Calculus 3 Lecture 14.7: TRIPLE Integrals Over Regions with CYLINDRICAL or SPHERICAL Coord.

[@josue4262]

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[@seenathgooding181]

Spherical Co-ordinates start @2:08:30 if anyone needed to know like I did.

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[@junior5146]

7:00 Ex 1 26:38 Ex 2 region bound by two functions (Good example to figure which is small or bigger function) 42:05 Ex 3 C.O.M 1:22:10 Ex 4 C.O.M 1:43:30 Ex 5 Moments of inertia problem 2:08:00 Intro spherical coordinates 2:19:00 Ex 6 Using spherical coordinates. p-simple instead of Z-simple. Note: p is ALWAYS positive. The angle is ALWAYS between 0 and pi. Thetha is ALWAYS 0 to 2pi. 2:36:20 Ex 7 spherical coordinates 2:50:00 Ex 8 Bound by two functions 3:02:00 Ex 9

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@@Deadhead710 is not getting a video video lol I see the story of video lol I think still a good taut Friend friend so I’m I’m trying lol sI I just got got ssstssstssststssssssstsssssstdssssssttss

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@@maitumeloteemane9593 is snot out of the bed that that I got to go eat it yyttttyytttdtttdttttdttttytd yo eteyttetsttttt d etyo yo ttettttteteytytyyeyyydtyeyyrtetttettettetetetetetettetetteytteeyteytyeyrreydttsydtdtttdtdtdttttttttttttdtttdtdtdttdtdttdtddtdtdteyetryyeteeyyyyteyteyteytetetryyyryyrryrrrytetyytetetettetryette

[@justinfish4747]

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[@keldonchase4492]

8:00 - Example 1 27:14 - Example 2 42:00 - Example 3 1:22:49 - Example 4 1:45:42 - Example 5 2:19:31 - Example 6 2:37:04 - Example 7 2:51:16 - Example 8 3:02:43 - Example 9

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Whats Up Professor Leonard! I just want to thank you sincerely . Its crazy to think that my whole Calculus journey was altered by a man I will probably never meet. You've allowed to me to actually understand the concepts, and now in multivariable, begin to tap into my creative side of math. Absolutely TREMENDOUS! Can't thank you enough... I can only live with your passion for math to truly show my appreciation. Cheers Prof. L. Hope everything is well!

[@KingMiiidas]

Reminds me of a Calc professor I had at Mesa Community; 3 hour classes taught by a professor who was both passionate and thorough with a conviction to give us an education worth the cost of tuition. He had a 2 hour drive to campus but showed up to class every day with a smile on his face. Hardest class I've ever taken but also the most satisfying. Glad to see there's more professors like him out there.

[@adamt6024]

Wait calc 3 was the hardest class you took what about physics lol that class to me is way harder

[@smoshfann24]

@@adamt6024 Yeah I'm in physics 2 right now and the time I spend on calc is negligible compared to the time I spend on physics. My class started with 22 people and were down to 12 after the second test

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@@smoshfann24 damm, dude my physics class at LBCC, CA started from 72 down to 33.

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These videos are so incredibly helpful, my entire calc class here in Pakistan relies on them to get through exams. I want to send you a big bouquet of flowers and a packet of tight t shirts to say thanks!!!

[@RafaSyed]

What are you studying ?

[@RafaSyed]

Im Doing BS in EEE

[@JawadAli-tj3ec]

Oh sorry rafa ... I thought thT u r wid me ,,,,,but u r with sakena.:-) :-)

[@RafaSyed]

Unfortuanately, Sukina can't reply to comments I guess so I'm with none. What I study was my reply to your question. It was a noob move to delete that reply Jawad

[@JawadAli-tj3ec]

Rafa Syed ,hmm noob ......I thought that when skina --------? Anyhow u looks so smart even u have noticed my cmnts,,,wao,,,,,,,From which university u r perusing EEE,,,,,,

[@obadam.6999]

Is it just me who got 8π/3 instead of π/3 @ 3:12:06

[@praisegeorge350]

yeah me too got 8pi/3

[@iamasian329]

