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Converting triple integrals to cylindrical coordinates (KristaKingMath)
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- เผยแพร่เมื่อ 17 ส.ค. 2024
- ► My Multiple Integrals course: www.kristaking...
Learn how to convert a triple integral from cartesian coordinates to cylindrical coordinates. You'll need to convert the function itself, the order of integration, and the limits of integration before evaluating the triple integral in cylindrical 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?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student-from basic middle school classes to advanced college calculus-figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: www.kristakingm...
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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.
It really freaked me out!
Do you still end up with 0?
@@Bestofchatgpt yes
why?i think it should be -2,2 and thetha should be 0 to pi...
@@anands9407 because r never be taking in negative
13 minutes of work for the answer of ZERO. Ladies and Gentlemen... Calculus.
+Philip Geraci LOL
LOL!
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.
Fantastic
Actually no, I did it and still, the answer is 0 :)
amazing explanation and video, however the bounds for r should be 0
yo math lady u ma nigga, save the day...erryday...
Facts
This really helped me but why did you make r from (-2,2) instead of from (0,2) ?
guys small mistake. r is from 0 to 2...
x=[0,2] its the radius not diameter
can you explain why r is from 0 to 2 instead of -2 to 2. #iambadatMath :)
@@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.
OMG! You have opened my eyes, this is easier than I expected.
this is amazing. i have been trying to learn this for hours, and now it all makes sense
...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.
+Thabang Joel LOL, I can't advocate skipping lectures, but I'm really glad the videos are helping!!
Perfectly taught! You make Calculus simple and understandable. Much better job than most professors. Thanks so much!
+Jennifer Flores I'm glad I could help!
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!!!
+Larry Gulliver So glad I could help! :D
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?
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
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.
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)
Congratulations on passing calc... your hard work paid off! And I'm so glad I was able to help along the way! :)
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!!
Amazing vid!
from the step where you get cos(theta) x [16/3 - 32/10 ],
simplified to 32/15 x cos(theta)
Thank you Ma'am...This video really helped a lot...👍👍👍
Elegant explanation- keep up the good work! Thank you, I appreciate your videos!
Thanks! I'm so glad you like them. :)
the point in dr, shouldn't it be from 0 to 2 instead of from -2 to 2??
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]
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?
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.
+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.
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.
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
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👍👍
Is theta always bounded by [0,2pi] because I am running into problems where it is bounded by [0,pi] or [0, pi/2]?
you're the best! This video is what i was looking for like 40 minutes!
+AJJRodyys I'm so glad it helped!!
Are the limits of theta always going to be 0 to 2pi unlesss specified that the volume is restricted in a quadrant or octant?
yup
It might not be zero as there might be a case of 2 quadrant
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
Shouldnt we take r from (0,2) ?
This is my exact homework problem that I was having problems with!! Thank you!!!
you're welcome, i'm so glad it helped!
wow amazing explanation, never knew this was this much easy Love from Pakistan
glad that i found your video, my exam is tomorrow 😭
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!
I think you don't need the answer anymore as you probably already graduated. :D
If it is just a coincidence then you can take other examples to prove or disprove yourself
Shoutout to Mrs. Kodan!
Small mistake, but still brillaint! Your voice for some reason helps me to absorb these info better
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.
this is literally the exact problem i was given
I hope it helped!
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.
aiugioa awdaw11 you obviously don’t understand calculus mate, go over simple integration then comeback
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.
You're welcome, I'm so glad this helped!
So when do you know when to use cylindrical coordinates and spherical coordinates?
The explanation of r=0,2 helped me a lot.
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
+ahmad hatem Aw thanks! I'm glad you're liking the videos!
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.
the bounds with respect to r would be 0->2 not -2->2. you can't have a negative r value.
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!
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.
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
Yes, you too.
I think for r its from 0 to 2. Not from -2 to 2. Thanks !
symmetry remember
@@sabelodavid3127 you can't have a negative radius
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?
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]
Yeah i was confused about that part too, and i have an exam tomorrow. so you sure about that ?
How can you get the limits for theta without doing a sketch???
