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
This man brought me back from a 56 to a 90. You know a person is special when they can revive your confidence in the face of something seemingly impossible. Forever grateful to you!
@@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
@@maitumeloteemane9593 is snot out of the bed that that I got to go eat it yyttttyytttdtttdttttdttttytd yo eteyttetsttttt d etyo yo ttettttteteytytyyeyyydtyeyyrtetttettettetetetetetettetetteytteeyteytyeyrreydttsydtdtttdtdtdttttttttttttdtttdtdtdttdtdttdtddtdtdteyetryyeteeyyyyteyteyteytetetryyyryyrryrrrytetyytetetettetryette
@@maitumeloteemane9593 is snot out of the bed that that I got to go eat it yyttttyytttdtttdttttdttttytd yo eteyttetsttttt d etyo yo ttettttteteytytyyeyyydtyeyyrtetttettettetetetetetettetetteytteeyteytyeyrreydttsydtdtttdtdtdttttttttttttdtttdtdtdttdtdttdtddtdtdteyetryyeteeyyyyteyteyteytetetryyyryyrryrrrytetyytetetettetryettet
Stopped going to class. One video from Professor Leonard is worth one hundred of those lectures I get from my college. We live in wonderful times when lectures like this are accessible to so many people all around the world.
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
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!
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 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
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!!!
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
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,,,,,,
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!
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.
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}]
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.
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..
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
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!!
> 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
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...
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).
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😉
@@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
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.
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.
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)
+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.
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.
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".
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...
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.
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.
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
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.
My university only seems to hire foreign nationals who are basically unintelligible but somehow supposed to teach me upper level math so I come here instead. Thanks Prof.
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
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??
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?
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.
Spherical Co-ordinates start @2:08:30 if anyone needed to know like I did.
Thanks fam
np
God among men.
Thank you!
Thank you
Professor Leonard: “Head nod if you’re ok with that”
Me at home: *nods*
skip to 1:10:17 for a blessing from Professor Leonard
I needed this in my life
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
Thanks alot
You’re a gem 🤎
.
.n
ooo poooooooopiiiyupouuioippooooopiuiopooooouopopopuoopiopooioloiopio7po777iou77777o67777ou7o77ooooooooouoooou
absolute GOAT
This man brought me back from a 56 to a 90. You know a person is special when they can revive your confidence in the face of something seemingly impossible. Forever grateful to you!
🥲
@@maitumeloteemane9593
@@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
@@maitumeloteemane9593 is snot out of the bed that that I got to go eat it yyttttyytttdtttdttttdttttytd yo eteyttetsttttt d etyo yo ttettttteteytytyyeyyydtyeyyrtetttettettetetetetetettetetteytteeyteytyeyrreydttsydtdtttdtdtdttttttttttttdtttdtdtdttdtdttdtddtdtdteyetryyeteeyyyyteyteyteytetetryyyryyrryrrrytetyytetetettetryette
@@maitumeloteemane9593 is snot out of the bed that that I got to go eat it yyttttyytttdtttdttttdttttytd yo eteyttetsttttt d etyo yo ttettttteteytytyyeyyydtyeyyrtetttettettetetetetetettetetteytteeyteytyeyrreydttsydtdtttdtdtdttttttttttttdtttdtdtdttdtdttdtddtdtdteyetryyeteeyyyyteyteyteytetetryyyryyrryrrrytetyytetetettetryettet
Stopped going to class. One video from Professor Leonard is worth one hundred of those lectures I get from my college. We live in wonderful times when lectures like this are accessible to so many people all around the world.
Graduated?
I totally agree with you. Thank you so much professor
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
Huge thanks
I wonder if theres a special place in Heaven fro people like u
I love you
@@d0rqu3 hhhhhhhhhhh hh hhhhhhhhhhhhh
O
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!
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.
Wait calc 3 was the hardest class you took what about physics lol that class to me is way harder
@@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
@@smoshfann24 damm, dude my physics class at LBCC, CA started from 72 down to 33.
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!!!
What are you studying ?
Im Doing BS in EEE
Oh sorry rafa ... I thought thT u r wid me ,,,,,but u r with sakena.:-) :-)
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
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,,,,,,
You help visualize and understand when most teachers just write down the formulas and do examples. And for that I applaud you sir
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!
When Prof. Leonard is the savior of your grade.
I was literally listening to the weeknd while taking these notes XD
@@jameschen2308
c c x cx🎉
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.
Is it just me who got 8π/3 instead of π/3 @ 3:12:06
yeah me too got 8pi/3
Checked using an online calculator (yep, calculator for triple integrals exist online) and it was 8pi/3
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}]
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.
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..
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
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!!
> 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
I'm over here nodding and raising my hands like I am physically your classroom haha =P... thanks for the knowledge, Professor! Super appreciated!
I passed calc 3 in the summer (1 month) and it was all thanks to professor leonard :D
I'm a colombian student, you just saved me Leonard, my note went from 32 to 80 thanks to you
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...
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).
