calculating area

Miike012

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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SammyS

That is not the graph of y=x² .

Miike012

1 - x^2, sorry lol

SammyS

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]

Mark44

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.