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
seriously you are very good......in 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..
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 :)
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!! :)
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. :)
hey love your teaching 10:25 is it really the area of that circle or is it the volume of some soild where its projection is the region represented by the circle i am not sure about the latter but pretty sure that the answer in not the area of that circle please explain love your videos thanks
3:24 If the limits of integration for x had been -2 to 2, how would you define your limits of integration for r and theta? I have a problem like that for homework, but I don't know if I am supposed to be using polar (or how to use polar for that).
Yes, but I knew I was going to back-substitute at the end of the problem, which is why I didn't bother. After back-substituting, I would have had to change the limits back to what they were originally.
LOL, I totally understand!! :) Unfortunately I'm going to have to ask you to ignore me for a few months, because I need to get these videos out so that they can help people in summer school, and so that they're ready for the fall when everybody gets back to school! :)
I have a question. You know how you found the new limits of integration in polar coordinates geometrically by drawing a picture? Is there a way to find the limits of integration analytically like how you substituted r^2 into the integrand? I can't seem to find a way to do it analytically.
integralCALC It was pretty rough but this stuff- I did well on.Hopefully there is a curve and I can keep my B+ haha You were really helpful though, I have you right there with PatrickJMT on the most helpful videos for Calculus!
So what if we were to tweak the bounds for example? Say from y = x to y = sqrt(9 - x^2) and from 0 to 1? I am curious as to what the picture would look like and how we would set up bounds for that...
You would keep everything exactly the same, except the limits of integration would be switched. So you'd still replace dx dy with r dr d(theta). You'd still replace the original integrand with sin(r^2). But because the order of integration was different, instead of int [0,pi] int [0,3], you'd end up with int [0,3] int [0,pi]. I hope that helps!
When you went into the "u-universe(as my great high school calculus teacher called it), and made u = r^2... Shouldn't the limits chang? So 0 would remain the same because (0)^2= 0... But the higher limit 3 would turn into 9 because (3)^2 equals 9...? So the new limits of integration should be from 0 to 9...right? Or am I being dumb?
jesus moreno you're not being dumb. i didn't change them because i was planning to back-substitute so that i could evaluate with respect to the original variable. but i should have indicated at that step that the limits were still related to the original variable. if you're not planning to back-substitute, then yes, you definitely have to change them so that the limits of integration are in terms of u, instead of r.
+TES HAI No, cos9 doesn't need to be integrated. cos9 is a constant, so when you integrate with respect to theta, you'd get (cos9)(theta). If you have cos(x), then you need to integrate, but cos of a constant is a constant, so it doesn't change like you're saying.
Hi Krista. Your videos are super useful. I like the presentation. I wanna know which software do you use which has black board like background and a nice cursor?
When r = 0 => u = 0, but when r = 3 => u = sqrt(3). But you kept the boundaries of integration from 0 to 3. I don't understand. Can you answer me please?
I'm slightly confused: for many of the problems I've been doing, I'm supposed to be expressing the limits of r as a function of θ, but here, you're expressing the limits of r as constants. I've tried this, but have gotten many answers wrong as a result. Thoughts?
+William Ford Even in this example, r is still a function of theta, it's just something like r=3+0theta, where the 0theta becomes 0, so you just end up with r=3. So it's not that r isn't a function of theta, just that the function happens to be a constant. And in this case, since we're looking at a circle around the origin, r will be a constant. As long as your function of r in terms of theta represents distance from the origin everywhere, r can be a function in terms of theta, or just a constant, like in this problem, and you should get the correct answer regardless.
I don't get what's the difference between an iterated integral and just a general double integral. They're both the same thing right? My textbook says that an iterated integral is a repeated integral which is just a double integral right..... Why would they label it double and iterated as if they're two different things?....
