Clarence
  • Clarence
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].
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
yea its false
Clarence
  • Clarence
Think you could help me prove it? I mean, it just sounds false in my head.
anonymous
  • anonymous
your right man it definently is false think i spelled that wrong

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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
anonymous
  • anonymous
so its deffinetly false
anonymous
  • anonymous
so if ur happy with my help click the best response
phi
  • phi
it sounds true to me
Clarence
  • Clarence
Think you can prove it phi?
phi
  • phi
Here is the plot of f(x)= √(r^2-x^2) , but with r=1
1 Attachment
phi
  • 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
phi
  • 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 \( \Delta x \)
phi
  • phi
Do you know what they mean by Riemann sum ?
Clarence
  • Clarence
Yeah, a while back some smart people walked me through it all... Actually had fun with maths that time :p
phi
  • phi
When people introduce Riemann sums, it is often for the "area under the curve" |dw:1444057014048:dw|
phi
  • 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
phi
  • phi
Or see some of Khan's videos https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/riemann-sums/v/simple-riemann-approximation-using-rectangles
phi
  • phi
in your problem the height of each recangle is f(x) - g(x) (as I defined them up above): \[ \sqrt{r^2 - x^2} - -\sqrt{r^2 - x^2} \\ =\sqrt{r^2 - x^2} +\sqrt{r^2 - x^2} \\ = 2\sqrt{r^2 - x^2} \]
Clarence
  • Clarence
That makes sense! Thanks!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.