Converting iterated integrals to polar coordinates (KristaKingMath)

Learn how to convert iterated integrals from cartesian coordinates to polar coordinates.

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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)

Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”

[kristakingmath]

NOT a silly question at all! :) The difference is that a double integral doesn't specify the order of integration, and you'll see integral notation like (\int\int)_R, which tells you that you're integrating over the region R, but doesn't tell you whether you should integrate first with respect to x or y. An iterated integral has already done the work for you and tells you the order of integration. You'll see integral notation like \int_0^1\int_-2^2. Hope that helps!! :D

[kristakingmath]

thank you very much! i'm so glad the videos are helping!! :)

[vickykong]

The final is just around the corner and I was glad to find your video. It works perfectly for me. Clear voice and steps. Keep it up !

[cielomartos]

I can not say this enough but thank you. You are the only reason I am doing so well in calc Please keep doing what you do.

[sofiarossi]

I know this was made in 2013 and you probably aren't looking at comments, but on the off chance that you do, thank you a lot for this video!! It was very clear and helped me a lot :)

[celsiusfahrenheit]

Thank you Krista for taking the time to explain.

[mayankjha]

seriously you are very all those year's I was always lack behind with basics....now you saved my life...after completing my exams. ..I will cover your all lectures...then move ahead. ...Thank you...you have very good ability to teach..

[kristakingmath]

no, because cos(9) is a constant. multiply that by -1/2, and you still have a constant. so you really have to look at that whole term, -1/2 cos(9), as a constant, just like if it were just 3, or 7. if you took the integral of 3, you'd get 3\theta, which means the integral of -1/2 cos(9) is (-1/2 cos(9))\theta. great question, and i hope that helps!! :)

[kristakingmath]

you're welcome!! hope you're having a great summer!! :)

[EslamMohamedX]

Your reasoning is perfectly logical. You make everything look zillion times simpler.

[wahidhamidy]

The best video of changing cart. integral to polar integral i 've found on youtube!
i appreciate your effort miss!

[trinistylez]

You are amazing!!..Im doing these stuff right now in Calc 3 and you teach and explain these topics so well..keep up the great work!.:)

[wenqingtoh]

Thank you so much for clearing my doubt on how to calculate the theta value which had bothered me for so long!! Thank you!!!!

[AaronFu]

best video on iterated integral so far. You even included you substitution. Thank you very much.

[csbnikhil]

The Colors used are very helpful, perfectly explained.

[kristakingmath]

thanks! i'm glad you liked it!! :D

[yvonnerodriguez]

You were really helpful can u post with more examples
Thank you

[kristakingmath]

The reason I didn't change the limits of integration is because I was planning to back-substitute for r before I evaluated over the interval. If you want to leave the function in terms of u, then you need to change the limits of integration so that they correspond to u. But if you back substitute and put the function back in terms of r, then you'd just need to change the limits of integration back anyway, so I didn't change them. :)

[.r.rrahul]

these videos are so amazing nd simple to understand..!

[kristakingmath]

i hope they help, and good luck on your test!! :D