Hi all! At 5:56 the bounds for r should be [0,2] instead of [-2,2]. I have an annotation for this, but sometimes annotations don't show up depending on your settings and where you're watching.
@amj_6
5 жыл бұрын
It really freaked me out!
@Bestofchatgpt
5 жыл бұрын
Do you still end up with 0?
@Daniel.Rosenthal
5 жыл бұрын
@@Bestofchatgpt yes
@anands9407
5 жыл бұрын
why?i think it should be -2,2 and thetha should be 0 to pi...
@ITACHIUCHIHA-yn1lo
4 жыл бұрын
@@anands9407 because r never be taking in negative
@ThePhillyg09
9 жыл бұрын
13 minutes of work for the answer of ZERO. Ladies and Gentlemen... Calculus.
@kristakingmath
9 жыл бұрын
+Philip Geraci LOL
@hg2.
7 жыл бұрын
LOL!
@KakinTseGamePlayer
7 жыл бұрын
I don't think the answer is 0, she made a major mistake that would change the entire answer of the question so you'd have to do it over to find the real answer.
@tiibrahim5714
6 жыл бұрын
Fantastic
@azizal-hunaiyyan5041
6 жыл бұрын
Actually no, I did it and still, the answer is 0 :)
@paclax7696
9 жыл бұрын
amazing explanation and video, however the bounds for r should be 0
@kristalenee8925
10 жыл бұрын
This really helped me but why did you make r from (-2,2) instead of from (0,2) ?
@harishbaskar2935
5 жыл бұрын
guys small mistake. r is from 0 to 2... x=[0,2] its the radius not diameter
@BruceWayne-zt7vt
3 жыл бұрын
can you explain why r is from 0 to 2 instead of -2 to 2. #iambadatMath :)
@alexanderbudianto7794
3 жыл бұрын
@@BruceWayne-zt7vt Because since it's a circle with r=2, the area is equal to the area of a circle with r=2 minus the area of a circle with r=0 (which has no area). There's no such thing as a circle with r=-2, since the radius of a circle is always positive.
@cloudsmasher69
Жыл бұрын
this is amazing. i have been trying to learn this for hours, and now it all makes sense
@DreamscaperTV
6 жыл бұрын
OMG! You have opened my eyes, this is easier than I expected.
@thabangmthethwa7233
9 жыл бұрын
...am a second year student studying Civil engineering. i don't go to lectures anymore. the videos u make are even better than going to lectures. you are a good teacher. thank you very much.
@kristakingmath
9 жыл бұрын
+Thabang Joel LOL, I can't advocate skipping lectures, but I'm really glad the videos are helping!!
@jenniferf6265
9 жыл бұрын
Perfectly taught! You make Calculus simple and understandable. Much better job than most professors. Thanks so much!
@kristakingmath
9 жыл бұрын
+Jennifer Flores I'm glad I could help!
@favio2491
9 жыл бұрын
yo math lady u ma nigga, save the day...erryday...
@beastmodefletcher
7 жыл бұрын
Facts
@SkatingToy
10 жыл бұрын
Thanks for the explanation. The only thing I'm confused about is theta's limits. When is it not from 0 to 2pi? Is it different if there are variables for the y limits? Great tutorial nonetheless, much better than my professor haha
@LeonTGBU
8 жыл бұрын
When converting to polar you have to set your r value to absolute value after conversion otherwise you will obtain a double of the actual volume.
@AnonymousIdiott
6 жыл бұрын
Hey Krista, why is the range for theta 0 to 2pi? Why is it not 0 to pi due to the fact the circle goes from 2 to -2?
@sirangus7571
9 жыл бұрын
This was almost an exact problem my prof went over in class...only you explained it so much better. Thanks CE!! You''re the bomb!!!
@kristakingmath
9 жыл бұрын
+Larry Gulliver So glad I could help! :D
@dstan16224
5 жыл бұрын
Thank you Ma'am...This video really helped a lot...👍👍👍
@ungell
10 жыл бұрын
Elegant explanation- keep up the good work! Thank you, I appreciate your videos!
@kristakingmath
10 жыл бұрын
Thanks! I'm so glad you like them. :)
@rodyys
8 жыл бұрын
you're the best! This video is what i was looking for like 40 minutes!
@kristakingmath
8 жыл бұрын
+AJJRodyys I'm so glad it helped!!
@JKhal
7 жыл бұрын
Amazing vid! from the step where you get cos(theta) x [16/3 - 32/10 ], simplified to 32/15 x cos(theta)
@mohannedalghazo7375
3 жыл бұрын
Thank you for pinning the correction because it was going to make me so confused!. Anyhow, it was a great explanation and thank you so much!!