Jimmy M Because they are slightly different, even though the difference is a little silly. A double integral has two integral signs, but the intervals for x and y (the limits of integration) haven't been specified yet. So a double integral would be something like int int_R xy dy dx, which means the double integral over the region R of the function xy. On the other hand, an iterated integral already has the limits of integration defined, so something like int_0^2 int_1^3 xy dy dx. You always end up turning a double integral into an iterated integral right away by adding limits of integration, which is why they seem so similar. I hope that helps!
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 !
Thanks! Good luck on your final!
thank you very much! i'm so glad the videos are helping!! :)
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
seriously you are very good......in 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..
Thank you so much! Good luck with your exams, and I hope you enjoy the rest of the videos!
I can not say this enough but thank you. You are the only reason I am doing so well in calc 3!!!!! Please keep doing what you do.
Thank you so much, Cielo! I'm so glad the videos have been helping! :D
Thank you Krista for taking the time to explain.
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 :)
I'm so glad it helped, Celestial! Thanks for letting me know! :D
you're welcome!! hope you're having a great summer!! :)
Your reasoning is perfectly logical. You make everything look zillion times simpler.
+Eslam Mohamed Thank you so much!
The best video of changing cart. integral to polar integral i 've found on youtube!
i appreciate your effort miss!
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!! :)
best video on iterated integral so far. You even included you substitution. Thank you very much.
I explain here :) integralcalc . com/how-I-create-my-videos
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!.:)
you've probably made it by now..im in calc 3 rn :(
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. :)
3:33 should be "entire half circle that is above x-axis " or ?
hey love your teaching
10:25 is it really the area of that circle or is it the volume of some soild where its projection is the region represented by the circle
i am not sure about the latter but pretty sure that the answer in not the area of that circle
please explain
love your videos
thanks
i hope they help, and good luck on your test!! :D
Thank you so much for clearing my doubt on how to calculate the theta value which had bothered me for so long!! Thank you!!!!
You're welcome, I'm so glad it helped! :)
thanks! i'm glad you liked it!! :D
You're so welcome! :)
Your videos make learning all this stuff so easy! Thank you so much for these, you've helped me do well on my final exam.
So glad I could help!
integralCALC thanks for that but in the end of the video you say it is the area but i think it is the volume of the function
saleh ali
Nope, it's the area. This is 2 dimensional function, not 3 dimensional.
I know I am saying always the same thing that you are awesome for your each video but I can't stop myself YOU ARE AWESOME !!!
+Emre Arslan Awww thank you so much!
The only thing that I can say is THANK YOU, you saved my day :)
You're welcome, I'm so glad it helped!!
Glad you liked it! :)
a tip : you can watch series at kaldroStream. Been using them for watching lots of of movies recently.
@Mustafa Merrick definitely, been using Kaldrostream for since november myself :D
@Mustafa Merrick Definitely, been using Kaldrostream for months myself =)
@Mustafa Merrick Yup, been using kaldroStream for months myself :)
@Mustafa Merrick yup, I have been using Kaldrostream for years myself :D
Krista U really are a KING
You definitely are the KING, thank you for your videos :)
So glad they're helping! :)
The Colors used are very helpful, perfectly explained.
+Nikhil CSB Thanks!
Absolutely in love with this! Life saver.
+TheDanischannel I'm so glad it helped!
Really nice video, I love how you break the question up so its nice and simple. :)
You were really helpful can u post with more examples
Thank you
this was the most helpful example i came across today! you rock!
Awesome, thank you, Ryke! :)
3:24 If the limits of integration for x had been -2 to 2, how would you define your limits of integration for r and theta? I have a problem like that for homework, but I don't know if I am supposed to be using polar (or how to use polar for that).
+xcrunner 0 to 2pi since you want the whole circle
I have a question at 8:40 don't we have to change the integral limits?
these videos are so amazing nd simple to understand..!
I understand! Thank you so much, you helped me pass this class!
Brilliant video. Cleared everything for me.
6:48 What if it's dxdy? Do you need to turn them around first?
since you are substituing u, shouldnt the range of u goes from zero to 9 since u = r^2 ?
