Volume in cylindrical coordinates | MIT 18.02SC Multivariable Calculus, Fall 2010

Volume in cylindrical coordinates
Instructor: Joel Lewis
View the complete course: ocw.mit.edu/18-...
License: Creative Commons BY-NC-SA
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[@robpatterson2861]

I love your videos! I'm so glad to have stumbled apon them. Examples are the best! To get that visualization of it really helps. I'm 57 and just having fun! Thanks!

[@prathikkannan3324]

Thanks Joel! I found the accompanying lecture pretty tough and this helped clarify some doubts.

[@ayhanbozkurt937]

the most useful lesson so far i encountered in youtube.

[@TheAhmedMAhmed]

u sound a bit like the joker from the dark night
:o

[@shuvbhowmickbestin]

Cant unnotice it anymore XD

[@harrydiv321]

if you parameterize the region as
x = r cos theta
y = 1 + r sin theta
z = z
where
0

[@middlevoids]

Be lazy!

[@mousavi128]

I wish I was at MIT

[@ricardo4fun922]

For those who care the final answer is pi/2.

[@pedroff_1]

Was a triple integral really required? Since the beginning, I was expecting it to result in a double integral where the function being integrated would be the difference in height between the parabola and the plane, which is the first thing you get when solving the innermost integral.

[@sathvikswaminathan7933]

Yep, it's not necessary to use a triple integral in this case. But it'll be when we are integrating the area element with some function dependent on the x,y,z variables.

[@nizarch22]

@@sathvikswaminathan7933 No. It is absolutely necessary -- you cannot solve the question without it. You're calculating volume, not surface area. And what he used was a volume element, not a surface element.

[@erikumble]

@@nizarch22 Actually, since the inner function is just 1, you do not need to use a triple integral, if you could instead define a function that gives the height of the solid at an (x,y) coordinate, which is what you get after computing the innermost integral. As Sathvik mentioned, when more complicated functions of x, y, and z are involved than simply 1, it is "necessary" to use triple integrals. In this case, the function is easy to practice setting limits of integration.
To note, however: multiple integrals simplify down to just one, so it is usually not "necessary" to use several, but it makes setting up the problem and remembering formulas much easier.

[@MarikoIchigo]

Things are making more sense now...

[@vincec894]

Its crazy how this is taught the same exact way at my small hometown university

[@springrollies]

the upper part's volume would be infinity, so it has to be the lower one

[@Anti-Prolixity]

@ny1fanta Depends on the shadow region. Shadow region (relative to origin) goes from 0 to pi.

[@shariffross3920]

"let me shade the region i like doing that"

[@anhtho3008]

so how do I know if it's not the upper part ( the remaning bigger part of the paraboloid after being cut by the plane ) but the lower one?

[@apb8813]

the upper part of the parabola would be infinite. We're interested in a finite volume.

[@jak712we]

thanks Joel

[@av733]

This guy is great

[@davidzhang8764]

Is the lower bound of z correct?

[@Devil-Jin-yt]

David Zhang yes brother, as the equation of plane is z=2y
And in polar form y should be written as r sin theta..
Hope it helps

[@shreyatomar1435]

Thank you 🙏🏻

[@bobkameron]

took some time but good exercise!

[@persianshawn92]

can someone explain why r goes from 0 to 2sin*teta* ? where does the 2 come from?

[@Speedfinders]

+Shahin Sateei I guess it's because he's talking about theta from 0 to pi and not 0 to 2*pi, that should be the same at last.

[@johannesnssethillestad3436]

+Shahin Sateei
The region R is described (in polar coordinates) by the equation r=2*sin(*theta*). Measured from the Cartesian origin, the lowest r possible is 0 (*theta* equals *Pi* or 0), and the largest r possible is 2, when *theta* equals *Pi*/2. Because the value of is dependent of *theta* the limits of integration for r is (0,2*sin(*theta*)). The number 2 comes from the maximum distance from the Cartesian origin to the circle C.
Hope this was helpful :)

[@urmitmahida5725]

use cartesian & polar relations x=rcos theta & y=rsin theta in X^2+y^2=2y u will get the limit of r=0 &2 sin theta

[@ny1fanta]

i thought theta is from 0 -> 2pi

[@T4l0nITA]

It's just pi because r starts from the origin (0), not from y=1 (the center of the circle), as you can see it doesn't go all around but just 180 degrees

[@bobkameron]

great!

[@joe_mama92]

Ambatukam

[@TheFarmanimalfriend]

adding r changes the value of the equation. Shame on you!

[@joe_mama92]

Ambatukam