Same, I've been checking for the ans many times but it's the same 8 pi/3. Thx for confirming.
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😉
@@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
You are right
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.
, off
I swear to god watching an hour of this the day of the exam saved my ass. Thanks Mr. Leonard lol
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.
Best teacher of Mathematics from whom I have learnt till date 🤩🤩🤩
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)
yep
ONE OF THE GREATEST LECTURE I HAVE TAKEN IN MY LIFETIME ❤🙌
STUDENTS! you are soo lucky with this profs
you're the shit man. guys if you want to pass calculus you cannot do it without Leonard. period
3:12:00 ---> V = 8pi/3
The algorithm auto-played this in my sleep and I woke up knowing triple integrals. Thanks Professor Leonard!
I have a test on this in 5 hours time. Watch video or practice examples? Life choices...
+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.
+Richard Were same here! D: lol
watch videos
good thing is that these videos go through example problems :), hope yalll did well
I wish all calculus professors were as enthusiastic as Professor Leonard
Hey prof. thanks for teaching almost all of my calc3 class at FSU :p
Hey I go to FSU too, stopped going to class and just started watching professor leonard's videos instead
@@jordansalcedo7390 I'm 4 years late, but I am also taking Calc III at FSU.
You are the reason I get up in the morning
you made spherical triple integrals look so easy lol .
YOU ARE THE BEST, MAN!
I CANNOT UNDERSTAND MY PROF. AND JUST COME TO SEE YOUR VIDEOS
I love u man thanks were studying for our final
Man I wish I had this guy for my calculus class, makes everything so easy to understand
This guy doesn't miss. Absolute wizard.
Thank you for being a wonderful person!
Alright Clark, you can't hide your identity forever. We know that you're Superman
2:16:40 that's a great way to explain the intuition behind where the rho sin phi comes from in spherical coordinates
I started nodding my head subconsciously.
You're the best professor ever!
Notes to self: 2:13:00 Spherical Coordinates
My professor just recites the textbook verbatim, which isn’t helpful. But you Professor Leonard explain everything so well, thank you.
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.
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".
Love your videos, they are great!! Awesome teacher!!
i think i'm in love... the most beautiful explanation i've ever seen!
I don't know how to show my appreciation
Thank you very much ❤
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...
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.
Great video, thank you Professor Leonard! :)
2:20:05 I totally like his body language.. it’s so energetic 😂😂😂
You saved my life Professor!!!
the superman, rescue to our exams
50:38 - Mass Density Directly Proportional (Distance)
thx for the awesome lecture vids.
watch these videos with playback speed of 1.5, it helps
That's what I do. Its perfectly understandable, and still at a pace you can soak the info in, but you cut the video time by about a third :)
Are you guys native english speakers?
yep, american.
Then i will get it too
Nyc
i just woke up to this. no idea what i fell asleep to that got me here, i hardly made it through math
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
Before the example, he stated that phi should always be in between 0 and pi which is why we have those bounds.
@@eyllengul
But the answer doesn't make sense in this case. We literally aren't finding the value over the same region
@@sharafmakk2936 it starts from top
You make me excited to learn math! Thank you!
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.
im not religious but god bless you you incredible human being you are my savior
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
i cant describe how helpful this is before my exam lol
Professor Leonard!!! I love you man.
@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.
The dude is a cool breeze! He's the best!
Love your videos, saved my life
Greatest of all time
Amazing I'm very grateful
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.
@TheLeonard... Announce Linear Algebra videos !!!
Might wanna double check example at 3:11:54, you said answer was pi/3, getting 8pi/3, double checked my work online, mistake?
+Adam Pluguez Thanks Adam, I put a blurb up there to note that.
Professor Leonard
Prof please do ODE pleeeeeeerereeez :)
I'm getting pi/3
oops I did this for z=1, which gives us pi/3.
My university only seems to hire foreign nationals who are basically unintelligible but somehow supposed to teach me upper level math so I come here instead. Thanks Prof.
when you pause it, work it out and actually get the answer. If it weren't for you, Leonard, I wouldn't know what to do!
Professor leonard is once in a life-time person thank you 🫶
best math prof!
how beautiful ur explanations are
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. :/
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
Hey Professor Leonard, this video has been really helpful and I aced my calc test thanks to you
Amazing guy. Love you!
17:25 important note regarding the function conversion
A fantastic resource!
Great video, thanks so much!
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??
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?
modulus of Y only on the +ve axis
Your EXCELLENT!!!
you are a great professor
maan dont know how to thank you ,you a real one
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.
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 :)
OOOOOHHH ok I was so confused with this. Thanks!
Professor Leonard you are a beast
You are great.
@2:52:43 WHY ARE WE ONLY DEALING WITH HALF CIRCLE?? CAN ANYONE PLEASE EXPLAIN ME?
You are the best, Thank you so much!
Went from a 67 to a 96; I owe this man 100% of my tuition
2:21:00 rho always positive, phi always b/w 0 and pi at most, and theta is as normal
rho and phi simples by taking x to be 0, theta simple as z being set to 0