What helped me understand the r boundaries was that you look for the intersection of the two boundary graphs. so z = 2 and other combine to make 4 = 1 + x^2 + y^2. After you substitute the values of x and y with polar coordinates you have a function of r. The range of this function of r gives you the boundaries (r>=0). In the video you told me what I was doing but not why (2:29). Thanks for the video though, great help!
@@raba2d723 4 years? That long? I haven't checked in with Krista for a good while (understatement), I headed for Physics-w-calculus, and Nuclear-Engineering-101-Berkeley. No - I didn't understand a lot of it, but Krista got me up and running on the math enough to hang in for the course lectures. Never once put pencil to paper -- LOL. Not exactly setting a good example, but it's a just a hobby. Thank you Krista!
Excellent explanation. Another way to do the u substitution is by evaluating what you equal u to with the limits of integration, for example if u = 1+r^2 and your limits of integration are 0 and sqrt(3), you evaluate that in u, getting 1 and 4 as the new limits of integration and that way you can evaluate the result of the integration.
I always run to this channel when my professors didn't explain clearly how they come up with their solution. Thanks a lot hope those adds and liking your videos could help.
Thanks a lot Krista.. u have taught this very well... Before watching Ur video that wasn't my cup of tea.. bt now I'm very happy and satisfied .. thanku once again 🙋🙋
PLS BE CAREFUL!!!! when u use u-subst., u have have pay attention to ur limits of int. because they will change...now it doesn't really matter in most cases because u know u'll eventually change back to ur original variables, but in an exam or in the case where u go ahead and work with 'u's, it could be very problematic. Just my two cents :)
ughhhhhhhhhhhhhhhhhh I have a similar problem just like this one. I was so happy cause I thought since they were similar I would get it but nope. Mine is " Find the area of surface. S is the part of the sphere x^2 + y^2 + z^2 = 9 that lies above the plane z=2." I obtain the answer (8/3)pi but the book says the correct answer is 6pi.
What helped me understand the r boundaries was that you look for the intersection of the two boundary graphs. so z = 2 and other combine to make 4 = 1 + x^2 + y^2. After you substitute the values of x and y with polar coordinates you have a function of r. The range of this function of r gives you the boundaries (r>=0). In the video you told me what I was doing but not why (2:29). Thanks for the video though, great help!
You're so good at explaining. If only you could explain to my professors how to properly teach calculus lol
I'm glad I can at least help through the videos. :)
.... They'd be out of a job... ha ha. All that tuition you pay and Krista runs circles around them :)
@@hg2. in polar coordinates!
@@raba2d723
4 years? That long?
I haven't checked in with Krista for a good while (understatement), I headed for Physics-w-calculus, and Nuclear-Engineering-101-Berkeley. No - I didn't understand a lot of it, but Krista got me up and running on the math enough to hang in for the course lectures. Never once put pencil to paper -- LOL. Not exactly setting a good example, but it's a just a hobby. Thank you Krista!
You've helped me so much in my calculus college classes ( calculus 1,2 &3). I'm so grateful. I can't say thank you enough .
You're welcome, laimino! I'm so glad I've been able to help along the way! :)
Hi! Glad you like it! I explain the process here :) integralcalc . com/how-i-create-my-videos/
Hope it went well! :D
Excellent explanation. Another way to do the u substitution is by evaluating what you equal u to with the limits of integration, for example if u = 1+r^2 and your limits of integration are 0 and sqrt(3), you evaluate that in u, getting 1 and 4 as the new limits of integration and that way you can evaluate the result of the integration.
Aww James c'mon!! This isn't too hard, you can totally do it! ;)
I always run to this channel when my professors didn't explain clearly how they come up with their solution. Thanks a lot hope those adds and liking your videos could help.
I'm so glad my videos can help fill the gap... thanks for letting me know that they're helping! :)
ur officially my favorite person . Calc quiz in7 hrs, might pull all nighter i dunno. I wish I found you earlier.
your explaining is very clearly, thanks a lot for your videos
Thank you! I'm glad you like them. :)
OMG! It was the exact exercise that I was having problems with! THANKS!!!!
+Pedro Vilaça You're welcome!
Thank you extremely
Why do you substract 2-(sqr1+r2) in min 5:24 apox.
Thank you, good videos.
Does anybody know the answer to this? I don't get it either, it seems arbitrary
Look here we're finding the volumne under z=2 line
So the height of the volume will be:
Total ht (z=2) - total ht of sphere (z in terns of x,y)
Brilliant, Thanks!
Awesome job with these videos :3
Thanks a lot Krista.. u have taught this very well... Before watching Ur video that wasn't my cup of tea.. bt now I'm very happy and satisfied .. thanku once again 🙋🙋
You're welcome, Prashank! I'm so glad it helped!
Great job. As a suggestion, you should add that the answer is actually 4 pi/3 cubic units because it's a volume.
this has been like super awesome thanx a lot
You're welcome, Kasansa, I'm so glad you liked it! :)
far out!!! great vid!!!
THANK YOU
thx for sharing
What program did you use to produce this video? I also teach math on-line and like this type of presentation.
thank you!
:D
Hang on, why can't you have a negative r? Isn't this just the same as a positive r flipped through pi radians?
Yes, technically that's true, but using a negative r value isn't helpful in this context.
PLS BE CAREFUL!!!! when u use u-subst., u have have pay attention to ur limits of int. because they will change...now it doesn't really matter in most cases because u know u'll eventually change back to ur original variables, but in an exam or in the case where u go ahead and work with 'u's, it could be very problematic. Just my two cents :)
the same answer ,bro .she made a fault in u subst so the result is wrong
Where did 2 - sqrt (1+r^2) come from? Isn't the function to be evaluated sqrt. (1+r^2)?
Why do we subtract sqrt(1-r^2) from 2? I need to understand why before I can apply it. I really appreciate your videos =)
amazing! easy to understand :)
+Estherr Barak I love that feedback! Exactly what I'm hoping for. :)
+CalculusExpert.com YOU THE REAL MVP
ughhhhhhhhhhhhhhhhhh I have a similar problem just like this one. I was so happy cause I thought since they were similar I would get it but nope. Mine is " Find the area of surface. S is the part of the sphere x^2 + y^2 + z^2 = 9 that lies above the plane z=2." I obtain the answer (8/3)pi but the book says the correct answer is 6pi.
OMG TY SO MUCH!!
You're very welcome!!
amazing
We could make the integral (1+r^2)^(.5) - 2 and if we had a negative answer we reverse it to the positive answer
can you please tell me the applications of double integral in engineering?
did you know you can calculate the circumference of a moose with quadruple integrals
How is it a volume integral if there is only two integral?Plz help me out
IT WORKEDDDDDD
Is it possible to solve this using a triple integral? and just have z be one of the bounds of integration?
thank you my lady
Very useful, thanks :-)
:D
isnt the integral of (x)^1/2 = (2(x)^3/2)/3?
How do you know that you are dealing with the full circle (2pi) ?
It's a sphere with centre on origin so 2 pie
Honestly i should be paying tuition to you not to my professors
why didnt you have z=+/- sqrt of ... and you only had z=+sqrt 1+x2+y2?
:)
You’re the best. I love you. I wish I could have you as my wife.
i know this has nothing to do with calculus but you are the sexiest tutor i´ve ever seen.....just keep doing what you are doing
Why can’t I find a girlfriend like you.
I'm gonna unsubscribe krista bcuz the calc level ur reviewing is too hard for me ! Jk with love