## Clarence one year ago This has to be false right? Suppose we are going to consider the disk of radius r as the region bounded between the graphs of the functions √(r^2-x^2) and -√(r^2-x^2). Then the area of the disk can be expressed as the limit of Riemann sums of rectangles of length Δx and height 2√(r^2-x^2) where xi are a partition of the interval [−r,r].

1. anonymous

yea its false

2. clarence

Think you could help me prove it? I mean, it just sounds false in my head.

3. anonymous

your right man it definently is false think i spelled that wrong

4. anonymous

Δx and height 2√(r^2-x^2) where xi are a partition of the interval the height is 3 then the check mark

5. anonymous

so its deffinetly false

6. anonymous

so if ur happy with my help click the best response

7. phi

it sounds true to me

8. clarence

Think you can prove it phi?

9. phi

Here is the plot of f(x)= √(r^2-x^2) , but with r=1

10. phi

f(x) in that graph is the top part of a circle the other graph g(x) = - sqrt(r^2-x^2) will be the bottom half of the circle

11. phi

*** Then the area of the disk can be expressed as the limit of Riemann sums of rectangles of length Δx and height 2√(r^2-x^2) where xi are a partition of the interval [−r,r]. *** This means we have lots of rectangles between -1 and 1 (let r=1 in this case) |dw:1444056657537:dw| the height of each thin rectangle is the top y value (i.e. f(x) ) minus the bottom y-value g(x) the width of each rectangle is $$\Delta x$$

12. phi

Do you know what they mean by Riemann sum ?

13. clarence

Yeah, a while back some smart people walked me through it all... Actually had fun with maths that time :p

14. phi

When people introduce Riemann sums, it is often for the "area under the curve" |dw:1444057014048:dw|

15. phi

in your problem, the bottom is not the x-axis, it is g(x)= - sqrt(r^2-x^2) but it is the same idea: add up the area of lots of thin rectangles (where we let their width become infinitely thin, and we have an infinite number of rectangles) https://en.wikipedia.org/wiki/Riemann_sum

16. phi
17. phi

in your problem the height of each recangle is f(x) - g(x) (as I defined them up above): $\sqrt{r^2 - x^2} - -\sqrt{r^2 - x^2} \\ =\sqrt{r^2 - x^2} +\sqrt{r^2 - x^2} \\ = 2\sqrt{r^2 - x^2}$

18. clarence

That makes sense! Thanks!!