Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Need help: use Green's Theorem to find the area under one arch of the cycloid: x= a*(zsin(z)), y = a*(1cos(z))...z is theta
 3 years ago
 3 years ago
Need help: use Green's Theorem to find the area under one arch of the cycloid: x= a*(zsin(z)), y = a*(1cos(z))...z is theta
 3 years ago
 3 years ago

This Question is Closed

ducthinhBest ResponseYou've already chosen the best response.0
\[\int\limits_{2\pi}^{0}ydxxdy+\int\limits_{0}^{2\pi}0dx =\int\limits_{2\pi}^{0}\lefta*a( 1\cos(\theta) (1+\cos(\theta)d \theta\a*a(\Theta si(\theta)(\sin(\theta)dtheta)\right)\]
 3 years ago

eletronic_rajeshBest ResponseYou've already chosen the best response.0
\[xdyydx=a*2(zsin z2+2\cos z)dt (now integrate \it with \in \limits 0 \to 2\pi ) then answer will be 3\pi z2\]
 3 years ago

ducthinhBest ResponseYou've already chosen the best response.0
let C1 x=a(zsinz) y=a(1cosz) C2 x=2piz y=0 so we have closed curve C =C1+C2 then area \[A=\int\limits_{0}^{2\pi}(1cosz)(1cosz)dz\int\limits_{0}^{2\pi}0d(z)=3\pi\]
 3 years ago

eletronic_rajeshBest ResponseYou've already chosen the best response.0
but in the above result the radius of the circle through which the cycloid is made is not present. in the integration the 'a' present in the equation should come. and the final answer should be 3pi a^2
 3 years ago
See more questions >>>
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.