What is the Method for Finding Area Between Curves?

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

I'm confused avout questions 2-3.
The answers for 2-2 is 1
So the answer for 2-3 is $$\frac{1}{3}$$
But, how the area looks like? Because $$ x^2 $$ will be an open curve upside? There's no boundary for above side.

 

[fresh_42]

Can you show us how you got to $x_0=1$ as intersection point? What did you get for the line equation $s(x)= m_{2.1}x +\dfrac{3}{2}$ and what for the tangent $t(x)=m_t + b_t\,?$ What is the slope $m_t$ of the tangent and where does it intersect the $y-$axis at $b_t\,$?

$s(x)$ should bound the open parabola from above. Try to draw your results.

 

[Helly123]

@fresh_42 ok.
L1 = line tangent to curve x^2
L2 = line orthogonal to L1

Gradient of L1 is the same as the curve that is 2x

L2 go through (0,3/2) and the point of L1 tangent to curve, we still don't know, let's refer it as (x, y ).
(x, y) on the curve too so it makes y = x^2. So the point is (x, x^2)

The tangent of L2 is orthogonal to tangent L1 that is -1/2x
L2 go through (0,3/2) and (x, x^2)

Using gradient = (y-y1/(x-x1)
I get x = 1, y = 1
So L2 = mx + c
When c = 3/2 and gradient = -1/2

For integral the boundaries are 0 and 1
The result is 1/3

 

[Helly123]

$s(x)$ should bound the open parabola from above. Try to draw your results.

Do you mean the boundaries are y coordinates? But the question is respected to x

 

[fresh_42]

@fresh_42 ok.
L1 = line tangent to curve x^2
L2 = line orthogonal to L1

Gradient of L1 is the same as the curve that is 2x

L2 go through (0,3/2) and the point of L1 tangent to curve, we still don't know, let's refer it as (x, y ).
(x, y) on the curve too so it makes y = x^2. So the point is (x, x^2)

The tangent of L2 is orthogonal to tangent L1 that is -1/2x
L2 go through (0,3/2) and (x, x^2)

Using gradient = (y-y1/(x-x1)
I get x = 1, y = 1
So L2 = mx + c
When c = 3/2 and gradient = -1/2

For integral the boundaries are 0 and 1
The result is 1/3

Yes, so $S_1=\frac{1}{3}$. The second area $S_2$, is now bounded by the $y-$axis and the line $L_2(x)=-\frac{1}{2}x+\frac{3}{2}$ which goes from $(0,\frac{3}{2})$ to $(3,0)$. The parabola crosses this area, so it's bounded from above. The other two vertices are $(0,0)$ and $(1,1)$. It looks like a piece of pie. Question 2-3 now asks to compute it's area $S_2$ and question 2-4 to calculate the ration between them. What is it, that you do not understand?

 

[fresh_42]

Do you mean the boundaries are y coordinates? But the question is respected to x

The boundaries are $x=0$ which is the $y-$axis, the line $L_2$ and the parabola. I would integrate the parabola from $x=0$ to $x=1$, the triangles from $x=0$ to $(3,0)$ and $x=1$ to $(3,0)$ and do some additions. Of course you can do it as well in one step.

 

[Helly123]

Yes, so $S_1=\frac{1}{3}$. The second area $S_2$, is now bounded by the $y-$axis and the line $L_2(x)=-\frac{1}{2}x+\frac{3}{2}$ which goes from $(0,\frac{3}{2})$ to $(3,0)$. The parabola crosses this area, so it's bounded from above. The other two vertices are $(0,0)$ and $(1,1)$. It looks like a piece of pie. Question 2-3 now asks to compute it's area $S_2$ and question 2-4 to calculate the ration between them. What is it, that you do not understand?

What I don't get is, how the area of S1(x) looks like?

I'm sorry, i cannot rotate it..

 

[fresh_42]

This is the area $S_2$ we are looking for, you only have to bound it from above by $L_2(x)=-\frac{1}{2}+\frac{3}{2}$ which goes from left to right and from $(0,\frac{3}{2})$ to $(3,0)$. $S_1$ is below the parabola, not above, but we already know $S_1=\frac{1}{3}$.

 

[Helly123]

This is the area $S_2$ we are looking for, you only have to bound it from above by $L_2(x)=-\frac{1}{2}+\frac{3}{2}$ which goes from left to right and from $(0,\frac{3}{2})$ to $(3,0)$. $S_1$ is below the parabola, not above, but we already know $S_1=\frac{1}{3}$.

View attachment 217874

So for example, we have to find integral of x^2 + 1 with vertices 1, 2
The area is below the parabola?

