Calculating Area: Horizontal or Vertical Rec?

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Using horizontal or vertical rectangles to approximate the area under a curve yields the same result, provided the limits of integration and the integrand accurately describe the same region. For the function y = 1 - x^2, both integration methods can be applied effectively, although one may be easier than the other depending on the specific function. When rotating the function about the x-axis, the same principle applies; both methods will produce the same volume result. The integral ∫(from 0 to 1) y^(1/2) dy is valid for calculating area, but the integrand must correspond correctly to the function being analyzed. Ultimately, the choice between horizontal and vertical approximations depends on convenience and the specific problem at hand.
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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 \displaystyle \int_{0}^{1}(1-x^2)\,dx\,.
 
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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