@JacobGetter
10 жыл бұрын
Let me know if I'm wrong or not, but the bounds of integration for (r) were [-2,2], which seems a bit strange as the radius can never be a negative value. Now I considered the bounds of integration to be [0,2] which actually turns out to give the same result (Zero), but I if i'm not mistaken this was only a coincidence that it turned out to give the same answer. So I believe in another situation this would have been incorrect as the bounds of integration for (r) must be non negetive. I am a student an want to varify my observation of these results. Thank you in advance!
@ammarshahid7792
5 жыл бұрын
I think you don't need the answer anymore as you probably already graduated. :D
@satpalkaushik8192
3 жыл бұрын
If it is just a coincidence then you can take other examples to prove or disprove yourself
@MisterBinx
9 жыл бұрын
I've just accepted that you add the extra r. I don't get it but for the sake of time I just assume there is a proof that explains that.
@lwfabsman
9 жыл бұрын
+MisterBinx It's called the Jacobian. Whenever you change the variable of integration, you get a Jacobian. If you google change of variable in triple integrals, or the Jacobian, you will know exactly what it is. Have a lovely day.
@MisterBinx
9 жыл бұрын
Funny thing is just last Sunday I spent all day figuring out what the Jacobian is lol. My book is really bad but I understand where the r comes from now. Same for change of variable with spherical coordinates.
@yasinosman9411
3 жыл бұрын
To be completely honest, with math, in the last 3 years I have been taking up level classes and moving up, the best thing is just to do it. No proofs or explanations because, at least for me I cant speak for others, its much easier just doing what is told versus knowing why everything is done. Sorry, I know I am 5 years late, wonder where you are with your math journey
@BigDbsk94
10 жыл бұрын
This is my exact homework problem that I was having problems with!! Thank you!!!
@kristakingmath
10 жыл бұрын
you're welcome, i'm so glad it helped!
@davidlegare5021
8 жыл бұрын
Are the limits of theta always going to be 0 to 2pi unlesss specified that the volume is restricted in a quadrant or octant?
@c82153
7 жыл бұрын
yup
@tanmay-jp7mt
6 жыл бұрын
It might not be zero as there might be a case of 2 quadrant
@ryan2229
2 жыл бұрын
the bounds with respect to r would be 0->2 not -2->2. you can't have a negative r value.
@almewai8008
2 жыл бұрын
OMG. You're genius!! I had alike question here and have been searching answers online and other sources for days. Then eventually, I ended up here. Thanks🙏🙏 Please make more videos on Calculus👍👍
@RuneScapeSteve
7 жыл бұрын
Is theta always bounded by [0,2pi] because I am running into problems where it is bounded by [0,pi] or [0, pi/2]?
@omyrazeem2571
2 жыл бұрын
wow amazing explanation, never knew this was this much easy Love from Pakistan
@observever7808
3 жыл бұрын
Small mistake, but still brillaint! Your voice for some reason helps me to absorb these info better
@miguelgomezescobar5701
9 жыл бұрын
the point in dr, shouldn't it be from 0 to 2 instead of from -2 to 2??
@Barreloffish
7 жыл бұрын
yes, r [0,2] since radius can't be negative. dxdy is made up of region R : x^2 + y^2 = 4, thus r[0,2], θ[0,2π]
@bonbonvrock84
7 жыл бұрын
Despite the fact r>=0, which means r[0,2], the answer still turns out to be zero. So is there sth still wrong or is zero the right answer nonetheless?
@aiugioaawdaw1161
5 жыл бұрын
The angle bounds should be from 0 to pi/2 then multiply the whole integral by 4 to prevent having a 0 answer and you would get an answer of 128/15.
@badbreedftw3337
4 жыл бұрын
aiugioa awdaw11 you obviously don’t understand calculus mate, go over simple integration then comeback
@0796675465
10 жыл бұрын
Shouldnt we take r from (0,2) ?
@TheFarmanimalfriend
4 жыл бұрын
some things. You can not determine a negative distance for r (math nerds excepted). To go from r to 2 is always zero as r is 2. It is incorrect to add an r to an equation that already has an r (rcosθ). Integrals sum from one number to the next. If the number is the same at the start of integration as at the end, the integration will always return zero.
@aspirepolitico724
7 жыл бұрын
The explanation of r=0,2 helped me a lot.
@Ibracadabra52
8 жыл бұрын
Very well explained video! You're explanations are perfect to understand. Unfortunately there was a lot of errors made in this video. r=2 not +-2 and divide by 160/30 not 160/3. Either way keep up the good work!
