We present an example of calculating a triple integral using spherical coordinates. www.michael-pen... www.randolphcol...
Пікірлер: 54
@tomatrix75253 жыл бұрын
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...
@Jschlick1002 жыл бұрын
this aint easy man lmao
@tanzilaraushan16022 жыл бұрын
@@Jschlick100 felt
@a.nelprober4971 Жыл бұрын
GOOD
@unclesam67643 жыл бұрын
I’m grateful for watching this video with such a lucid explanation. God Bless you sir
@avdrago71704 жыл бұрын
Cospi/6 = sqrt3/2, not 1/2
@rebucato31422 жыл бұрын
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.
@alexalbors72132 жыл бұрын
Just realized the same thing
@scoobydoo35578 ай бұрын
Yeah ur right, but I think the process and the sketches are all correct
@rookiemvp20082 жыл бұрын
I hate calc 3.
@f.osborn15798 ай бұрын
I took multi variable calc and passed but it made no sense…this video is good though.
@prathikkannan332428 күн бұрын
Exceptionally clear! Thanks Michael.
@maxyao47092 жыл бұрын
but how can the radius be 4 if y is bounded by [0,2]
@savangolakiya19262 ай бұрын
magic
@betramlalusha70602 жыл бұрын
I hope you are doing well and everything in your life is going great! Thank you for this! Wow!
@Carusot5 күн бұрын
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
@seanfischler13208 ай бұрын
Thanks for the video! For some reason I had more success understanding you than my professor.
@Martin-iw1ll Жыл бұрын
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
@userHamza3 ай бұрын
You explain this problem very smoothly
@noahifiv5 ай бұрын
Michael Penn has awesome teaching skills
@dalibormaksimovic63992 жыл бұрын
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.
@dilsedhoni92292 жыл бұрын
Wonderfully explained
@umalog1433 жыл бұрын
It is really nice sir
@jaineshmachhi1982 жыл бұрын
Coolest way to end the video saying" Good now it's a good place to stop"😂😂
@walter90292 ай бұрын
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 ?
@sarykhalaf2 жыл бұрын
how to determine exactly what’s the equation is????
@f.r.y58572 ай бұрын
good job sir
@maxwellsoko98032 жыл бұрын
You're great Sjr
@james-md1cf9 ай бұрын
Very comprehensive problem
@eulzzzz76826 ай бұрын
Cosine of π/6 is √3/2 14:25
@alidaqa273821 күн бұрын
9:46 should be sqrt 6p^2 sin^2
@majidrazavi94672 жыл бұрын
Excelant i realy enjoyed god bless you
@user-dl4mq1zv8n8 ай бұрын
You are a legend
@thatthila2 жыл бұрын
Thank you :3. It's so niceee!
@yangfiona86498 ай бұрын
Thank you!
@Laerteufv3 жыл бұрын
cos (pi/6)=(3^1/2)/2
@fredderf20682 жыл бұрын
True dat
@ev4_gaming3 жыл бұрын
very helpful
@7_str_76 ай бұрын
Nice 👍
@noelani9764 жыл бұрын
4^5=1024 and not 2048
@pingpongfulldh23084 жыл бұрын
Nice
@EfirDopАй бұрын
❤❤❤❤❤
@yBazo823 жыл бұрын
quality
@unconscious56305 ай бұрын
4^5 is not 2048 and and cos(pi/6) is not 1/2
@mateja901 Жыл бұрын
Where can I find problems like this?
@moondrop32354 ай бұрын
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.
@lopa7974 жыл бұрын
love you🤗
@elf_someone Жыл бұрын
لحسة مخ يارب فكني من الرياضيات
@endaleyohannes71063 жыл бұрын
cool
@ThAlEdison3 жыл бұрын
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).
@darksoles13052 жыл бұрын
I know I'm late, but set the equations equal to each other and you will find the circle where they intersect