Finding the area between y = 2sqrt(x), y = 4, and y = -2x + 4

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In summary: Yes, that looks correct. You can split the region into two parts and sum the areas of each part, or you can use a single integral with the appropriate limits of integration to find the total area.
  • #1
s3a
818
8

Homework Statement


Problem:
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the
area of the region.

Given curves:
y = 2sqrt(x), y = 4, and y = -2x + 4

Technically, the curves are given as follows (but, I simplified them):
2y = 4sqrt(x), y = 4, 2y + 4x = 8

This ( http://www.wolframalpha.com/input/?i=find+area+between+y+=+2sqrt(x),+y+=+4,+y+=+-2x+++4 ) Wolfram Alpha link shows the plot for the area wanted.

Homework Equations


Definite integration and drawing functions.

The Attempt at a Solution


Could someone please tell me why that part that looks like a triangle which is from x = 0 to x = 4 and y = 2 to y = 4 is not shaded?
 
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  • #2
Because it doesn't fall within the region of integration.

You have specified three curves. The Wolfram Alpha apparently chose the region where y is >= 2*sqrt(x) and y <= 4-2x. If you want a different region, you will have to give WA a different specification on what to integrate.

BTW, please follow the HW template. Stuffing the problem statement into the thread title is not recommended.
 
  • #3
BTW, please follow the HW template. Stuffing the problem statement into the thread title is not recommended.
Sorry, I assumed people would see the equations in the Wolfram Alpha link as they look at the plot; I added it now in the opening post.

If you want a different region, you will have to give WA a different specification on what to integrate.
I'm just trying to answer a problem that says to find the area between those three functions. In other words, I want to do what the problem intends for me to do. I thought this meant to find ONLY the part that looks like a triangle which is from x = 0 to x = 4 and y = 2 to y = 4 and is not shaded in Wolfram Alpha's plot. Could you please tell me what exactly the problem intends for me to do?
 
  • #4
BTW, please follow the HW template. Stuffing the problem statement into the thread title is not recommended.
Sorry, I assumed people would see the equations in the Wolfram Alpha link as they look at the plot; I added it now in the opening post.

If you want a different region, you will have to give WA a different specification on what to integrate.
I'm just trying to answer a problem that says to find the area between those three functions. In other words, I want to do what the problem intends for me to do. I thought this meant to find ONLY the part that looks like a triangle which is from x = 0 to x = 4 and y = 2 to y = 4 and is not shaded in Wolfram Alpha's plot. Could you please tell me what exactly the problem intends for me to do?
 
  • #5
s3a said:

Homework Statement


Problem:
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the
area of the region.

Given curves:
y = 2sqrt(x), y = 4, and y = -2x + 4

Technically, the curves are given as follows (but, I simplified them):
2y = 4sqrt(x), y = 4, 2y + 4x = 8

This ( http://www.wolframalpha.com/input/?i=find+area+between+y+=+2sqrt(x),+y+=+4,+y+=+-2x+++4 ) Wolfram Alpha link shows the plot for the area wanted.

Homework Equations


Definite integration and drawing functions.

The Attempt at a Solution


Could someone please tell me why that part that looks like a triangle which is from x = 0 to x = 4 and y = 2 to y = 4 is not shaded?

For this type of problem I strongly urge you to avoid the use of Wolfram Alpha: It is important to know how to do these things manually. Besides that, it is much easier to do it by hand than to try to finesse Alpha to do what you want. Of course, a hand-drawn sketch will be crude and approximate, but that should be enough to be getting on with.
 
  • #6
Ray Vickson said:
For this type of problem I strongly urge you to avoid the use of Wolfram Alpha: It is important to know how to do these things manually. Besides that, it is much easier to do it by hand than to try to finesse Alpha to do what you want. Of course, a hand-drawn sketch will be crude and approximate, but that should be enough to be getting on with.
I DID draw it manually for the sake of knowing how to do it but, I didn't want to show you guys my piece-of-garbage drawing. ;) (In other words, it's the same thing as Wolfram Alpha's drawing, just uglier.)

So ... was I right (and Wolfram Alpha wrong) about what area I need to compute?
 
  • #7
s3a said:
Sorry, I assumed people would see the equations in the Wolfram Alpha link as they look at the plot; I added it now in the opening post.


I'm just trying to answer a problem that says to find the area between those three functions. In other words, I want to do what the problem intends for me to do. I thought this meant to find ONLY the part that looks like a triangle which is from x = 0 to x = 4 and y = 2 to y = 4 and is not shaded in Wolfram Alpha's plot. Could you please tell me what exactly the problem intends for me to do?

OK, so now you need to compute the area of that almost-triangular region. You are intended to apply one of the formulas you have probably learned for doing that.

Just as a reminder, though: you can split the area up into a large number of narrow vertical rectangles or a large number of narrow horizontal rectangles. Choose whichever method you like best.
 
  • #8
Okay so, this (check the TheAreasToCompute.jpg attachment) is how I compute the area(s) I want, right?
 

Attachments

  • TheAreasToCompute.jpg
    TheAreasToCompute.jpg
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1. How do I find the area between these three curves?

To find the area between these curves, you will need to first graph them to determine the region of interest. Then, you can use the integral of the top curve minus the integral of the bottom curve within the boundaries of the region to find the area.

2. What is the equation for y = 2sqrt(x)?

The equation for y = 2sqrt(x) is a curve that is in the shape of a parabola, with the vertex at (0,0) and increasing as x increases. The function represents a square root of x multiplied by 2, which is why it is called "2sqrt(x)".

3. How do I graph y = -2x + 4?

To graph y = -2x + 4, you can use the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is -2 and the y-intercept is 4. Plot the y-intercept on the y-axis and use the slope to find other points on the line. Then, connect the points to graph the line.

4. What is the significance of finding the area between these curves?

Finding the area between these curves can have various applications in real life situations. It can be used to calculate the volume of a three-dimensional shape, determine the amount of a certain substance in a mixture, or approximate the distance traveled by an object with a changing velocity. It is an important concept in calculus and has numerous practical uses in science and engineering.

5. Can I use a calculator to find the area between these curves?

Yes, you can use a calculator to find the area between these curves. However, it is important to note that calculators can only provide an approximation of the area. For a more accurate result, it is recommended to use calculus techniques to find the exact value of the area.

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