A community for students.
Here's the question you clicked on:
 0 viewing
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^2x^2) and √(r^2x^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^2x^2) where xi are a partition of the interval [−r,r].
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^2x^2) and √(r^2x^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^2x^2) where xi are a partition of the interval [−r,r].

This Question is Closed

clarence
 one year ago
Best ResponseYou've already chosen the best response.0Think you could help me prove it? I mean, it just sounds false in my head.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your right man it definently is false think i spelled that wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Δx and height 2√(r^2x^2) where xi are a partition of the interval the height is 3 then the check mark

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so its deffinetly false

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if ur happy with my help click the best response

clarence
 one year ago
Best ResponseYou've already chosen the best response.0Think you can prove it phi?

phi
 one year ago
Best ResponseYou've already chosen the best response.1Here is the plot of f(x)= √(r^2x^2) , but with r=1

phi
 one year ago
Best ResponseYou've already chosen the best response.1f(x) in that graph is the top part of a circle the other graph g(x) =  sqrt(r^2x^2) will be the bottom half of the circle

phi
 one year ago
Best ResponseYou've already chosen the best response.1*** Then the area of the disk can be expressed as the limit of Riemann sums of rectangles of length Δx and height 2√(r^2x^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 yvalue g(x) the width of each rectangle is \( \Delta x \)

phi
 one year ago
Best ResponseYou've already chosen the best response.1Do you know what they mean by Riemann sum ?

clarence
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, a while back some smart people walked me through it all... Actually had fun with maths that time :p

phi
 one year ago
Best ResponseYou've already chosen the best response.1When people introduce Riemann sums, it is often for the "area under the curve" dw:1444057014048:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.1in your problem, the bottom is not the xaxis, it is g(x)=  sqrt(r^2x^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
 one year ago
Best ResponseYou've already chosen the best response.1Or see some of Khan's videos https://www.khanacademy.org/math/integralcalculus/indefinitedefiniteintegrals/riemannsums/v/simpleriemannapproximationusingrectangles

phi
 one year ago
Best ResponseYou've already chosen the best response.1in 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
 one year ago
Best ResponseYou've already chosen the best response.0That makes sense! Thanks!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.