A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
How do I use integration to prove that the area of a circle radius r is equal to pi r^2
anonymous
 5 years ago
How do I use integration to prove that the area of a circle radius r is equal to pi r^2

This Question is Closed

Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The key is to use trigonometric substitution. If you start with: \[4\int\limits_{0}^{r}\sqrt{r^2x^{2}}dx\] Then make the following substitutions based on the angle:\[x=r\cos \theta, \sqrt{r^2x ^{2}}=r\sin \theta, dx=r\sin \theta d\theta\] You'll get: \[4\int\limits_{\pi/2}^{0}r^2\sin^2 \theta d \theta\] or \[4r^2\int\limits_{0}^{\pi/2}\sin^2 \theta d \theta\] I'll leave it to you to work out the definite integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so how did you get the 4 in front of the integral?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The integral represents the area of a quarter of a circle with radius r. I could have put 2 in front and worked out the sum from r to r instead of 4 times 0 to r.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Come to think of it, there's a much easier solution:\[\int\limits_{0}^{r}2 \pi xdx\]The integrand represents the circumference of a circle with radius x. The area of the circle is the cumulative sum of the circumference of the concentric circles from 0 to r.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Theres a simpler method If u divide the circle into many concentric rings, each at a distance r, and having thickness dr the area of one disc is \[dA=2{\pi}rdr\] \[\int\limits_{}^{}dA = \int\limits_{0}^{R}2{\pi}rdr = {\pi}R ^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2\pi}\int\limits_{0}^{r}rdrd \Theta\] \[\int\limits_{0}^{2\pi}[r^2/2][0,r]d \Theta=\int\limits_{0}^{2\pi}r^2/2d \Theta\] \[\int\limits_{0}^{2\pi}r^2/2d \Theta=r^2/2(\Theta)[0,2\pi]=\pi r^2\]
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.