Question

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].

Answer 1

yea its false

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

so its deffinetly false

Answer 6

so if ur happy with my help click the best response

Answer 7

it sounds true to me

Answer 8

Think you can prove it phi?

Answer 9

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

Answer 10

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

Answer 11

*** 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)

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 \( \Delta x \)