@jessica8789
3 жыл бұрын
Correct me if I’m wrong, but I thought it would be +-2 because you need two values for your integral, an upper and a lower, and when you take the sqrt of both sides of something, you will usually end up with 2 answers, one positive and one negative of the value on the other side.
@gkbeastboy
7 жыл бұрын
Shoutout to Mrs. Kodan!
@courierjaune1791
8 жыл бұрын
Thank you so much! I spent hours trying to figure out how to do these types of problems and your video made it make sense after only the second time through.
@kristakingmath
8 жыл бұрын
You're welcome, I'm so glad this helped!
@badradish2116
7 жыл бұрын
this is literally the exact problem i was given
@kristakingmath
7 жыл бұрын
I hope it helped!
@reggiecook8923
8 жыл бұрын
So when do you know when to use cylindrical coordinates and spherical coordinates?
@hamadahatem98
8 жыл бұрын
I wanna thank you for the amazing effort that you have made , not only in this video but the whole channel is impressive . and iam not overestimating but you are actually better than the doctors in my colledge . thanks alot and good luck . keep it on
@kristakingmath
8 жыл бұрын
+ahmad hatem Aw thanks! I'm glad you're liking the videos!
@jembo2000a464
3 жыл бұрын
I beliebe that r should be 0 to 2 and not -2 to 2 thats why you got 0 as your final answer
@Lozantrack
9 жыл бұрын
You explained it so much easier then my professor.. thank you!!
@kristakingmath
9 жыл бұрын
Lozantrack Glad I could help!
@rivalo5
8 жыл бұрын
Theta doesn't have to be from 0 to 2pi. It's better to make a sketch. For instance integrating 'y' from [0,2] instead of [-2,2] would give a theta of [-1/2pi,1/2pi]
@Ayberkoski365
8 жыл бұрын
Yeah i was confused about that part too, and i have an exam tomorrow. so you sure about that ?
@TonyFangtf245yay
8 жыл бұрын
How can you get the limits for theta without doing a sketch???
@wendydelgado3972
9 жыл бұрын
I have a question: doesn't r have to be: r>or equal to 0????? there cannot be negative radius right? that s what our professor taught us... so why do u have -2 to 2
@TheFreezingTuberJosh
6 жыл бұрын
Yes, you too.
@ckong25
3 жыл бұрын
You explained it very well. Thanks!
@kristakingmath
2 жыл бұрын
Thank you so much, I'm glad you liked it! :)
@ernest0508
7 жыл бұрын
Thank you for the video. You explain the problem so well!
@kristakingmath
7 жыл бұрын
Thanks, I'm so glad you liked it! :D
@sambhavsingh5088
2 жыл бұрын
thanks a lot. best video yet
@hassamrajpoot8397
4 жыл бұрын
I don't know if it's too late or not , I wanted to ask about theta , the domain of theta would be [0 ,2π] , that's understandable . But the limits will always be from 0 to 2π ? Every time?
@relaxingzone3265
10 жыл бұрын
I think for r its from 0 to 2. Not from -2 to 2. Thanks !
@sabelodavid3127
6 жыл бұрын
symmetry remember
@tomislavstipancik2322
5 жыл бұрын
@@sabelodavid3127 you can't have a negative radius
@thomasshelby5286
2 жыл бұрын
Got 59/60 in my Calculus exam.....
@SPECHALAIGENT
9 жыл бұрын
The limits of r should be from 0 till 2. Cuz if we graph x=root(4-y^2) and x=(-)root(4-y^2) then we simplify them to x^2+y^2=4. Which is the equation of a circle. If we want to evaluate the radius or simply r of the circle, we start from 0 and and up at 2. However, thank you for explaining this. It really helped me understand this and hopefully I'll do great in my exam. :)
@wynetvngw691
2 жыл бұрын
u"r right
@livingalife0171
8 жыл бұрын
Quick question about order of integration. If the question was ordered so that it was: dz dy dx, does that mean after the transformation it would be: dz d(theta) dr?
@grantauletta4324
9 жыл бұрын
Is the domain for theta always 0 to 2pi for these?
@aiugioaawdaw1161
5 жыл бұрын
No. It would be better if you integrate from 0 to pi/2 then multiply the integral by 4. Theta 0 and 2pi have the same value so there are times that they would cancel each other out.
@sadowon2002
3 ай бұрын
glad that i found your video, my exam is tomorrow 😭
@MarkNealJr
6 жыл бұрын
In your example the order of integration was dzdxdy and that was converted to rdzdrd(theta). If the order changes to lets say, dxdydz, would that convert to rdrd(theta)dz?
