Multivariable Calculus | Triple integral with spherical coordinates: Example.

We present an example of calculating a triple integral using spherical coordinates.
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[@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

[@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.

[@prathikkannan3324]

Exceptionally clear! Thanks Michael.

[@maxyao4709]

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

[@savangolakiya1926]

magic

[@betramlalusha7060]

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

[@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

[@seanfischler1320]

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

[@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

[@userHamza]

You explain this problem very smoothly

[@noahifiv]

Michael Penn has awesome teaching skills

[@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

[@umalog143]

It is really nice sir

[@jaineshmachhi198]

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

[@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 ?

[@sarykhalaf]

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

[@f.r.y5857]

good job sir

[@maxwellsoko9803]

You're great Sjr

[@james-md1cf]

Very comprehensive problem

[@eulzzzz7682]

Cosine of π/6 is √3/2 14:25

[@alidaqa2738]

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

[@majidrazavi9467]

Excelant i realy enjoyed god bless you

[@user-dl4mq1zv8n]

You are a legend

[@thatthila]

Thank you :3. It's so niceee!

[@yangfiona8649]

Thank you!

[@Laerteufv]

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

[@fredderf2068]

True dat

[@ev4_gaming]

very helpful

[@7_str_7]

Nice 👍

[@noelani976]

4^5=1024 and not 2048

[@pingpongfulldh2308]

Nice

[@EfirDop]

❤❤❤❤❤

[@yBazo82]

quality

[@unconscious5630]

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

[@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.

[@lopa797]

love you🤗

[@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 🥵