For anyone wondering about the derivation of the unit normal vector for the sphere: The normal vector for a sphere centered at the origin is always , as this is the vector field that points radially outward from the origin. To derive a unit vector, it is the vector itself divided by its magnitude. The magnitude of the vector is found by taking the square root of x^2 + y^2 + z^2, with the x,y, and z coming from the x,y, and z components of the vector we are finding the magnitude of, which is just . As given in the equation for the sphere, x^2 + y^2 + z^2 =4. Since the magnitude of the vector is the square root of the quantity x^2 + y^2 + z^2, which is equal to 4 by the equation for the sphere, the magnitude is found to be equal to 2. The original vector then just needs to be divided by the magnitude, hence the 1/2 out in front of the for the unit normal vector.
If we use geometric element like square or circle like pixel in computer term on one dimension then intergate on a limited range and domain and error trapping method
BEST EXPLANATION EVER OD DOUBLE AND SURFACE INTEGRALS> I LOVE THAT YOU COMPARED BOTH OF THEM TO SHOW THE DIFFERENCE AND SIMILARITIES AS WELL AS GAVE ALL THE POSSIBLE CASE. AMAZING VIDEO!!
Doubt this will be seen before my final lol, but can anyone help explain how he derives the normal vector in the second example? Since he's using the formula form that converts the vector field into a scalar of two variables, I want to think we need to some rphi X rtheta cross product, but I'm sure I'm just thinking about getting normal vectors wrong.
Are you sure you explained double-integrals correctly? As far as I know, double-integrals are integrals of 3-dimensional functions and they give you the volume of the function in some defined region. The way you say it, they don't appear to be any different from single integrals.
"There is a flux form which looks like double integral over a surface of some vector fields dotted with the normal vector times the surface element" I don't understand this sentence
Remy Gelenidze the flux form basically shows the flux through the entire area. The double integral is just multiplying the summed dx and dy, so length times width should ring a bell here. And the vector function is just your function f, the normal vector is the direction perpendicular to that area given by dxdy, so you just want to find how much of your vector function is in the direction of the surface dxdy in the direction perpendicular to it (normal vector). Hope this helps
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
For anyone wondering about the derivation of the unit normal vector for the sphere: The normal vector for a sphere centered at the origin is always , as this is the vector field that points radially outward from the origin. To derive a unit vector, it is the vector itself divided by its magnitude. The magnitude of the vector is found by taking the square root of x^2 + y^2 + z^2, with the x,y, and z coming from the x,y, and z components of the vector we are finding the magnitude of, which is just . As given in the equation for the sphere, x^2 + y^2 + z^2 =4. Since the magnitude of the vector is the square root of the quantity x^2 + y^2 + z^2, which is equal to 4 by the equation for the sphere, the magnitude is found to be equal to 2. The original vector then just needs to be divided by the magnitude, hence the 1/2 out in front of the for the unit normal vector.
That was a crystal clear explanation, thank you!
why ds=dv/dρ and not ds=dv/dθ or ds=dv/dφ?
Thanks. I didn't know how he got the unit normal vector of the sphere. You are very thoughtful to provide this explanation.
Oh my god thank youuu💞💞💞
i don't know who spread the word that calc3 was easier than calc2 but no
seriously. Calc II like 1/10th the visuospatial skills and critical thinking
The concepts in Calc3 are definitely more difficult, however, the computations are harder in Calc2.
Both easy if done with enough practice.
@@Name-jw4sj no
It depends on the professor. Calc 3 was easier for me because my professor was so amazing
Very organized and consciousness towards different audiences. Color coded, preview, nice! Well done!
change speed to 1.25
Your are welcome!
try 2
haha funny
If we use geometric element like square or circle like pixel in computer term on one dimension then intergate on a limited range and domain and error trapping method
I'm a bit confused about how you are calculating n hat dot dS. Why ? did you take a derivative some where?
GOSH THAT IS AWESOME. PLEASE MAKE THAT EXCELLENT WORK CONTINUE
amazing video, simply amazing!
BEST EXPLANATION EVER OD DOUBLE AND SURFACE INTEGRALS> I LOVE THAT YOU COMPARED BOTH OF THEM TO SHOW THE DIFFERENCE AND SIMILARITIES AS WELL AS GAVE ALL THE POSSIBLE CASE. AMAZING VIDEO!!
I disagree, it's too wordy and confusing
Excelllent video!!!! Thanks!
very helpful drawing! thank you!
Excellent presentation!!!
the best explanation of surface integrals
thank you sir
r u a robot ?
awesome video, quite clear
Awesome job thanks a lot!!!!!
you solved my problem 👍
Great work, very helpful, Thanks
Very informative! Thank you so much!
Bravo! I love this systimatization(regimentation)
Not bad! But our sirs explanation is top class n he recommended everyone to check this vid out. He has 22 research papers!
Who's your sir?!
super useful and well organized
When F= (yz + zx + xy ) through an area S which is part of a circle with radius a in the first quadrant of the xy plane. Calculate the flux?
Nice
good movie thanks you very mutch
Doubt this will be seen before my final lol, but can anyone help explain how he derives the normal vector in the second example?
Since he's using the formula form that converts the vector field into a scalar of two variables, I want to think we need to some rphi X rtheta cross product, but I'm sure I'm just thinking about getting normal vectors wrong.
Maybe the gradient?
You print, how cute
When youre doing the unit vector, why dont you divide it by its magnitude?
I don't get this either!
because it cancels out
the magnitude of any unit vector is 1.
Any body got the answer
@@ssctarget7953 magnitude of a unit vector is 1
Surface area of a sphere = 4pir^2 which would be 16pi for this problem. How did he get 32pi?
Are you sure you explained double-integrals correctly?
As far as I know, double-integrals are integrals of 3-dimensional functions and they give you the volume of the function in some defined region. The way you say it, they don't appear to be any different from single integrals.
No. Double integrals are double integrals
why ds=dv/dρ and not ds=dv/dθ or ds=dv/dφ?
Wonderfull lf you have problem for sleeping !
😂😂😂😂😂😂
"There is a flux form which looks like double integral over a surface of some vector fields dotted with the normal vector times the surface element"
I don't understand this sentence
Remy Gelenidze the flux form basically shows the flux through the entire area. The double integral is just multiplying the summed dx and dy, so length times width should ring a bell here. And the vector function is just your function f, the normal vector is the direction perpendicular to that area given by dxdy, so you just want to find how much of your vector function is in the direction of the surface dxdy in the direction perpendicular to it (normal vector). Hope this helps
Very Helpfull.Thanks.
thank you very much
solve one question by taking surface of ellipsoid.
You glossed over too many details in the sphere example. Could use more detail on normal vector determination, too. Overall, very helpful.
thanks that very helpful
Thanks!
Using Gauss divergence theorem ,the last sum answer is 3×volume of sphere=256π
Is it correct or not,anyone guide
very difficult to understand
Holy shit! This guy put me to sleep in exactly 7 minutes. I am gonna play him every night to doze off. Very very monotonic.
شكرا
Thank uuuuuu
dont go speed in subjects
varibles
yes
ㄹㅇ 설명 개찰지네
Wow!
wow...very dry..........
The voice sounds synthesised or electronically masked, which I find extremely hard on the ear.
Please speak slowly i can't hear you properly
horrible
His voice is unbearable