@Jordie389
9 жыл бұрын
Thank you so much! You are awesome! Keep up the great work. You are one of the best math teachers on youtube :D
@kristakingmath
9 жыл бұрын
Enrica Montez Thank you very much!
@c82153
7 жыл бұрын
very clear and concise, nice video!
@kristakingmath
7 жыл бұрын
Thanks Chris!
@DavidRodriguez-ul3ib
8 жыл бұрын
you ma'am are a god send
@ngacala1
5 жыл бұрын
What video or note app are you using in making your presentation?
@stuartward1357
3 жыл бұрын
square root of r^2 should be +r and -r so why do you automatically assume it is +r as lower bound and not -r?
@somalethapn4583
8 жыл бұрын
Thnks a lot... watching this an hour before exam, and I feel Im saved...
@kristakingmath
8 жыл бұрын
+Hrithu O A You're welcome, I hope the exam went great!
@securevulnerability2331
6 жыл бұрын
Your videos are life saving :)
@Hephaestus_God
3 жыл бұрын
Does anyone know how to go from a "Double integral in rectangular coordinates to a triple integral in cylindrical coordinates?"... I can't find anywhere how to go from a double integral to a triple in another coordinate system. And I ended up with this question on a test.
@anonymousn1nja
8 жыл бұрын
from 2:35 to 3:35 you could've just used r^2=x^2+y^2 to substitute
@dayeongE
10 жыл бұрын
if x is given as constant, (so dzdydx) do I set the constant equal to rcostheta ? and ignore equations for y boundary and just call it 0 to 2pi?
@3ashe8able
9 жыл бұрын
I will pass this course because of you ,,, Thanks alot
@kristakingmath
9 жыл бұрын
That's awesome, you're welcome!
@xxdriftking027xx
8 жыл бұрын
Since we've integrated along the bounds of [-2, 2] in the R plane when the bounds should have been [0, 2], i assume then that the calculations done after 11:02 are incorrect? despite incorrect calculations we should still reach the same final value of zero (0) as 2pi and 0 of sin is still equal to 0 correct? As stated in other comments, the evaluation of these cylindrical coordinates is finding the volume of the solid? How is it possible to have a volume equal to 0?
@mrsebakuna
8 жыл бұрын
if the integrant xz would be 1 then it would be volum
@Ensign_Cthulhu
8 жыл бұрын
My 3D geometry is shaky, but taking a guess, if the thing being integrated for volume is in fact an open surface within the given bounds, then it has no volume.
@devotion_tws
2 жыл бұрын
What if its dz dy dx instead? Would the teta still be 2pi and 0?
@poppyblop484
7 жыл бұрын
Hi, im a little confuse as to why we always need to convert them to their respective coordinate system. why cant we just integrate them using the cartesian coordinate system? Thanks
@drewgraham1482
9 жыл бұрын
Thank you for this video. Very well explained!
@kristakingmath
9 жыл бұрын
+Drew Graham Thanks!
@freaky504
10 жыл бұрын
The "extra" r you multiplied in @ 8:10 is derived from the determinant of the cylindrical Jacobian matrix I guess. Basically, what you're actually doing is applying the integration transformation formula for cylindrical coordinates. Now I get it! :D
@swizzbeats1212
9 жыл бұрын
Hey just a slight mistake, at 12:30 you said and wrote 160/3 when it's 160/30 :)
@alexpalacios4767
7 жыл бұрын
how do you find the bounds if we are only given the bounds for z?
@kojowiafe6574
8 жыл бұрын
In this problem, instead of r < 2 I don't understand why -2< r >2. I thought r cannot be negative.
@karanrawat1659
5 жыл бұрын
So theta always goes from 0 to 2π until unless there is some bound in cylindrical coordinate system
@spencerantoniomarlen-starr3069
8 жыл бұрын
Around time 12:32, the third component of the integrand should probably be 160/30cos(theta) not 160/3cos(theta).
@ItsTopCat
8 жыл бұрын
I caught that error too
@SmileEVERYDAYlaugh
9 жыл бұрын
A triple integral is calculating mass right? So does this means the mass of this cylinder is 0? I think I understand why we get zero, because we evaluated the integral from -2 to 2 in the Z plane, so the volume is essentially just cancelling itself out, but how can the mass be zero? Is it just the effect of cylindrical coordinates? Or am I over thinking this?
@adeedaas
9 жыл бұрын
Pretty sure it's volume
@jazmingtz9463
5 жыл бұрын
I finally understood!! thank you so much!!!
