Finding the Area Bounded by Two Functions

[gruba]

Homework Statement

Find area bounded by parabola [itex]y^2=2px,p\in\mathbb R[/itex] and normal to parabola that closes an angle [itex]\alpha=\frac{3\pi}{4}[/itex] with the positive [itex]Ox[/itex] axis.

Homework Equations

-Area
-Integration
-Analytic geometry

The Attempt at a Solution

For [itex]p>0[/itex] we can find the normal to parabola such that it closes an angle [itex]\alpha[/itex] with positive [itex]Ox[/itex] axis. For [itex]p<0[/itex] we can't find a normal on parabola for given condition.
Point [itex]A[/itex] in first quadrant where the angle between positive [itex]Ox[/itex] axis and parabola is found by
[itex]y=\sqrt{2px},y'=1\Rightarrow A\left(\frac{p}{2},p\right),p>0[/itex].
Now we can find the line that contains point [itex]A[/itex]:
[itex]y-p=-x+\frac{p}{2}\Rightarrow y=-x+\frac{3p}{2},p>0[/itex].
This is the normal on parabola that satisfies the condition for [itex]\alpha[/itex].
Second point of the normal is found by solving the system:
[itex]y=-x+\frac{3p}{2},y^2=2px\Rightarrow x=\frac{3p}{2}-y\Rightarrow y_1=-3p,y_2=p[/itex]. Point [itex]B[/itex] has to be in the fourth quadrant, so [itex]B=\left(\frac{9p}{2},-3p\right),p>0[/itex].

Now the problem is reduced: Find the area bounded by [itex]y^2=2px[/itex] and [itex]y=-x+\frac{3p}{2}[/itex] where [itex]p>0[/itex] which is the area [itex]AOB[/itex].

How to integrate these functions to get the area?

 

[geoffrey159]

For this kind of problem, you'll need a parametrization of the parabola ##f(t) = ( \frac{t^2}{2p} , t ) ##.
Then you must find ##t## such that the normal vector ##N(t)## forms an at most ##\frac{3\pi}{4}\ \text{mod } \pi## angle with ##\vec i##. If my trigonometry is right (not guaranteed) you must find ## |t| \le p ##.

Then the area under this curve is ##{\cal A} = -\int_{(C)} y \ dx = \int_{-p}^p y(t) x'(t) \ dt ##.

 

[epenguin]

The part after the intersection of the normal is just a triangle, so the area of that is easy.

The part from x = 0 till the intersection involves essentially the integration ∫x^½dx - not all that difficult I hope.

But if it is, or if you prefer, you can turn your graph on its side and flip it, then you are you doing a (definite) integration of y²dy .. Subject that area from the rectangle. The idea of integration by parts, if and when you learn or have learned that, is nothing but this.

 

[gruba]

I am getting the negative result for total area. Here is how I integrated:

[itex]A=A_1-A_2[/itex] where [itex]A_1[/itex] is the area below [itex]y^2=2px,p>0[/itex] on interval [itex][-3p,p][/itex] integrating on [itex]y[/itex] axis.
[itex]A_2[/itex] is the area below [itex]y=-x+\frac{3p}{2},p>0[/itex] on interval [itex][-3p,p][/itex] integrating on [itex]y[/itex] axis.

[itex]\int\limits_{-3p}^p \left(\frac{y^2}{2}-\frac{3p-2y}{2}\right)\mathrm dy[/itex]
[itex]A_1=\int\limits_{-3p}^{p} \frac{y^2}{2}\mathrm dy=\frac{14p^3}{3},p>0[/itex]
[itex]A_2=\int\limits_{-3p}^{p}\frac{3p-2y}{2} \mathrm dy=10p^2[/itex]

[itex]A=A_1-A_2=\frac{2p^2(7p-15)}{2}[/itex]

Why is this wrong? It should be positive for every [itex]p>0[/itex], right?

 

[geoffrey159]

I don't know what is wrong in your calculations because I find difficult to follow what you are thinking. However, if you are open to discussion, you can think of the problem this way:

Parametrize your hyperbola with ##f(t) = (t^2 / 2p, t) ##, which is a nice ##{ \cal C }^1## function. At point ##t##, a tangent vector to the curve is ##f'(t) = (t/p,1)## and a normal vector is ##N(t) = (-1,t/p)##.

Now call ##\alpha(t) = \angle (\vec i, N(t)) ## the angle between the x-axis and the normal to the parabola. You want to find to find ##t## such that ##(3\pi /4 \le \alpha(t) \le \pi) \text{ mod } \pi ##, because the angle between two lines is defined mod ##\pi##.
The cosine function is decreasing on ##[0,\pi]##, so you have ##-1 \le \cos(\alpha(t)) \le -1/\sqrt{2} ##.

Evaluate ##\cos(\alpha(t)) = \frac{\vec i . N(t)}{ ||\vec i || \ || N(t) ||} = \frac{-p}{\sqrt{p^2 + t^2 }}##

and sorting these inequalities, you have ##|t| \le p##.

Now you have the formula for calculating the area of a plane domain enclosed in a curve : ##{\cal A} = -\int_{(C)} y \ dx ##

 

[epenguin]

Yours is somewhat the same idea as I suggested, actually neater if you can carry it off.