Checked using an online calculator (yep, calculator for triple integrals exist online) and it was 8pi/3

[@angrycat2789]

I chcked with wolframalphaand yes I also got 8pi/3. You can paste this on wolframalpha to see for yourself: Integrate[ Integrate[ Integrate[(rho^2 ) (Sin[phi]), {rho, 0, 2*Sec[phi]}], {phi, 0, Pi/4}], {theta, 0, 2 Pi}]

[@kangspike7408]

Yuop I wasted an hour recalculating this.... I got 8pi/3, My roommate got 8pi/3 and the online calculator also got 8pi/ 3. Please like Obada M's comment so it can be shown on the top for future potential victims.

[@ganeshramaswamy911]

we are essentially calculating the volume of a cone here.. V = 1/3 pi r^2 h.. you can also check it simply by plugging r = 2 and h = 2 from the figure..

[@globalevents0]

This is my final section of Calculus 3! Thank you for the work, and dedication you put in to these videos. As you know, you have helped thousands of students achieve mathematical greatness, and every one of us will say that you are the best to do it.

[@lukedimitriades9835]

You help visualize and understand when most teachers just write down the formulas and do examples. And for that I applaud you sir

[@graceless000]

Literally got more from you than my own professor. He just doesn't seem to get as excited as you. He also doesn't tell us these little tricks. Now it may be because we're online and it's covid season and my lectures are only 50 minutes but this has been so helpful and makes me feel less dumb. Thank you!

[@Omegas88]

When Prof. Leonard is the savior of your grade.

[@jameschen2308]

I was literally listening to the weeknd while taking these notes XD

[@calinciocanea8548]

@@jameschen2308 c c x cx🎉

[@GenesisWithin]

Hey Professor Leonard, you've been invaluable to me with my Calc 3 and 2 studies. Thank you so much! :) Also, at 3:12:13 I got 8pi/3. Not sure if it's correct...

[@ayeshaalhosani6358]

Yes I got 8pi/3 too. I guess he missed the 2 besides sec(phi) when substituting the upper function of Rho (i.e., after integrating with respect to Rho).

[@anand26shweta]

Same, I've been checking for the ans many times but it's the same 8 pi/3. Thx for confirming.

[@alessandrovecchi5147]

Ofc it's 8π/3. It's a cone whose base is a circle of radius 2 and whose height is 2. Thus the volume is Ab*h/3=π*2^2*2/3=8π/3 No need to do the integral😉

[@JM-ty6uq]

@@alessandrovecchi5147 Yes if you just want the answer to the volume, you don't need the integral! Now without doing an integral, give me the mass of this cone when the density is proportional to phi*theta

[@rangerbeast]

You are right

[@ahmedrateb1259]

at 3:12:09, the correct answer for the volume is 8pi/3 not pi/3, this can be proven by the fact that the desired object is a true cone with a circle base at z=2 with a radius of 2 and the apex of the cone is at the origin so the height of the cone is 2. using the formula for the volume of the cone v=1/3*pi*r^2*h, you get 8pi'3

[@pastelshoal]

This is my last week of calc 3 and god do I wish I just skipped out on lectures and watched these. The examples he does are not nearly as complex as what we do at my uni, but the way the material is presented is super informative and really gives me an idea of how to approach these problems and what they mean. Thank you!!

[@liantiu8185]

> Ex 1 7:00 Ex 2 region bound by two functions (Good example to figure which is small or bigger function) 26:38 Ex 3 C.O.M 42:05 Ex 4 C.O.M 1:22:10 Ex 5 Moments of inertia problem 1:43:30 - Intro spherical coordinates 2:08:00 > Ex 6 Using spherical coordinates. p-simple instead of Z-simple. > Note: p is ALWAYS positive. The angle is ALWAYS between 0 and pi. Thetha is ALWAYS 0 to 2pi. 2:19:00 > Ex 7 spherical coordinates 2:36:20 > Ex 8 Bound by two functions 2:50:00 > Ex 9 3:02:00

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I passed calc 3 in the summer (1 month) and it was all thanks to professor leonard :D

[@cubby091398]

I am setting at an 86% in my Multivariable Calculus class. Hopefully, I can maintain that or maybe go higher. This has been a really tough class. The videos are really helpful and I wouldn't be doing this well if it wasn't for them. Thank you very much.

