Multivariable Calculus | Triple integral with spherical coordinates: Example.

[@tomatrix7525]

Every time I see some beautiful multivariable calculus like this I am in awe. These techniques really allow us to do hard stuff so easily...

[@Jschlick100]

this aint easy man lmao

[@tanzilaraushan1602]

@@Jschlick100 felt

[@a.nelprober4971]

GOOD

[@unclesam6764]

I’m grateful for watching this video with such a lucid explanation. God Bless you sir

[@gosports8586]

Not really trippin about the integral mistakes, especially since they were done at the top of your head. Just grateful for showing how to set up spherical from rectangular triple integrals! Thanks!

[@avdrago7170]

Cospi/6 = sqrt3/2, not 1/2

[@rebucato3142]

Pretty sure the answer is more complicated than that... The first integral evaluates to (4^5)/5 = 1024/5, not 2048/5 the second integral is fine The third integral evaluates to (1-sqrt(3)/2), not 1/2, as cos(pi/6) = sqrt(3)/2, not 1/2 The answer is 512pi/5*(1-sqrt(3)/2) Making mistakes on the blackboard is fine as long as they are corrected. However, I am shocked that there are no amendments to this video 2 years after it has been released. This shouldn't happen and I hope Penn can correct this so students would not be confused.

[@alexalbors7213]

Just realized the same thing

[@scoobydoo3557]

Yeah ur right, but I think the process and the sketches are all correct

[@rookiemvp2008]

I hate calc 3.

[@f.osborn1579]

I took multi variable calc and passed but it made no sense…this video is good though.

[@davidsteiget4433]

Same and I thought it was supposedly the easiest of the sequence

[@Endgator]

​@davidsteiget4433 the equations are easy, the conceptualization is what gets most people.

[@noahifiv]

Michael Penn has awesome teaching skills

[@betramlalusha7060]

I hope you are doing well and everything in your life is going great! Thank you for this! Wow!

[@prathikkannan3324]

Exceptionally clear! Thanks Michael.

[@maxyao4709]

but how can the radius be 4 if y is bounded by [0,2]

[@savangolakiya1926]

magic

[@paulnokleberg5188]

The spherical portion of the volume is above the xy plane, so the y bounds don't define it. The x bounds initially limit theta to pi, a half circle, and the y bounds further limit theta to pi/2, a quarter circle. When you plot it on desmos, it's easier to see. Good question. I can see how that might be confusing at first. A picture is worth a thousand words in calc 3.😊

[@Martin-iw1ll]

I think it is necessary to check whether the top sphere actually touches the bottom cone, cause if i change the limits for y to from 0 to sqrt(1-y^2), then the answer would be very different and also 0

[@seanfischler1320]

Thanks for the video! For some reason I had more success understanding you than my professor.

[@userHamza]

You explain this problem very smoothly

[@Carusot]

Consider the hot air balloon with equation 9x^2 +9y^2 +4z^2 = 100. The temperature in the hot air balloon is given by the following function: f(x,y,z) = 18x^2 + 18y^2 + 8z^2 − 20 • Convert the regular area into appropriate spherical coordinates. • Calculate the volume of the balloon. • Calculate the average temperature in the Balloon

[@dalibormaksimovic6399]

Hi. I am interested in how one can determine boundaries of integration when there is no a explicit function for z in terms of y, or y in terms of x. For instance, calculate the volume of body bounded by following surfaces: x^2+y^2 = cz, x^4+y^4=a^2(x^2+y^2) and z=0.

[@dilsedhoni9229]

Wonderfully explained

[@sarykhalaf]

how to determine exactly what’s the equation is????

[@katlegodonald243]

Thanks for this video ,it's appreciated.

[@umalog143]

It is really nice sir

[@walter9029]

Maybe somebody can help me out: I know you can see the bounds for theta on the board, but is there an algebraic way to derive them from the equations and the xyz bounds ?

[@maxwellsoko9803]

You're great Sjr

[@james-md1cf]

Very comprehensive problem

[@majidrazavi9467]

Excelant i realy enjoyed god bless you

[@DhruvaDevaraaj]

You are a legend

[@thatthila]

Thank you :3. It's so niceee!

[@f.r.y5857]

good job sir

[@eulzzzz7682]

Cosine of π/6 is √3/2 14:25

[@jaineshmachhi198]

Coolest way to end the video saying" Good now it's a good place to stop"😂😂

[@yangfiona8649]

Thank you!

[@mateja901]

Where can I find problems like this?

[@moondrop3235]

If you're taking Calc 3, it's rather common. Probably learn it near the end after learning how to convert bounds to polar and cylindrical coordinates and during triple integration.

[@ev4_gaming]

very helpful

[@Laerteufv]

cos (pi/6)=(3^1/2)/2

[@fredderf2068]

True dat

[@pingpongfulldh2308]

Nice

[@7_str_7]

Nice 👍

[@yBazo82]

quality

[@alidaqa2738]

9:46 should be sqrt 6p^2 sin^2

[@lopa797]

love you🤗

[@unconscious5630]

4^5 is not 2048 and and cos(pi/6) is not 1/2

[@EfirDop]

❤❤❤❤❤

[@noelani976]

4^5=1024 and not 2048

[@elf_someone]

لحسة مخ يارب فكني من الرياضيات

[@endaleyohannes7106]

cool

[@ThAlEdison]

In this problem sqrt(16-x^2-y^2)=sqrt(3(x^2+y^2)) becomes x^2+y^2=4, which means that the intersection of the cone and sphere are at the same as the limits on the x-y plane, but if the bounds on x had been 0, sqrt(1-y^2), then you would've had a spherical section on top of a cylinder on top of a cone. I'm not 100% sure how I would convert the equation if that was the case. It may require changing it to two separate integrals, one where phi goes from 0 to csc^-1(4) and rho goes from 0 to 4, and a second where phi goes from csc^-1(4) to pi/6 and rho goes from 0 to csc(phi).

[@darksoles1305]

I know I'm late, but set the equations equal to each other and you will find the circle where they intersect

[@armnhammer733]

This is wrong

[@nikkk709]

I love you

[@nikkk709]

Hot teacher 🥵