Calculating Area: Horizontal or Vertical Rec?

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Homework Help Overview

The discussion revolves around calculating the area under the curve of the function y = 1 - x^2, specifically exploring the use of horizontal versus vertical approximation rectangles. Participants are considering the implications of these choices on the results of integration and volume of revolution calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning whether using horizontal approximation rectangles would yield the same area as vertical ones. They are also exploring the correctness of their integrals and the implications of rotating the function about the x-axis.

Discussion Status

Some participants have provided insights suggesting that both horizontal and vertical slices can yield the same results, depending on the limits of integration and the integrand. There is an ongoing exploration of the ease of integration with different techniques, but no consensus has been reached.

Contextual Notes

Participants have noted potential confusion regarding the correct function representation and the limits of integration. There is also a mention of the need to solve for x in the context of the integration process.

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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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 .
 
1 - x^2, sorry lol
 
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]
 
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.
 

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