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, off

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I'm a colombian student, you just saved me Leonard, my note went from 32 to 80 thanks to you

[@georgesadler7830]

Professor Leonard ,thank you for another tremendous and long video/lecture on Triple Integrals Over Regions with Cylindrical and Spherical Coordinates in Multivariable Calculus. This is material that I saw in multiple courses in college, however the explanation was hard to understand.. Although the material in Calculus Three continues to be massive, this type of teaching style really helps me absorb the material.

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I swear to god watching an hour of this the day of the exam saved my ass. Thanks Mr. Leonard lol

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[@tauceti8341]

0:59. Remember that X's & Y's can be re-written as x=rcos y=rsin. Why? This is cylindrical form. What is cylindrical form? Cylindrical form is also polar form with a height z, where z=z (height). Thus the function for tripple integrals f(x,y,z) -> f(rcos,rsin,z)

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yep

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you're the shit man. guys if you want to pass calculus you cannot do it without Leonard. period

[@markkennedy9767]

2:16:40 that's a great way to explain the intuition behind where the rho sin phi comes from in spherical coordinates

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Hey I go to FSU too, stopped going to class and just started watching professor leonard's videos instead

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[@jgc9199]

you made spherical triple integrals look so easy lol .

[@richardwere8000]

I have a test on this in 5 hours time. Watch video or practice examples? Life choices...

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+Richard Were That's what I really like about Leonard though, he usually has a lot of worked examples. At some points you can pause and try to work through it yourself as well.

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+Richard Were same here! D: lol

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watch videos

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good thing is that these videos go through example problems :), hope yalll did well

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I am kinda sad after my finals on monday, I'll miss this great teacher. If I forgot after the final to say, just thank you prof. You really are the teacher here. Everyone can be a prof but not most of them can be a teacher. We had a prof that just does torture his students and try to bully them. I kinda get used to be seen as failure though, those old farts are above the skies in their perspective probably. However, just thanks.

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Man I wish I had this guy for my calculus class, makes everything so easy to understand

[@victorh780]

3:12:00 ---> V = 8pi/3

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This guy doesn't miss. Absolute wizard.

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I started nodding my head subconsciously.

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honestly your lectures are amazing, at which university do you teach, because I would move there just to be in your class, and I live in South Africa so...

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The algorithm auto-played this in my sleep and I woke up knowing triple integrals. Thanks Professor Leonard!

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[@Peter_1986]

There is a correction annotation at 3:12:20, which says *_"Mistake!!!! You should get 8pi/3 .... I changed the function on the fly and forgot to change the answer!! My bad :)"._* You can read it if you download the extension "Annotations Restored".

[@shawnarmstrong6824]

Math notation can be displayed at: try.websharper.com/embed/setr/0000Du [If not rendering notation, refresh page] CH 14.7 00:00:00 INFO: Cylindrical Review *) Cylindrical conversion *)*) x = rcos(theta) *)*) y = rsin(theta) *)*) z = z *)*) x^2+y^2 = r^2 *)*) dV = r dz dr d theta *) Blue print: int_(theta_1)^(theta_2)int_(r_1)^(r_2)int_(z_1)^(z_2)[f(rcos(theta),rsin(theta),z)]r dzdrd theta *)*) 'z' must always be first 00:07:05 INFO: Cylindrical Triple Integral Example 1 *) int_()^()int_()^()int_(T)^()[y]dV, T: Solid bound by z = 4-x^2-y^2 in first octant *) STEP1: Find z-simple *)*) 0 2pi *)*) 0 < theta < 2pi *) After setup: int_(0)^(2pi)int_(0)^(pi/4)int_(0)^(2)[rho*sin(phi)cos(theta)*rho cos(phi)]rho^2 sin(phi) d rho d phi d theta 03:03:00 INFO: Spherical Triple Integral Example 4 *) Volume of a solid bounded by z = 2 & z = sqrt[x^2+y^2] *) Solve for rho-simple *)*) Graph it *)*) Convert to spherical: z = 2 -> rho*cos(phi) = 2 -> rho = 2/cos(phi) -> rho = 2sec(phi) *)*) 0 z = y, z = y is line with slope 1 -> tan(phi) = 1 therefore, phi = pi/4 *)*) 0