Yes, but I knew I was going to back-substitute at the end of the problem, which is why I didn't bother. After back-substituting, I would have had to change the limits back to what they were originally.
LOL, I totally understand!! :) Unfortunately I'm going to have to ask you to ignore me for a few months, because I need to get these videos out so that they can help people in summer school, and so that they're ready for the fall when everybody gets back to school! :)
You've helped me a lot. Greetings from Mexico city.
+LALO365 Glad I could help!
you are the best .really love the way u teach
Glad you like it!
You are AMAZING! Thanks so so much for the help!
Do you have any videos regarding integrals in cylindrical and spherical coordinates???
Your videos are awesomeee! :) Thank you so much!
tnx a LOT, you explain very easy/detailed and your voice is nice.
tnx ;)
awesome Voice!
just made my day!
Thanks Krista!
Easy to understand! Thank you so much 😊
Thank you so much! I forgot everything the next day when I actually tried to do my homework!
I had a problem just like this I had to figure out! Thanks so much!
You're welcome, Rory! I'm so glad it helped! :)
This was extremely helpful, thank you!
Thanks for letting me know! :)
Hi . Please can the max value of r be r=3cos¢ ? I'm perplex on this please for an explanation if no
I have a question. You know how you found the new limits of integration in polar coordinates geometrically by drawing a picture? Is there a way to find the limits of integration analytically like how you substituted r^2 into the integrand? I can't seem to find a way to do it analytically.
A life saver T-13 hours before my Calculus 3 Final!
teleton11 Good luck on your final, I hope you do great!!
integralCALC It was pretty rough but this stuff- I did well on.Hopefully there is a curve and I can keep my B+ haha
You were really helpful though, I have you right there with PatrickJMT on the most helpful videos for Calculus!
integralCALC BTW, just curious- what is your math education?
teleton11 Thank you very much! I hope there's a curve too!
integralCALC I didn't study math, I've just been a tutor for a long time.
Thank you a lotttt please do more examples about two intersecting circles
That was some great explanation... thanks!
you're welcome!
Isn't the answer should be pi/2 (1-sin9)? double integration of sin is still sin right?
it is clear and straight forward
Thank you so much Krista King .. Help me thank you
This really helped me out. cheers
I appreciate your effort. Thanks again:)
You're welcome! :)
So what if we were to tweak the bounds for example? Say from y = x to y = sqrt(9 - x^2) and from 0 to 1? I am curious as to what the picture would look like and how we would set up bounds for that...
+Luca Capobianchi In that case you wouldnt use polar coordinates lol
There's Chain Rule "hidden" in the r substitution for sin(r^2) right?
thank you so much we need more exemples and please if you can the triple integral
+imad liani You're welcome, and I will definitely continue to add more videos.
Do you have more videos explaining this or a nice website please put it in description
NICE,,JUST NEED 10 MIN TO STUDY.
THERE IS BETTER THAN MY TEACHER TEACHING 2 HR..orz
Do you have any vid where you use pts (x,y,z) to find surface area?
Thanks! By the 5 minute mark everything from class clicked in lol.
Oh good! I'm so glad it made sense! :)
This video was amazing. Thank you!!!!!
Thanks! :D
thank you so much, the explanation was clearly
thanks so much, vivid explanation!
You're so welcome! I'm glad you liked it. :D
you are my hero thanks,,, krista king
what if the differentials were dxdy would you still use the same format or would you switch them and make it r dr d(theta)?
You would keep everything exactly the same, except the limits of integration would be switched. So you'd still replace dx dy with r dr d(theta). You'd still replace the original integrand with sin(r^2). But because the order of integration was different, instead of int [0,pi] int [0,3], you'd end up with int [0,3] int [0,pi]. I hope that helps!
This helps a lot.Thank you
+eddy lee You're welcome, I'm so glad it helped!