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?
Is the domain for theta always 0 to 2pi for these?
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.
Hey just a slight mistake, at 12:30 you said and wrote 160/3 when it's 160/30 :)
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
What if its dz dy dx instead? Would the teta still be 2pi and 0?
In this problem, instead of r < 2 I don't understand why -2< r >2. I thought r cannot be negative.
You explained it so much easier then my professor.. thank you!!
Lozantrack Glad I could help!
So theta always goes from 0 to 2π until unless there is some bound in cylindrical coordinate system
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?
madam your work on triple integration great
Thnks a lot... watching this an hour before exam, and I feel Im saved...
+Hrithu O A You're welcome, I hope the exam went great!
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. :)
u"r right
how do you find the bounds if we are only given the bounds for z?
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?
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?
if the integrant xz would be 1 then it would be volum
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.
Thank you so much! You are awesome! Keep up the great work. You are one of the best math teachers on youtube :D
Enrica Montez Thank you very much!
thanks a lot.
best video yet
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?
you ma'am are a god send
Thank you for the video. You explain the problem so well!
Thanks, I'm so glad you liked it! :D
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.
No, -2,2 is an integration over the radius and 0 to 2pi integrates over the circumference
is there an example where theta's domain is not 0 to 2pi
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.
thanks man, that helps a lot.
from 2:35 to 3:35 you could've just used r^2=x^2+y^2 to substitute
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
Mohammad Nadeem r must be from 0 to 2. [-2,2] is the diameter of the circle and its not the r.
I will pass this course because of you ,,, Thanks alot
That's awesome, you're welcome!
You explained it very well. Thanks!
Thank you so much, I'm glad you liked it! :)
I beliebe that r should be 0 to 2 and not -2 to 2 thats why you got 0 as your final answer
how come you didnt replace the z in the integral by r?
What video or note app are you using in making your presentation?
wow! wow!... this saves me a hell lot of time... be writing test in 2hours... thanks for the simple explanation without having to draw all the crazy 3D graphs...
Luci van tangent I'm glad it helped, and I hope your test went great!!
It went all well ☺ 😀
Luci van tangent That's great news! Congratulations on being done!!
Thanks.
hie can anyone please explain to me how the limts from 0 to 2pi were obtained...
very clear and concise, nice video!
Thanks Chris!
Lol professor gave this as homework. Thank you so much
Your videos are life saving :)
Why no one mentions the jacobian?
Thank you for this video. Very well explained!
+Drew Graham Thanks!
but why r starts from -2? It should not!
What type of software you use for teaching?ty
Hey, Venkatraman! It's called Sketchbook.
Around time 12:32, the third component of the integrand should probably be 160/30cos(theta) not 160/3cos(theta).
I caught that error too
I finally understood!! thank you so much!!!
You're welcome, Jazmin, I'm so glad it made sense! :D
Super dooper hellpful!
couldn't you have just put in R^2 under the radical instead of rcostheta and rsintheta, since it equals x^2+y^2?
yes, she just took the longer approach
Phew..thanks a lot..saved my ass in college Multi variable
You're welcome, glad it helped! :)
thank you very much it's helped me a lot I have exam tomorrow and I study on your videos thaaanx 😁
You're welcome, I'm so glad it helped! Good luck on your exam, I hope it goes great! :D
all of this work for just 0 - 0 = 0 :), but i understand, so Thanks!
Good explanations
Thank you Munashe! :)
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.
defiantly better to change to r^2
Got 59/60 in my Calculus exam.....
can anyone tell me how she got the teta?
shouldn't the bounds for Theta be -Pi/2 to pi/2??
Jonathan Medrano How did you work out the bounds for theta?
instead of doing all that extra work for converting the limits for 'x' couldn't you just leave the variables in terms of x and y?
you would get
x = sqrt(4 - y^2)
*square both sides*
x^2 = 4 - y^2
x^2 + y^2 = 4
*change (x^2 + y^2) into r^2*
r^2 = 4
r = +- 2
Really helping me!!!!!🔥🔥🔥🔥🔥