[@ppgamerr3897]

2:20:05 I totally like his body language.. it’s so energetic 😂😂😂

[@devinwalker4161]

My professor just recites the textbook verbatim, which isn’t helpful. But you Professor Leonard explain everything so well, thank you.

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Exactly. The value in having a professor, is that they can do what the textbook cannot do, which is to give you the confidence that they have an understanding of the subject material, as motivation for you to understand it too.

[@dzdzdzdzdzmr8696]

For Ex at 2:52:33 is the reason why we don't draw the full circle y^2+z^2=4 when doing rho-simple because phi only goes from 0 to pi starting from the z-axis and therefore touches z=y first? And is that the same reason why he only included the right side of the half circle. Thanks for answering

[@lindah.9002]

Notes to self: 2:13:00 Spherical Coordinates

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thx for the awesome lecture vids.

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You saved my life Professor!!!

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Love your videos, saved my life

[@AP-ek6mg]

Might wanna double check example at 3:11:54, you said answer was pi/3, getting 8pi/3, double checked my work online, mistake?

[@ProfessorLeonard]

+Adam Pluguez Thanks Adam, I put a blurb up there to note that.

[@muathalha1532]

Professor Leonard Prof please do ODE pleeeeeeerereeez :)

[@emilianorodriguez807]

I'm getting pi/3

[@emilianorodriguez807]

oops I did this for z=1, which gives us pi/3.

[@sarahmejia2220]

Does anyone understand what he means by the disc being heavier at the edge of the cylinder when r gets bigger?? @1:14:35 Just some clarification. So does each disc still have uniform density at every point of "z"="r"= whatever number??

[@brandonlui]

I love u man thanks were studying for our final

[@geraldochavez4988]

damn watching your videos makes me feel like im cheating. i feel like your holing my hand the whole way, but im able to understand and prep myself for questions. Thank you for making the material less difficult.

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Professor Leonard!!! I love you man.

[@joshuasizer1709]

You make me excited to learn math! Thank you!

[@graceless000]

At 3:12:23 I didn't get pi/3. I got (4pi((sqrt2) - 1))/(3). I'm not quite sure how the answer is just pi/3. :/

[@shantelaloe2557]

I'm not sure if you saw the other comments, but the answer should be 8pi/3. After you use u-sub to integrate phi you will end up with (8/3) u^-2/2 with bounds sqrt(2)/2 to 1. Your problem most likely came from evaluating the bounds. When you plug in sqrt(2)/2 and raise it to the -2 power, you end up with 2, not sqrt(2). Hope this helps

[@captainprice8886]

the superman, rescue to our exams

[@markkennedy9767]

At 32:12 he gives the reason for not taking z= -9 as"getting the volume above the xy plane". But wouldn't z = -9 never be a solution anyway since 8z= x^2+y^2. And x^2+y^2 is always positive so z would have to be positive.

[@buddydog1956]

The dude is a cool breeze! He's the best!

[@rkitlee]

at 39:34 why is the graph of the region have a radius of 2 root 2, shouldn't it just be a radius of 3? This is the first time i've been confused in one of these videos.

[@welais]

We're looking at the level curve when z=1, so x^2+z^2=8. THerefore, the radius is sqrt(8) i.e. 2 root 2 :)

[@superharryboy]

OOOOOHHH ok I was so confused with this. Thanks!