When you went into the "u-universe(as my great high school calculus teacher called it), and made u = r^2... Shouldn't the limits chang? So 0 would remain the same because (0)^2= 0... But the higher limit 3 would turn into 9 because (3)^2 equals 9...? So the new limits of integration should be from 0 to 9...right? Or am I being dumb?
jesus moreno you're not being dumb. i didn't change them because i was planning to back-substitute so that i could evaluate with respect to the original variable. but i should have indicated at that step that the limits were still related to the original variable. if you're not planning to back-substitute, then yes, you definitely have to change them so that the limits of integration are in terms of u, instead of r.
thankgod you dont have a robotic voice like other similar tutorials
Thanks a lot for such a good explanation :)
+The Unofficials You're welcome, I'm so glad it helped!
Silly question - what is the difference between an iterated and a double integral (if any)?
thank u so much....my all douts are clear...❤😘😍😍😍😍😍❤
Great video!! I just want to mention that you forgot to integrate second integral at the end. It should be Pi/2(1-sin(9))
+TES HAI No, cos9 doesn't need to be integrated. cos9 is a constant, so when you integrate with respect to theta, you'd get (cos9)(theta). If you have cos(x), then you need to integrate, but cos of a constant is a constant, so it doesn't change like you're saying.
+Krista King | CalculusExpert.com That's correct.
Aw thanks! :D
Hi Krista. Your videos are super useful. I like the presentation. I wanna know which software do you use which has black board like background and a nice cursor?
Hi Tianhong! I use Sketchbook, which is made by Autodesk. :)
Nice explanation
Thanks, now I won't fail my test tomorrow :')
+Minh Tri I hope it goes well!
If it is cos in place of sin,what is the outer integral limits
At 9:20, shouldn't that have been -1/2 sin(9)?
Awesome! That makes me so happy. :D
When r = 0 => u = 0, but when r = 3 => u = sqrt(3). But you kept the boundaries of integration from 0 to 3. I don't understand. Can you answer me please?
wish I knew this channel existed before the semester started haha...
OMG! LIFE SAVER!
I'm slightly confused: for many of the problems I've been doing, I'm supposed to be expressing the limits of r as a function of θ, but here, you're expressing the limits of r as constants. I've tried this, but have gotten many answers wrong as a result. Thoughts?
+William Ford Even in this example, r is still a function of theta, it's just something like r=3+0theta, where the 0theta becomes 0, so you just end up with r=3. So it's not that r isn't a function of theta, just that the function happens to be a constant. And in this case, since we're looking at a circle around the origin, r will be a constant. As long as your function of r in terms of theta represents distance from the origin everywhere, r can be a function in terms of theta, or just a constant, like in this problem, and you should get the correct answer regardless.
y are we not including the lower part of the circle or rater why did we only use half circle
Darling would you be kind enough to tell ,e what app and device you use to make these videos?
very helpful thank you, wish you were my professor. I'm sure many others would agree XD
I don't get what's the difference between an iterated integral and just a general double integral. They're both the same thing right? My textbook says that an iterated integral is a repeated integral which is just a double integral right..... Why would they label it double and iterated as if they're two different things?....
Jimmy M Because they are slightly different, even though the difference is a little silly. A double integral has two integral signs, but the intervals for x and y (the limits of integration) haven't been specified yet. So a double integral would be something like int int_R xy dy dx, which means the double integral over the region R of the function xy. On the other hand, an iterated integral already has the limits of integration defined, so something like int_0^2 int_1^3 xy dy dx. You always end up turning a double integral into an iterated integral right away by adding limits of integration, which is why they seem so similar. I hope that helps!
integralCALC Thanks for the reply. I swear these darn mathematicians just want to make our lives harder.
I love your videos but can u slow down with the videos during summer break?! Lol im really not tryna see calc in my subscription list lol
What if the inner integral is from some value to 0? Do we just reverse it?
for example from -sqrt(a^2-y^2) to 0
puu.sh/smeHy/1f2d9f5a39.png
Make more video on polar coordinate changes
Thank you so much.
thanks make my life so easy
You're welcome, farrukh! I'm so glad it helped!