Recent content by physicss

I think I found my mistake, thanks for the help

thank you for your answer. could H(1−r/R) be right ?

I give up but still thank you for your time and help

Hello, I used (0,R), (0,2pi), (0,H/R*z) as the limits and I get pi*H^4*r^4/4R^2. is my solution correct? thanks in advance ( the function I used: r^3 z)

wouldn´t the general limits be (0,h), (0,2pi) and (0, r-rz/h) rdrdφdz?

could I also use the limits from 0 to r, from 0 to h and from 0 r/h (h - √(x^2 + y^2))? (in cartesian) and would 0 to r, 0 to 2pi and 0 to z = h - (h/r) * sqrt(x^2 + y^2) in cylindrical coordinates be right for the limits? thank you in advance

yes, I more or less know how to convert Cartesian coordinates into cylindrical coordinates

like a rectangular prism

x from 0 to r y from 0 to r z from 0 to h ∫0h ∫0r ∫0r z(x^2 + y^2) dx dy dz would that be right?

the inner integral has to go from x to 0.5x+1 xy dy and the outer from 0 to 2 dx I guess?

Thanks, while writing down I swapped x and y. 0.5x+1 is AB

x and 2x-2

Hello, I recalculated it ( in a shorter way). I still get 2: ∫(0 to 2) (∫(0 to y) xy dx) dy = ∫(0 to 2) (y^3)/2 dy =2 what am I doing wrong? thanks in advance

could I also calculate it by forming a rectangle? ∫∫R xy dA = ∫∫S u(u+v) dudv ∫∫S u(u+v) dudv = ∫0^2 ∫0^1 u(u+v) dvdu = ∫0^2 [(u^2v/2) + (uv^2/2)]_0^1 du = ∫0^2 (u^2/2 + u/2) du = [(u^3/6) + (u^2/4)]_0^2 = 2.