# Area enclosed between y^2=4-x and y=x-2

• Bill_Nye_Fan
In summary: So if I were to set up the integral to find the area enclosed between two curves, would I just subtract the function on the bottom from the function on top?
Bill_Nye_Fan

## Homework Equations

[/B]
Definite integrals and area between two curves

## The Attempt at a Solution

[/B]
Also find attached.

The answer in the back of the book says that part c should be 10/3, but that would mean that (8sqrt (2))/3 would need to cancel out in both area 2.1 and area 2.2, whichdoesn't work because areas are always positive.

Why didn't you calculate the area under the parabola and only subtract the triangle? It seems you used more areas than necessary. But I'm not good at reading on the side.

fresh_42 said:
Why didn't you calculate the area under the parabola and only subtract the triangle? It seems you used more areas than necessary. But I'm not good at reading on the side.
Ah yes that is a much better idea thank you. But still, my method should still be getting the same answer but it's not. Could you please explain to me where I'm going wrong?

Bill_Nye_Fan said:

## Homework Equations

[/B]
Definite integrals and area between two curves

## The Attempt at a Solution

[/B]
Also find attached.

The answer in the back of the book says that part c should be 10/3, but that would mean that (8sqrt (2))/3 would need to cancel out in both area 2.1 and area 2.2, whichdoesn't work because areas are always positive.
View attachment 101868 View attachment 101869
Areas may always be positive, but definite integrals are not required to be positive. To calculate an area using a definite integral sometimes requires some interpretation of what the definite integral means in that case.

http://tutorial.math.lamar.edu/Classes/CalcI/AreaBetweenCurves.aspx

Going off fresh_42's comment, you can also evaluate the definite integral to calculate the area under a curve using dy instead of dx as the variable of integration.

Bill_Nye_Fan said:
Ah yes that is a much better idea thank you. But still, my method should still be getting the same answer but it's not. Could you please explain to me where I'm going wrong?
To be honest: I didn't like the root expressions. First thing I did was to exchange x and y.

SteamKing said:
Going off fresh_42's comment, you can also evaluate the definite integral to calculate the area under a curve using dy instead of dx as the variable of integration.
In other words, using horizontal strips instead of vertical strips.

Thank you all for your help, but I feel like some of this stuff is a little beyond my current course. We've only just started looking at definite integrals, and have spent even less time with areas between two curves. I highly doubt we're expected to be using horizontal strips instead of vertical ones when we haven't even learned about conics or anything really.

But again thank you all for your help still

Bill_Nye_Fan said:
Thank you all for your help, but I feel like some of this stuff is a little beyond my current course. We've only just started looking at definite integrals, and have spent even less time with areas between two curves. I highly doubt we're expected to be using horizontal strips instead of vertical ones when we haven't even learned about conics or anything really.

But again thank you all for your help still
You don't need to know anything about conics to switch the variable of integration. Often times, switching is done as a matter of convenience to simplify the integral.

I think my main problem is that I still don't know how to calculate the area between two curves properly.

Say I was just given y=x-2 and y=-sqrt(4-x) and had to find the area enclosed between them between x=0 and x=2
Isn't the correct way of doing this to just set up an integral over the given domain and subtract the function on the bottom from the function in top? So integral(2, 0, (x-2-(-sqrt(4-x)))dx

SteamKing said:
You don't need to know anything about conics to switch the variable of integration. Often times, switching is done as a matter of convenience to simplify the integral.
... or simply exchange x and y and integrate as you are used to, which is the same.

fresh_42 said:
... or simply exchange x and y and integrate as you are used to, which is the same.
Okay thank you I'll just do it that way

## 1. What is the significance of the area enclosed by the two curves?

The area enclosed between the curves y^2=4-x and y=x-2 is important in calculus and geometry. It represents the definite integral of the two functions, which can be used to calculate the area under a curve.

## 2. How do I find the equations of the two curves?

The equations of the two curves can be found by solving for y in each equation. In this case, y^2=4-x can be rewritten as y = ±√(4-x) and y=x-2 can be rewritten as y=x-2. These equations can then be graphed to determine the shape of the curves.

## 3. What is the relationship between the two curves?

The two curves intersect at two points, (0,-2) and (3,1). These points represent the boundaries of the enclosed area. The curves are also symmetrical about the line y=-1, meaning that the area enclosed between them is the same on either side of this line.

## 4. How can I calculate the area enclosed between the two curves?

To calculate the area enclosed, you can use the definite integral of the two functions over the given interval. In this case, the area can be calculated by finding the definite integral of y^2=4-x and y=x-2 from x=0 to x=3. This can be done using integration techniques such as the substitution method.

## 5. Are there any real-world applications of this area enclosed by the two curves?

Yes, the concept of calculating the area under a curve is used in many real-world applications. For example, it can be used to calculate the volume of irregularly shaped objects, the area of land for agricultural purposes, and the total amount of rainfall over a given area. It is also used in physics and engineering to find the work done by a variable force and to calculate the center of mass of a system.

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