Objective: 6. Applications with double & triple integrals - center of mass
Пікірлер: 26
@arsalanmalik3801 Жыл бұрын
This video singlehandedly helped me get 20% of my entire EM course and I passed. Love you and thank you 🖤
@Mary-mj7le3 жыл бұрын
Thanks! My professor in covid has never posted a lecture so this helps a lot... She just has "office hours"
@eiao994 жыл бұрын
Good explanation for the book's concepts, thanks for sharing!
@yaredyirgalem31532 жыл бұрын
You have no idea how much ur helping me out Tnx alot from 🇪🇹 ethiopia
@ahihinah97263 жыл бұрын
excuse , on the #3 , the center of mass on x is 0 because they are already balance on x-axis since (-a,0) and (a,0) right ? ( hope you reply my question , you are a great teacher )
@alexandraniedden53373 жыл бұрын
Yes, x = 0 because the figure is symmetric over the y-axis. It's more than the fact that the points are (-a,0) and (a,0) - the entire figure needs to be symmetric with respect to x, not just those two points. I hope that makes sense.
@ahihinah97263 жыл бұрын
@@alexandraniedden5337 i got it , thank you very much
@enriquesoler31503 ай бұрын
in 23:59, when doing the double integral for the y coordinate of center of mass, where is r*sin(theta) coming from?
@alexandraniedden53373 ай бұрын
r*sin(theta) is the polar conversion for y (when converting from rectangular to polar form). r*cos(theta) is used to convert x from rectangular to polar.
@adrianramirez88323 жыл бұрын
Shouldn't y bar for the last example be 3a/4pi? How did it simplify to 2pi when the only thing that gets crossed out is the a^3? (a^4)/4 * 3/a^3pi = 3a/4pi
@alexandraniedden53373 жыл бұрын
The sin(theta) integral gives us a two (-cos(pi) + cos(0) = 2), so we end up with (a^4/4)*2 = (a^2/2). The answer is correct as given.
@beastvenom41932 жыл бұрын
I would love to learn entire maths from you😁😁
@juliak98993 жыл бұрын
Hi, at around 18:48, you say that we don't need to calculate both x bar and y bar for example 1 because they will be the same. Why is that and how could I know for future examples when to only check for one (i.e. how can I know for future problems if x bar and y bar will give the same result)?
@alexandraniedden53373 жыл бұрын
Because the figure is symmetric with respect to the line y = x, we know that the center of mass will have the same values for the x- and y-coordinates. This should make sense looking at the figure - if we had to approximate the center of mass, we can guess that the x- and y-coordinates will be the same.
Hello how did u get the mass 1/24? Need solution please
@benjaminbenjamin88342 жыл бұрын
Please tell which book you were using for these lecture videos?
@aadityasuthar2199 Жыл бұрын
Thomas Calculus
@wisdomwayeducation Жыл бұрын
THANKS SIS VERY MUCH
@abdulhadifayad46713 жыл бұрын
Sorry to bother you Ms. but at 21:20 how did you find the equation of density?
@alexandraniedden53373 жыл бұрын
Questions are never (or rarely) a bother! The first sentence of the problem helped me write the equation for density. We know that density is proportional to distance from the center. Center is (0, 0), so distance from center is d = sqrt[(x - 0)^2 + (y - 0)^2] which simplifies to sqrt(x^2 + y^2). For the proportional part then, we get density = k*sqrt(x^2 + y^2). From the equation, as density increases, so does the distance (hence proportional).
@geraldcalero41793 жыл бұрын
At 14:32 how do you find the eq y=2-2x?
@alexandraniedden53373 жыл бұрын
I started with slope = (change in y) / (change in x). In this case to go from the top point to the bottom point, y changes by 2 and x changes by 1, with a downward slope, so m = -2. The y-intercept is 2 (the point (0, 2)). I then used slope-intercept form: y = mx+b, so y = -2x+2.