[@sambhavjain6766]

My doubt is in example 2 i think we can find volume below x-y plane which will be considered as depth but in example z not equals to -9 because it does not satisfy 8Z^2=X^2+Y^2.

[@Festus2022]

At 2:54:48 you shade in "our region" on the trace drawing for positive Y-values but NOT the mirror image region corresponding to negative Y-values. Wouldn't the negative Y-values be included in our region?

[@uraibasif8053]

modulus of Y only on the +ve axis

[@bogboyz]

i just woke up to this. no idea what i fell asleep to that got me here, i hardly made it through math

[@jameshanna2027]

Greatest of all time

[@jasonfrimpong7208]

Amazing I'm very grateful

[@goodc.2853]

Is the shortcut method that Professor explains at timestamp 1:00:00 for the center of mass applicable to any problem involving cylinder/spherical coordinates as long as the mass is evenly spread out in the center of z?

[@khayatu6272]

@TheLeonard... Announce Linear Algebra videos !!!

[@VolkanAydngul]

@2:52:43 WHY ARE WE ONLY DEALING WITH HALF CIRCLE?? CAN ANYONE PLEASE EXPLAIN ME?

[@brookskennedy9487]

im not religious but god bless you you incredible human being you are my savior

[@hussamabdulghani6088]

how beautiful ur explanations are

[@sharafmakk2936]

2:43:00 I think the range of phi should be -(pi/2) to pi/2 since it has to cover the whole upper half of the sphere

[@eyllengul]

Before the example, he stated that phi should always be in between 0 and pi which is why we have those bounds.

[@sharafmakk2936]

​@@eyllengul But the answer doesn't make sense in this case. We literally aren't finding the value over the same region

[@bakdauletabdikhan]

@@sharafmakk2936 it starts from top

[@rahulshrestha8007]

Amazing guy. Love you!

[@Scott123180]

Great video, thanks so much!

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A fantastic resource!

[@isabellaz1035]

best math prof!

[@tiaanjansevanrensburg2762]

You are the best, Thank you so much!

[@phambee]

3:08:39 you mean 2pi, not pi/2. The students corrected you the first time you said that on accident at 2:59:59 but they didn't tell you the 2nd time. lol Just a heads up in case you wanted to add a note for future viewers/students :)

[@Heran-h2g]

Professor leonard is once in a life-time person thank you 🫶

[@kimboslice63]

@41.29 Integration is right but substitution of lower bound isn't. I think it should be (1/64 r^4)*r. Since the integral with respect to z is 1/2 z^2.

[@anonymank]

i cant describe how helpful this is before my exam lol

[@uxoy]

50:38 - Mass Density Directly Proportional (Distance)

[@evertonsantosdeandradejuni3787]

1:10:11 - He Talked to us :D - Blessed us

[@Kudravets-Diana]

Why it is not Krsin(some angle) In 1:15.05 ?

[@tannerberry8133]

Professor Leonard: " 2:18:34 " Me *softly sobing*: ohhhkay professor leonard your right

[@mxlexrd]

The explanation that starts around 2:47:17 is quite misleading. The professor states correctly that the integral will equal zero, but then implies that this is somehow incorrect. It isn't incorrect, zero is the correct result for this integral. The reason that this integral equals zero is that our original mass density function was 'y', which is negative when y

[@DivineGenetics]

+Professor Leonard (or anyone else), at 34:45 are you graphing the region at the xy-plane (z=0) or are you graphing at z=1?

[@DivineGenetics]

Or would it make more sense to say that he is drawing the level curve at z=1 as projected onto the xy-plane?

[@Maistora11]

Yeah, the intersection is above the xy-plane at z=1 but we want the projection on the xy-plane for our region.

[@mbongiseniherbertmagutshwa9354]

maan dont know how to thank you ,you a real one

[@zakooralbity6265]

you are a great professor

[@cameronadams4366]

@33:26, 1) you say there is a cylinder, that's not true, there is a sphere and a paraboloid 2) question: z = -9 does not make sense as a solution to the intersection at z = -9 does not lie on either surfaces? confused