 4$

Find the intersection of these points, i.e. when y satisfies

$\displaystyle y^2 +3y -40 = 0$

I hope I expanded that right of Mr. Fantastic is going to jump on me!

Then simply integrate from the bounds that you find.

A little tip, if you notice the above is a nice quadratic function! We can then find the roots by factoring.

$\displaystyle y^2 +3y -40 = 0 = (y-5)(y+8)$
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