Help Find the Area Between 3 Curves

[swiftmatt]

1. Homework Statement

Find the area bounded by the following curves:
---------
y = x
y = -x + 6
y = (x / 2)
---------


2. Homework Equations

N/A

3. The Attempt at a Solution

Here is a link to what the graph looks like:
http://www.wolframalpha.com/input/?i=plot+y%3Dx+%3B+x%2By%3D6+%3B+2y%3Dx

So far, I have found the Intersections of the three lines at (0,0), (3,3), and (4,2) by setting 2 of the equations at a time, equal to each other. Now, I am trying to setup the problem where I actually calculate the area between the lines. I know I need to find the Integral of the equations at some point, but I don't know what the equations are supposed to be. Any and all help is appreciated!

~Matt

 

[Mark44]

1. Homework Statement

Find the area bounded by the following curves:
---------
y = x
y = -x + 6
y = (x / 2)
---------


2. Homework Equations

N/A

3. The Attempt at a Solution

Here is a link to what the graph looks like:
http://www.wolframalpha.com/input/?i=plot+y%3Dx+%3B+x%2By%3D6+%3B+2y%3Dx

So far, I have found the Intersections of the three lines at (0,0), (3,3), and (4,2) by setting 2 of the equations at a time, equal to each other. Now, I am trying to setup the problem where I actually calculate the area between the lines. I know I need to find the Integral of the equations at some point, but I don't know what the equations are supposed to be. Any and all help is appreciated!

~Matt

Whether you use vertical strips or horizontal strips, you're going to need two integrals. If you use vertical strips, the integrands will change at x = 3. For each integral, the area of the typical area element will be (y_upper - y_lower)Δx, but the expressions will be different on the two intervals.

 

[swiftmatt]

Mark,

Thanks for your quick reply. Our instructor never taught us anything about using strips. All the homework problems up to this one in this section involved a few steps (for 2 curves).

1. Find Intersecting Points.
2. Use the x-coordinates as the upper and lower x that will be substituted into the problem.
3. Integrate the formula that will be used to find the area.
4. Subtract the Integrated Formula (substituting lower x) from the Integrated Formula (substituting upper x).
5. Answer is the area.

Again, this was the process for 2 curves. Finding the area with a 3rd curve was never explained to us.
Also, sorry in advance to the way I explained the steps. I don;t have all the terminology down, so I hope I explained it as best as possible :)

 

[Mark44]

Mark,

Thanks for your quick reply. Our instructor never taught us anything about using strips. All the homework problems up to this one in this section involved a few steps (for 2 curves).

1. Find Intersecting Points.
2. Use the x-coordinates as the upper and lower x that will be substituted into the problem.

Above, I think this means the integration bounds. For this problem, one interval is [0, 3] and the other is [3, 4], again assuming that you will be using vertical strips, as described in my previous post.

3. Integrate the formula that will be used to find the area.

The "formula" (i.e., the integrand) is almost identical to the area of the typical area element.

4. Subtract the Integrated Formula (substituting lower x) from the Integrated Formula (substituting upper x).

Integrated Formula = antiderivative, a much more commonly used term.

5. Answer is the area.

Again, this was the process for 2 curves. Finding the area with a 3rd curve was never explained to us.

Each integral will involve only two curves (lines in this problem).

Also, sorry in advance to the way I explained the steps. I don;t have all the terminology down, so I hope I explained it as best as possible :)

Here's an example that's similar to your problem: Find the area bounded by the curves y = x, y = 1, and x= 2.

The region is a triangle with vertices at (1, 1), (2, 2), and (2, 1). This is slightly simpler than your problem, since only one integral is needed.

Typical area element: ΔA ≈ (y_upper - y_lower)Δx = (x - 1)Δx[/color], with Δx ranging from 1 to 2.

Integral:
$$ A~=~\int_1^2 x - 1~dx = \left. (1/2)x^2 - x \right|_1^2 = (2 - 2) - (1/2 - 1) = 1/2$$

Notice the similarity between my expression for the typical area element and the integrand.