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.
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
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.
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 ?
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.
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).
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...
this aint easy man lmao
@@Jschlick100 felt
GOOD
I’m grateful for watching this video with such a lucid explanation. God Bless you sir
Cospi/6 = sqrt3/2, not 1/2
I hope you are doing well and everything in your life is going great! Thank you for this! Wow!
I hate calc 3.
I took multi variable calc and passed but it made no sense…this video is good though.
You explain this problem very smoothly
Michael Penn has awesome teaching skills
Wonderfully explained
Thanks for the video! For some reason I had more success understanding you than my professor.
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.
Just realized the same thing
Yeah ur right, but I think the process and the sketches are all correct
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
It is really nice sir
You're great Sjr
Excelant i realy enjoyed god bless you
Thank you :3. It's so niceee!
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.
Very comprehensive problem
but how can the radius be 4 if y is bounded by [0,2]
magic
very helpful
good job sir
You are a legend
Thank you!
how to determine exactly what’s the equation is????
Cosine of π/6 is √3/2 14:25
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 ?
Nice 👍
Nice
Coolest way to end the video saying" Good now it's a good place to stop"😂😂
love you🤗
quality
Where can I find problems like this?
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.
❤❤❤❤❤
cos (pi/6)=(3^1/2)/2
True dat
cool
4^5=1024 and not 2048
لحسة مخ
يارب فكني من الرياضيات
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).
I know I'm late, but set the equations equal to each other and you will find the circle where they intersect
4^5 is not 2048 and and cos(pi/6) is not 1/2
This is wrong
I love you
Hot teacher 🥵