Calculating Area: Horizontal or Vertical Rec?

  • Thread starter Miike012
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In summary, horizontal approx. rectangles and vertical approx. rectangles will give the same result when evaluating the area of a function.
  • #1
Miike012
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I added a picture of the function y = x^2 with given bounds... My question is, for functions such as the following one in the paint doc, I know its recommended to use vertical approx. rectangles when evaluating the area,.. Would I also get the same result if I used horizontal approx rec.?

The Same question goes for rotating the function about the x-axis... Can I get the same result if I used horizontal approx rec.?

If I decided to use horizontal rec, would the following be correct...



∫(from 0 to 1) y^1/2dy in terms of calculating area.
 

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  • #2
Miike012 said:
I added a picture of the function y = x^2 with given bounds... My question is, for functions such as the following one in the paint doc, I know its recommended to use vertical approx. rectangles when evaluating the area,.. Would I also get the same result if I used horizontal approx rec.?

The Same question goes for rotating the function about the x-axis... Can I get the same result if I used horizontal approx rec.?

If I decided to use horizontal rec, would the following be correct...

∫(from 0 to 1) y^1/2dy in terms of calculating area.
attachment.php?attachmentid=42849&d=1327009006.jpg


That is not the graph of y=x2 .
 
  • #3
1 - x^2, sorry lol
 
  • #4
Miike012 said:
I added a picture of the function y = x^2 with given bounds... My question is, for functions such as the following one in the paint doc, I know its recommended to use vertical approx. rectangles when evaluating the area,.. Would I also get the same result if I used horizontal approx rec.?

The Same question goes for rotating the function about the x-axis... Can I get the same result if I used horizontal approx rec.?

If I decided to use horizontal rec, would the following be correct...



∫(from 0 to 1) y^1/2dy in terms of calculating area.

Solve y = 1 - x2 for x, where x ≥ 0 to get your integrand.

... although it seems to me that it's at least as easy to evaluate [itex]\displaystyle \int_{0}^{1}(1-x^2)\,dx\,.[/itex]
 
  • #5
In answer to your questions, it doesn't matter whether you use horizontal slices or vertical slices, as long as your limits of integration and integrand describe the same region.

The same is true if you have a volume of revolution. You should get the same number with any technique. However, the integration is sometimes easier to perform when you use one technique.
 

1. What is the difference between horizontal and vertical rectangle?

Horizontal rectangles have a longer base and a shorter height, while vertical rectangles have a shorter base and a longer height.

2. Does the orientation of the rectangle affect the area calculation?

Yes, the orientation of the rectangle does affect the area calculation. The formula for calculating the area of a rectangle is length x width, and the length and width will be different depending on the orientation.

3. How do I determine if a shape should be measured horizontally or vertically?

This depends on the orientation of the shape. If the shape is taller than it is wide, it should be measured vertically. If the shape is wider than it is tall, it should be measured horizontally.

4. Can the area of a rectangle be calculated using only one dimension?

No, the area of a rectangle cannot be calculated using only one dimension. Both the length and width are needed to accurately calculate the area.

5. Are there any special considerations when calculating the area of irregular shapes?

Yes, when calculating the area of irregular shapes, it is important to break the shape into smaller, simpler shapes (such as rectangles or triangles) and calculate their areas separately. Then, add these areas together to get the total area of the irregular shape.

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