@kristakingmath
5 жыл бұрын
You're welcome, Jazmin, I'm so glad it made sense! :D
@trgarg542
2 жыл бұрын
madam your work on triple integration great
@venkatramansampath
5 жыл бұрын
What type of software you use for teaching?ty
@kristakingmath
5 жыл бұрын
Hey, Venkatraman! It's called Sketchbook.
@littlebits6231
7 жыл бұрын
how come you didnt replace the z in the integral by r?
@TheFallAcademy
10 жыл бұрын
Wait if you integrate from 0 to 2pi with the radius going from -2 to 2 wouldn't you integrate over the circle twice, so shouldn't the radius go from 0 to 2pi.
@jamesharden8061
9 жыл бұрын
No, -2,2 is an integration over the radius and 0 to 2pi integrates over the circumference
@haydenadamson556
8 жыл бұрын
is there an example where theta's domain is not 0 to 2pi
@conradschmidt5608
8 жыл бұрын
If they ask you to do it over the first quadrant of the xy plane then it is from 0 to pi/2, if the angle is from 180 degrees to 290 then it is from pi to (2pi)*(290/360). Just think about which angles on the xy plane the integration covers and make sure it's in radians. I hope that helps.
@haydenadamson556
8 жыл бұрын
thanks man, that helps a lot.
@shuvbhowmickbestin
5 жыл бұрын
Why no one mentions the jacobian?
@thomasshelby5286
2 жыл бұрын
Really helping me!!!!!🔥🔥🔥🔥🔥
@aliyashah4484
9 жыл бұрын
Super dooper hellpful!
@CHEAVICHET
7 жыл бұрын
Why need to add extra r?
@jeremiahstevens6944
8 жыл бұрын
What program are you using to do the writing?
@kristakingmath
8 жыл бұрын
I explain here: www.kristakingmath.com/blog/how-i-create-my-videos
@tonyten8912
3 жыл бұрын
Good explanations
@kristakingmath
2 жыл бұрын
Thank you Munashe! :)
@Techformative557
7 жыл бұрын
Phew..thanks a lot..saved my ass in college Multi variable
@kristakingmath
7 жыл бұрын
You're welcome, glad it helped! :)
@baselbasels3321
8 жыл бұрын
thank you very much it's helped me a lot I have exam tomorrow and I study on your videos thaaanx 😁
@kristakingmath
8 жыл бұрын
You're welcome, I'm so glad it helped! Good luck on your exam, I hope it goes great! :D
@legoyoda9
6 жыл бұрын
Lol professor gave this as homework. Thank you so much
@mrdragon8760
9 жыл бұрын
I am confused about r being -2 to 2.. it does'nt make sense...some ppl are saying its 0 to 2 is'nt that correct ? I think that the answer came 0 because of that
@louiecarmen7050
9 жыл бұрын
Mohammad Nadeem r must be from 0 to 2. [-2,2] is the diameter of the circle and its not the r.
@iZenZation
5 жыл бұрын
Thanks.
@devrajnaik1856
7 ай бұрын
but why r starts from -2? It should not!
@decideanozivamugwagwa798
3 жыл бұрын
hie can anyone please explain to me how the limts from 0 to 2pi were obtained...
@xJY369x
7 жыл бұрын
shouldn't the bounds for Theta be -Pi/2 to pi/2??
@bonbonvrock84
7 жыл бұрын
Jonathan Medrano How did you work out the bounds for theta?
@ashtonkrause7211
4 жыл бұрын
Isn't it simpler to recognize that z=sqrt(x^2+y^2) looks a lot like if you solved for radius. r=sqrt(x^2+y^2). The same goes for x=sqrt(4-y^2) which is what would happen if you solved for x using r^2=x^2+y^2. That means 4 is r^2, or in this case x^2, so x=sqrt(4)= + or - 2. I know this video is really old but this makes it a bit easier if the limits can be solved using r^2=x^2+y^2.
@f-22raptor25
4 жыл бұрын
defiantly better to change to r^2
@drewksidetour
8 жыл бұрын
couldn't you have just put in R^2 under the radical instead of rcostheta and rsintheta, since it equals x^2+y^2?
@ItsTopCat
8 жыл бұрын
yes, she just took the longer approach
@jorgejeronimo7421
8 жыл бұрын
does dy always convert to d(theta ) to (0,2pi) ?
@kristakingmath
8 жыл бұрын
Yes, unless there's something that limits the interval for theta. For example, if the lower limit for y in this problem had be 0, that would tell you that you're only dealing with the half circle that sits above the x-axis, not the full circle. In that case, the half circle above the x-axis only encompasses angles [0,pi], not [0,2pi] like for the full circle. So your limits of integration for theta would then have been only [0,pi]. I hope that makes sense!
Пікірлер: 233