Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

hmm Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{0}^{2\pi} \sqrt{2 \cos \theta + 2} d \theta\]
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
Find the area shared by the polar equations \[r = 2 \cos \theta\] \[r = 2 \sin \theta\] dw:1327853300525:dw
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
polar integration eh ....
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
A = 1/2 .... something or another, id have to refresh my memory
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
I asked the same question earlier but nobody could answer me and I felt pretty smart for that. :P
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
im guessing that since this is 2 parts that it might be a double integral tho, sound about right?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1327854057363:dw or maybe even this rendition?
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
The first question I did a double integral and got \[\theta \sqrt{2 \cos \theta + 2} \theta \sin \theta  \cos \theta  \theta^2/2\] from 2 pi to zero. Since this was originally \[\int\limits_{0}^{2\pi} \sqrt{(1 + \cos^2 \theta) + \sin^2 \theta}\] for the arc length of the cardioid. It's not supposed to be negative, which I got.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
negative just means you got the bottom over the top i believe; just drop the sign and it should be good
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
The second questions about the polar area is the area between two circles whihc I drew on the top. I used the calculator to find the bound to be 0 and pi, but it might be wrong.
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
bound is 0 and pie/4
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
Nope, I got  11.57 but the answer is 8.
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
Oh, I mgiht have gotten the upper and lower bound wrong. But I was trying to find the area without the shared part first then use that to be subtracted from the area of the entire circle but I am wondering if there's a better idea. I'll try the pi over 4 bound.
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
I still get a negative number for the polar area with the new upper and lower bound.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
if we move the circles around we might get an easier read? or at least I do
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
The answer is pi/2  1
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
this is just the quarter of a circle  the cresent part then
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
it intersects at tan(thetha) = 1 which means at pie/4 and 3*pie/4 ..
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
Ah, I don't understand what you just said. :O
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
(pi/4)(1pi/4) = 2pi/4  1 pi/2  1
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
How did you get that?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1327854888415:dwyou end up with a 1x1 square with an area of 1;
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
dw:1327854924962:dw
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
How do I prove that in equations?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
id move the circles around to equavelent positions and do:\[\int_{0}^{pi/2}\frac{1}{2}(R^2r^2)dr\]
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
whre R = (x^2+y^2=1) and r = ((x1)^2+(y1)^2=1)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
but state it in sin cos stuff maybe lol
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
I tried that but I must have made a mistake in the integration because I got \[(\sin2 \theta )/2 3 \theta\] at the end.
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
I integrated \[ (2\cos \theta) ^2  (2 \sin \theta )^2\] /2
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
keep the same equations perhaps and chnge integraion to pi/2 to 0
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
Nope, still doesn't work.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
yeah, polars im still a bit rusty on, got the concept but terrible at polarizing it
 2 years ago

hmm Group TitleBest ResponseYou've already chosen the best response.1
Yeah, I don't polar stuff either.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
ill reread my book at the house and know this stuff better tomorrow :) i cant seem to make sense of what I find online ....
 2 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.