Integration in polar coordinates | MIT 18.02SC Multivariable Calculus, Fall 2010

[lphiljipl]

THANK YOUUUUUUUUUUUUU

[maplestorypl]

I don't know if someone already caught this but at 7:08, it's supposed to be 1/r^3, so the answer turns out to be a bit more complicated than the one he ended up with.

[jbonn5]

@maplestorypl He wrote it correctly. dA becomes r dr dtheta so the integral does become 1 / r^2.

[FirstGradeCalculus]

Great videos, David. Thanks kindly!

[TDOTRIX]

great video

[PeaceUdo]

I just wish someone can explain how you know what theta is. How do you know its pi over four? Ugh

[Jameel263]

@PeaceUdo

Question a and b) The upper bound for y is y=x.

The line y = x is always at a 45 degree (pi/4) angle with the x axis.

If you dont get why, then for example lets say y = x = n (as y=x)

then

tan θ = n/n

tan θ = 1

therefore θ=45 degree (pi/4)

[Pandarchon]

@PeaceUdo I may be too late, but the angle is based on those special triangles, and specifically in this case it's the 1-1-root2 triangle. Since you know its slope is 1, the triangle can be drawn and then you will notice that the maximum angle is pi/4

Hope it helped.

[liveabovethecrowd]

lets be real, no one did it on their own

[cvchaparro]

This was a very helpful video, I now have a better understanding of changing the limits of integration for converting from rectangular to polar coordinates. Thanks.

[MyChatwin]

my god ur so young!

[gigel2006]

@MyChatwin He is 28, but looks like 20! XD

[moasfco11]

thanks very helpful examples

[marcuswauson]

cos of 0 is not 0, cos of 0 is 1, your answer to a should be (squareroot of 2) minus 2 all over 4. As cos of 0 is one, and one minus a half should give you squareroot of 2 over four minus a half.

[gnauhandy]

@marcuswauson you're evaluating 1/2 sin theta from 0-Pi/4.

[Alderslodge]

@marcuswauson

The integrand is [cos(theta)-(1/2)*cos(theta)] but when integrated, it becomes [sin(theta)-(1/2)*sin(theta)] from zero to pi/4. When evaluated equals sqrt(2)/2 - sqrt2)/4 = sqrt(2)/4

[mo3ius1]

nice shirt where did you get i? very impressive

[Mrhoe31]

thank u very much

[jsm666]

At the point where r = tan theta.sec theta, I began to understand the point behind some of the more mind-numbingly boring exercises we had rammed down our throats early in high-school trig.I suspect early trig would be a lot more fun if we could be shown (not necessarily with a full understanding!) where the usefulness comes in. I suspect, too, that some of the seemingly pointless exercises you're grinding through here have practical applications in some branch or other of technology, yes?

[rain74925]

Thanks! I think you articulate it really well. That's awsome! Wish you're my prof.

Not to be disrespectful, but are you an undergraduate? You look like a peer.

[beercity123]

Worth the watch! Very helpful!

[flukeekulf]

Thankyou vr much^^ Ur vr helpful....ur more detailed than my cal teacher hehe ^^