Converting triple integrals to cylindrical coordinates (KristaKingMath)

[@kristakingmath]

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]

It really freaked me out!

[@Bestofchatgpt]

Do you still end up with 0?

[@Daniel.Rosenthal]

@@Bestofchatgpt yes

[@anands9407]

why?i think it should be -2,2 and thetha should be 0 to pi...

[@ITACHIUCHIHA-yn1lo]

@@anands9407 because r never be taking in negative

[@ThePhillyg09]

13 minutes of work for the answer of ZERO. Ladies and Gentlemen... Calculus.

[@kristakingmath]

+Philip Geraci LOL

[@hg2.]

LOL!

[@KakinTseGamePlayer]

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]

Fantastic

[@azizal-hunaiyyan5041]

Actually no, I did it and still, the answer is 0 :)

[@paclax7696]

amazing explanation and video, however the bounds for r should be 0

[@harishbaskar2935]

guys small mistake. r is from 0 to 2... x=[0,2] its the radius not diameter

[@BruceWayne-zt7vt]

can you explain why r is from 0 to 2 instead of -2 to 2. #iambadatMath :)

[@alexanderbudianto7794]

@@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.

[@kristalenee8925]

This really helped me but why did you make r from (-2,2) instead of from (0,2) ?

[@favio2491]

yo math lady u ma nigga, save the day...erryday...

[@beastmodefletcher]

Facts

[@cloudsmasher69]

this is amazing. i have been trying to learn this for hours, and now it all makes sense

[@DreamscaperTV]

OMG! You have opened my eyes, this is easier than I expected.

[@thabangmthethwa7233]

...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]

+Thabang Joel LOL, I can't advocate skipping lectures, but I'm really glad the videos are helping!!

[@SkatingToy]

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

[@jenniferf6265]

Perfectly taught! You make Calculus simple and understandable. Much better job than most professors. Thanks so much!

[@kristakingmath]

+Jennifer Flores I'm glad I could help!

[@LeonTGBU]

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.

[@vor946]

damn you are even explaining common denominator in triple integrals in cylindrical coords video thats why you are great teacher, i passed calculus with high marks thanks to you (and pauls notes website)

[@kristakingmath]

Congratulations on passing calc... your hard work paid off! And I'm so glad I was able to help along the way! :)

[@dstan16224]

Thank you Ma'am...This video really helped a lot...👍👍👍

[@JKhal]

Amazing vid! from the step where you get cos(theta) x [16/3 - 32/10 ], simplified to 32/15 x cos(theta)

[@ungell]

Elegant explanation- keep up the good work! Thank you, I appreciate your videos!

[@kristakingmath]

Thanks! I'm so glad you like them. :)

[@MisterBinx]

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]

+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]

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]

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

[@AnonymousIdiott]

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?

[@miguelgomezescobar5701]

the point in dr, shouldn't it be from 0 to 2 instead of from -2 to 2??

[@Barreloffish]

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]

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?

[@BigDbsk94]

This is my exact homework problem that I was having problems with!! Thank you!!!

[@kristakingmath]

you're welcome, i'm so glad it helped!

[@JacobGetter]

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]

I think you don't need the answer anymore as you probably already graduated. :D

[@satpalkaushik8192]

If it is just a coincidence then you can take other examples to prove or disprove yourself

[@sirangus7571]

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]

+Larry Gulliver So glad I could help! :D

[@almewai8008]

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👍👍

[@mohannedalghazo7375]

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!!

[@davidlegare5021]

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]

yup

[@tanmay-jp7mt]

It might not be zero as there might be a case of 2 quadrant

[@0796675465]

Shouldnt we take r from (0,2) ?

[@badradish2116]

this is literally the exact problem i was given

[@kristakingmath]

I hope it helped!

[@Ibracadabra52]

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]

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.

[@rodyys]

you're the best! This video is what i was looking for like 40 minutes!

[@kristakingmath]

+AJJRodyys I'm so glad it helped!!

[@courierjaune1791]

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]

You're welcome, I'm so glad this helped!

[@TheFarmanimalfriend]

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.

[@gkbeastboy]

Shoutout to Mrs. Kodan!

[@aiugioaawdaw1161]

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]

aiugioa awdaw11 you obviously don’t understand calculus mate, go over simple integration then comeback

[@rivalo5]

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]

Yeah i was confused about that part too, and i have an exam tomorrow. so you sure about that ?

[@TonyFangtf245yay]

How can you get the limits for theta without doing a sketch???

[@observever7808]

Small mistake, but still brillaint! Your voice for some reason helps me to absorb these info better

[@RuneScapeSteve]

Is theta always bounded by [0,2pi] because I am running into problems where it is bounded by [0,pi] or [0, pi/2]?

[@relaxingzone3265]

I think for r its from 0 to 2. Not from -2 to 2. Thanks !

[@sabelodavid3127]

symmetry remember

[@tomislavstipancik2322]

@@sabelodavid3127 you can't have a negative radius

[@reggiecook8923]

So when do you know when to use cylindrical coordinates and spherical coordinates?

[@sadowon2002]

glad that i found your video, my exam is tomorrow 😭

[@aspirepolitico724]

The explanation of r=0,2 helped me a lot.

[@ryan2229]

the bounds with respect to r would be 0->2 not -2->2. you can't have a negative r value.

[@hamadahatem98]

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]

+ahmad hatem Aw thanks! I'm glad you're liking the videos!