 

[fresh_42]

$S_1$ is below (outside, green) the parabola, $S_2$ is above (inside, red). We already know that $S_1=\dfrac{1}{3}$.
Now there are three possibilities to calculate $S_2$:

• in three steps with the help of two triangles and $S_1$
• in two steps with $\int_0^1 L_2(x)dx$ and $S_1$
• directly in one step with the difference function of $L_2$ and the parabola.

I don't see why you move the parabola upwards with new boundaries. If you know the formula of the area of triangles, you can use the first method, if you like to integrate, then the other two are fine. The second explicitly uses the previous result of $S_1$ and the third calculates everything new in one step. It's used here, that integrals as well as areas can be added: $\int f(x)dx \pm \int g(x)dx = \int (f(x) \pm g(x)) dx$.

Of course one has to be cautious as an integral describes an oriented volume (area), i.e. it can be negative whereas an area is not, but this doesn't play a role here. I only mention it, to prevent others from over-correcting me, and for you to be aware of the difference between an area and an integral.

 

[Helly123]

$S_1$ is below (outside, green) the parabola, $S_2$ is above (inside, red). We already know that $S_1=\dfrac{1}{3}$.
Now there are three possibilities to calculate $S_2$:

• in three steps with the help of two triangles and $S_1$
• in two steps with $\int_0^1 L_2(x)dx$ and $S_1$
• directly in one step with the difference function of $L_2$ and the parabola.

I don't see why you move the parabola upwards with new boundaries. If you know the formula of the area of triangles, you can use the first method, if you like to integrate, then the other two are fine. The second explicitly uses the previous result of $S_1$ and the third calculates everything new in one step. It's used here, that integrals as well as areas can be added: $\int f(x)dx \pm \int g(x)dx = \int (f(x) \pm g(x)) dx$.

Of course one has to be cautious as an integral describes an oriented volume (area), i.e. it can be negative whereas an area is not, but this doesn't play a role here. I only mention it, to prevent others from over-correcting me, and for you to be aware of the difference between an area and an integral.

i am sorry, i uploaded wrong picture. Is this right? I understand your explanation. If the area is negative (below x coordinates) there are moment necessary to give negative sign.

 

[fresh_42]

Yes, that has been the function and integral in your previous post. I don't understand where it comes from and what it should be. I read question 2-3 as follows (although the word "implies" is a bit strange here):

"Letting $S_2$ be the value of the region with the line (*), this parabola and the line $x=0$ implies $S_2=[{2-4}]$"

Letting $S_2$ be the value of the region $\leftrightarrow$ area
with the line (*) $\leftrightarrow$ $L_2(x)=y= -\dfrac{1}{2} \cdot x+\dfrac{3}{2}$, i.e. the red line in my picture
this parabola $\leftrightarrow$ $p(x)=y=x^2$. i.e. the blue curve in my picture
and the line $x=0$ $\leftrightarrow$ i.e. the $y-$axis
implies $S_2=[{2-4}]$ $\leftrightarrow$ i.e. the red area labeled with $S_2$ in my picture

There are several ways to calculate $S_2$ (see post #9). As we already know the area of $S_1$ (green area in my picture labeled with $S_1$) it might be easiest to use $S_1$ and some triangles and additions / subtractions. But as an area you might as well write it as an integral $S_2= \int_0^1 f(x)dx$ where $f(x)$ is composed by the line $L_2(x)$ and the parabola $p(x)$.

 

[Helly123]

Yes, that has been the function and integral in your previous post. I don't understand where it comes from and what it should be. I read question 2-3 as follows (although the word "implies" is a bit strange here):

"Letting $S_2$ be the value of the region with the line (*), this parabola and the line $x=0$ implies $S_2=[{2-4}]$"

Letting $S_2$ be the value of the region $\leftrightarrow$ area
with the line (*) $\leftrightarrow$ $L_2(x)=y= -\dfrac{1}{2} \cdot x+\dfrac{3}{2}$, i.e. the red line in my picture
this parabola $\leftrightarrow$ $p(x)=y=x^2$. i.e. the blue curve in my picture
and the line $x=0$ $\leftrightarrow$ i.e. the $y-$axis
implies $S_2=[{2-4}]$ $\leftrightarrow$ i.e. the red area labeled with $S_2$ in my picture

There are several ways to calculate $S_2$ (see post #9). As we already know the area of $S_1$ (green area in my picture labeled with $S_1$) it might be easiest to use $S_1$ and some triangles and additions / subtractions. But as an area you might as well write it as an integral $S_2= \int_0^1 f(x)dx$ where $f(x)$ is composed by the line $L_2(x)$ and the parabola $p(x)$.

Sorry, i meant as an example that to find the area of x^2 + 1. So, the parabola i drew is right?