I have the feeling that you're making some mistakes that you wouldn't make if you if you drew a diagram of it. In the 'strips' that you are ssumming by integration you should find no negative lengths.

I don't know if there are any other mistakes, but I can see your two terms in the integral are the wrong way round, or wrong signs.

 

[gruba]

Parametrize your hyperbola with ##f(t) = (t^2 / 2p, t) ##, which is a nice ##{ \cal C }^1## function. At point ##t##, a tangent vector to the curve is ##f'(t) = (t/p,1)## and a normal vector is ##N(t) = (-1,t/p)##.

Thanks for the explanation. I tried to understand the method with parametrization, but could not figure it out.
I will however study it further.

I don't know what is wrong in your calculations because I find difficult to follow what you are thinking

In my last post where I explained how I integrated, it is just the basic definition of an integral, which is the area under a curve.
So the area under parabola minus the area under orthogonal line would give the total area.
However, it seems that this gives wrong result because it is negative for some [itex]p>0[/itex].

Could you elaborate on this basic method?

 

[gruba]

I have the feeling that you're making some mistakes that you wouldn't make if you if you drew a diagram of it. In the 'strips' that you are ssumming by integration you should find no negative lengths.
I don't know if there are any other mistakes, but I can see your two terms in the integral are the wrong way round, or wrong signs.

Could you point out the errors you have found?

 

[geoffrey159]

You have a curve ##F(x,y) = 0##, where ##F## is defined by ##F(x,y) = y^2 - 2px ##.
A normal vector to this curve is ##\text{grad } F ## at every point ##(x,y)##.

The same reasoning I tried to explain will apply here: the normal line is directed by the gradient, and the question is at what condition does the angle between the x-axis and the gradient is included in ##[3\pi /4, \pi ]##. You will find ##|y| \le p## which is equivalent to ##0 \le x \le p/2##.

I find in both cases the same answer : ##{\cal A} = \frac{2}{3} p^2 ##

 

[gruba]

I find in both cases the same answer : ##{\cal A} = \frac{2}{3} p^2 ##

So, there is a solution for both [itex]p>0[/itex] and [itex]p<0[/itex]?

 

[geoffrey159]

By definition of a parabola, the parameter is > 0.

 

[epenguin]

Could you point out the errors you have found?

I already did one.

And the limits are wrong I think.

You had a good idea and I'm doing quite well, it is just a matter of careful execution.

Do that diagram and I'm sure you will see these things.

If not, post it up here and we will see if there is any misconception.

 

[gruba]

Do that diagram and I'm sure you will see these things.
If not, post it up here and we will see if there is any misconception.

I will again explain how the easy method should work (but I think I get the wrong area result):

We look at a graph for [itex]p>0[/itex]. From my original post, the coordinates of [itex]A[/itex] and [itex]B[/itex] are [itex]A=\left(\frac{p}{2},p\right),B\left(\frac{9p}{2},-3p\right)[/itex]. We need to find the area [itex]AOB[/itex]. If we rotate coordinate plane by [itex]\pi/2[/itex], we will see that the area below parabola minus the area below orthogonal line will give the area [itex]AOB[/itex].

Integrating along [itex]y[/itex] axis gives area below parabola: [itex]A_1=\int_{-3p}^p \frac{y^2}{2}\mathrm dx[/itex].
Integrating along [itex]y[/itex] axis gives area below orthogonal line: [itex]A_2=\int_{-3p}^p \frac{3p-2y}{2}\mathrm dx[/itex].

Subtracting [itex]A_1[/itex] from [itex]A_2[/itex] gives the result from my previous post.

I still don't understand why this is not working?

 

[epenguin]

The diagram reveals I had maybe misunderstood the question. I understood

Homework Statement

Find area bounded by parabola [itex]y^2=2px,p\in\mathbb R[/itex] and normal to parabola that closes an angle [itex]\alpha=\frac{3\pi}{4}[/itex] with the positive [itex]Ox[/itex] axis.

Which does not seem to me very good English, to mean the area cut off by the parabola, the normal, and the x-axis.

But assuming your understanding is what was meant then your limits are OK, then what do you have shown look to be all OK. And calculating it I get a positive quantity, 2/3 p² (though I easily make errors of calculation). So maybe it is some error in the calculation.)

(Have to close for tonight now).

 

[epenguin]

"Subtracting A₁ from A₂" is the opposite of what you do in #4, I think. Mentioned before.
I see Geoffrey gets same result as I do.

 

[gruba]

"Subtracting A₁ from A₂" is the opposite of what you do in #4, I think. Mentioned before.
I see Geoffrey gets same result as I do.

Could you show how you are getting the result?

 

[geoffrey159]

It bugs me that you have explained things so many times but still, it's not crystal clear for me what you are doing. I can't track your ideas in details and it bugs me. Maybe the problem is in communication of ideas.
Have you ever observed the precision of the staff mentors on this forum? In a couple sentences, they are able to explain what they think very clearly. You can track their thinking, and thanks to their expression, people manage to understand things they couldn't by themselves.
Read a few of them to understand what I mean: Haruspex, Ray Vickson, Krylov, Samy_A, mfb, micromass ... and many others. You should try to immitate the good exemples, it's free :-)

 

[epenguin]

Could you show how you are getting the result?

Easier you tell us the result you have got! If it is just the same as I and geoffrey got except with reversed sign then the explanation surely must be what I have already given.