[@omyrazeem2571]

wow amazing explanation, never knew this was this much easy Love from Pakistan

[@wendydelgado3972]

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]

Yes, you too.

[@sambhavsingh5088]

thanks a lot. best video yet

[@Lozantrack]

You explained it so much easier then my professor.. thank you!!

[@kristakingmath]

Lozantrack Glad I could help!

[@ckong25]

You explained it very well. Thanks!

[@kristakingmath]

Thank you so much, I'm glad you liked it! :)

[@Hephaestus_God]

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.

[@ernest0508]

Thank you for the video. You explain the problem so well!

[@kristakingmath]

Thanks, I'm so glad you liked it! :D

[@swizzbeats1212]

Hey just a slight mistake, at 12:30 you said and wrote 160/3 when it's 160/30 :)

[@anonymousn1nja]

from 2:35 to 3:35 you could've just used r^2=x^2+y^2 to substitute

[@SPECHALAIGENT]

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]

u"r right

[@trgarg542]

madam your work on triple integration great

[@aliyashah4484]

Super dooper hellpful!

[@stuartward1357]

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?

[@DavidRodriguez-ul3ib]

you ma'am are a god send

[@c82153]

very clear and concise, nice video!

[@kristakingmath]

Thanks Chris!

[@kojowiafe6574]

In this problem, instead of r < 2 I don't understand why -2< r >2. I thought r cannot be negative.

[@somalethapn4583]

Thnks a lot... watching this an hour before exam, and I feel Im saved...

[@kristakingmath]

+Hrithu O A You're welcome, I hope the exam went great!

[@3ashe8able]

I will pass this course because of you ,,, Thanks alot

[@kristakingmath]

That's awesome, you're welcome!

[@spencerantoniomarlen-starr3069]

Around time 12:32, the third component of the integrand should probably be 160/30cos(theta) not 160/3cos(theta).

[@ItsTopCat]

I caught that error too

[@livingalife0171]

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?

[@karanrawat1659]

So theta always goes from 0 to 2π until unless there is some bound in cylindrical coordinate system

[@grantauletta4324]

Is the domain for theta always 0 to 2pi for these?

[@aiugioaawdaw1161]

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.

[@securevulnerability2331]

Your videos are life saving :)

[@freaky504]

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

[@haydenadamson556]

is there an example where theta's domain is not 0 to 2pi

[@conradschmidt5608]

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]

thanks man, that helps a lot.

[@jembo2000a464]

I beliebe that r should be 0 to 2 and not -2 to 2 thats why you got 0 as your final answer

[@jazmingtz9463]

I finally understood!! thank you so much!!!

[@kristakingmath]

You're welcome, Jazmin, I'm so glad it made sense! :D

[@drewgraham1482]

Thank you for this video. Very well explained!

[@kristakingmath]

+Drew Graham Thanks!

[@poppyblop484]

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

[@Jordie389]

Thank you so much! You are awesome! Keep up the great work. You are one of the best math teachers on youtube :D

[@kristakingmath]

Enrica Montez Thank you very much!

[@hassamrajpoot8397]

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?

[@dayeongE]

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?

[@thomasshelby5286]

Really helping me!!!!!🔥🔥🔥🔥🔥

[@baselbasels3321]

thank you very much it's helped me a lot I have exam tomorrow and I study on your videos thaaanx 😁

[@kristakingmath]

You're welcome, I'm so glad it helped! Good luck on your exam, I hope it goes great! :D

[@SmileEVERYDAYlaugh]

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]

Pretty sure it's volume

[@ashtonkrause7211]

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]

defiantly better to change to r^2

[@MarkNealJr]

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?

[@tonyten8912]

Good explanations

[@kristakingmath]

Thank you Munashe! :)

[@legoyoda9]

Lol professor gave this as homework. Thank you so much

[@Techformative557]

Phew..thanks a lot..saved my ass in college Multi variable

[@kristakingmath]

You're welcome, glad it helped! :)

[@alexpalacios4767]

how do you find the bounds if we are only given the bounds for z?

[@iZenZation]

Thanks.

[@devotion_tws]

What if its dz dy dx instead? Would the teta still be 2pi and 0?

[@littlebits6231]

how come you didnt replace the z in the integral by r?

[@mrdragon8760]

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]

Mohammad Nadeem r must be from 0 to 2. [-2,2] is the diameter of the circle and its not the r.

[@TheFallAcademy]

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]

No, -2,2 is an integration over the radius and 0 to 2pi integrates over the circumference

[@xxdriftking027xx]

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]

if the integrant xz would be 1 then it would be volum

[@Ensign_Cthulhu]

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.

[@Eng-Amer-Al-Najjar]

thank you very very much, amazing

[@devrajnaik1856]

but why r starts from -2? It should not!

[@drewksidetour]

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]

yes, she just took the longer approach

[@thomasshelby5286]

Got 59/60 in my Calculus exam.....

[@leylaselman9776]

you are amazing

[@CHEAVICHET]

Why need to add extra r?

[@shuvbhowmickbestin]

Why no one mentions the jacobian?

[@sudhakarmathsacademy]

It is really great mathematician

[@kristakingmath]

Thank you, Sudhakar! :)

[@xJY369x]

shouldn't the bounds for Theta be -Pi/2 to pi/2??

[@bonbonvrock84]

Jonathan Medrano How did you work out the bounds for theta?

[@bartroovers834]

all of this work for just 0 - 0 = 0 :), but i understand, so Thanks!

[@jayyupinhu8710]

Awesome video!

[@kristakingmath]

Thank you!