 

[fresh_42]

Sorry, i meant as an example that to find the area of x^2 + 1. So, the parabola i drew is right?

Yes.

And as a side note for what I've meant by an oriented volume: In your example you get $\int_1^2 (x^2+1)dx= \frac{10}{3}$ and $\int_2^1 (x^2+1)dx= -\frac{10}{3}$ although the area is the same, namely $|\,\int_2^1 (x^2+1)dx \,| = + \frac{10}{3}$. The integral distinguishes the direction in which the area is measured. It is called orientation or oriented volume (here area).

 

[Helly123]

$S_1$ is below (outside, green) the parabola, $S_2$ is above (inside, red). We already know that $S_1=\dfrac{1}{3}$.
Now there are three possibilities to calculate $S_2$:

• in three steps with the help of two triangles and $S_1$
• in two steps with $\int_0^1 L_2(x)dx$ and $S_1$
• directly in one step with the difference function of $L_2$ and the parabola.

Three possibilities
the 3rd step "directly in one step with the difference function of L2 and parabola" is the most efficient way for me.

but for the "three steps with two triangles", looks like a lot more work to do.
Do you mean that we first find integral within x = 0 to x = 3 (let's refer it as A)
then, minus A with S1
also, minus A with integral within x = 1 to x = 3 of L2 (let's refer it as B)
so, S2 = A - S1 - B ?

for "two steps with the help of L2 and S1"
is that we find integral of L2 within x = 0 to x = 1 (let's refer it as C)
then we minus C with S1
so, S2 = C - S1 ?

 

[fresh_42]

Three possibilities
the 3rd step "directly in one step with the difference function of L2 and parabola" is the most efficient way for me.

Well, it's the most compact way. Whether it's efficient is a bit of a matter of taste. I find the methods below faster, because we don't have to integrate the $L_2$ part.

but for the "three steps with two triangles", looks like a lot more work to do.
Do you mean that we first find integral within x = 0 to x = 3 (let's refer it as A)
then, minus A with S1
also, minus A with integral within x = 1 to x = 3 of L2 (let's refer it as B)
so, S2 = A - S1 - B ?

Yes, but I won't integrate $L_2$. The area formula for a right triangle is simply half the product of its catheti (legs), so you can more or less immediately tell the number and $S_1$ is already known. Here we have $A=\operatorname{vol}(\{(0,\frac{3}{2}),(0,0),(3,0)\})=\frac{1}{2} \cdot \frac{3}{2} \cdot 3$ and $B=\operatorname{vol}(\{(1,1),(1,0),(3,0)\})= \frac{1}{2} \cdot 1 \cdot 2$. No integration needed.

for "two steps with the help of L2 and S1"
is that we find integral of L2 within x = 0 to x = 1 (let's refer it as C)
then we minus C with S1
so, S2 = C - S1 ?

Yes. Here we calculate $C=\operatorname{vol}(\{(0,\frac{3}{2}),(0,0),(1,0),(1,1)\})$ as area below $L_2$. And again we can integrate $L_2$ or add the square with length $1$ to the area of the triangle above the square. I find those elementary geometric figures easier than an integral, but as I said, it's a matter of taste.

I hope I got all the coordinates of the points correct.

 

[Helly123]

yes, you are right. that way is easier. I didn't realized the triangles exactly before.

Yes. Here we calculate $C=\operatorname{vol}(\{(0,\frac{3}{2}),(0,0),(1,0),(1,1)\})$ as area below $L_2$. And again we can integrate $L_2$ or add the square with length $1$ to the area of the triangle above the square.
.

we can add triangle with square also then to minus S1. to find S2 , now the others possibilities are quite effective.

 

[fresh_42]

yes, you are right. that way is easier. I didn't realized the triangles exactly before.

we can add triangle with square also then to minus S1. to find S2 , now the others possibilities are quite effective.

There is one advantage of the geometric approach over the integral: in case the entire figure is intersected by the x-axis, i.e. with areas below and above the x-axis, we have to be very cautious with the integrals, as areas below the x-axis yield negative results and areas above positive. So we'll have to make sure, in case we are interested in a positive area, that nothing cancels out if written in one integral. Here it doesn't make a difference, since all areas are in a nicely behaving sector.

 

[Helly123]

There is one advantage of the geometric approach over the integral: in case the entire figure is intersected by the x-axis, i.e. with areas below and above the x-axis, we have to be very cautious with the integrals, as areas below the x-axis yield negative results and areas above positive. So we'll have to make sure, in case we are interested in a positive area, that nothing cancels out if written in one integral. Here it doesn't make a difference, since all areas are in a nicely behaving sector.

Yes, thank you