Double Integral Using Polar Coordinates - Part 2 of 3

[lucaciandrei]

you're such a lefty...:D great vid man, but i was hoping for something more challenging...:D

[pharmacy20101]

you are really great teacher , thanks alot :)

[TheFlowerofFaith]

omg, your already married! Darn!

[TheFlowerofFaith]

omg, marry me!

[jazzlover06]

correct me if I'm wrong, but I don't see why the identity is squared twice at 5:01. I believe it should be the /int{_0}{^pi}(/frac{32}{3}*[/f­rac{1}{2} (1-cos2/theta)]^2)

[yucky1336]

when you grunted. i did too and laughed. lol thanks for doing this problem

[Sbskyfire]

Anyone else think he sounds abit like Doug from the Hangover?

[chaffinc09]

after 4 minutes when you're using identities, couldn't you factor out a (2sintheta)^2

[OntologicalQuandary]

isnt cos(0) = 1 -___-

[bnssapp]

lol............integral from 0 to 0 is 0...thats kinda cute...lol

[seiko14]

thank you:)

[PouyaPouya]

Thanks very much for the video.

Just one thing, It didn't seen that tedious until you started keep saying it was. I was actually enjoying the video.

[RoboTekno]

then the TI-89 was invented.

[Josh42222]

" so hey this is kinda cute" lol. i don't know why but that made me laugh. Nice vid 5*

[JRajashekar]

Thx

[alcideslugo]

Excellent way to teach calculus, If I just figured out that youtube is not just for jokes and music, I wouldn't repeat that awful class last semester!

Thanks and congratulations from Venezuela

[swingsetbucket]

One thing you could've done is use some subbing while you were simplifying like: sin(x)=u cos(x)=v than let w equal something and then maybe you can just use greek letters for the rest.

[patrickJMT]

yes, but then you have to be careful not to integrate them using the power rule though!

[swingsetbucket]

true

[bordenw]

why do you have (1/2(1-cos2x)^2) twice?

[BelikeObama]

at 5:06 ?

I believe that it shouldn't be squared twice also. He just wrote it down but when he solved the integral, he completley ignored it which is correct check 6:45. So he computed the problem correctly. Just a little mistake I believe.

I used a calculator and the solution should be = [ (8* (3π -8)) / 3 ] or simplified into 3.79941