Recent content by CollinsArg

Why is this way of thinking wrong?. can't I assume that when Δx tends to zero is a sufficient approximation of what I want to get? It confuses me with the basic idea of integrating a function to get the area beneath a curve of a function (which isn't also as perfect) . PD: I put Δx tends to...

Thank you :)

Hi! I've found some excercices and expanation of Work always consider a constant velocity, this is a net Force equal to cero. Like spring or gravity excercices related against a force applied. Does this relation with constant velocity has some usefull explanation why? Should I always assume...

Oh yes, I just forgot a four,because I had taken r2 as common factor. Then it seems it's right :)

I got 8π/3 with that integral, but doing it the way I told you before I got 32π/3. I'm not sure if it's right, anyways I still don't understand, why the function is the difference of z's, though I understood the procedure you did afterwards, but I can't connect why should the function be that...

I don't understand why the function would be a difference between the two z's, is there a visual way to see this?

Nice! Honestly, I've never used double integral with polar coordinates to resolve a solid of revolution. But, what I would do, is what you just said, I would first find a way to find the cone and paraboloid equation (because you are not given them) the integrate them separately. It seems hard...

Yes, that's what I meant, remember that the function you plug in there is the function you are integrating (as a "roof") limited by the integral limits you put from each point. I recommend you watching MIT Calculus courses, they are in youtube and are very good ones :) btw, vela is right, note...

Why did you put ƒ(r,θ)=6-r.cos(θ) as the function? Isn't it just the function of the stright line? And if you put it there it would become a three dimensional equation like y=6-x + 0.z, there you get a plane instead of a cone as you described.

If I'm not mistaken, the gradien of an equation is normal to the level of superficies/curves of that function, (that is, a dimension less than the original function), but the gradient of a function is not normal to the function where it comes from, as a gradient will be one dimension smaller...

Thank you so much!

Homework Statement I'm given the integral show in the adjunct picture, in the same one there is my attempt at a solution. Homework Equations x = r.cos(Θ) y = r.sin(Θ) dA = r.dr.dΘ The Attempt at a Solution [/B] I tried to do it in polar coordinates, so I substituted x=r.cos(Θ) y=r.sin(Θ) in...

I recommend you Calculus of one variable - Stewart

Adding to what was already said, suppose I add a force on the book pressing on it with my hand against the table. Then the normal force of the table on the book becomes greater than before, but the weight of the book will stay the same, so, not always the normal force will be necessarily equal...

Wound't it mean also that she could slow down the tangential velocity too? depending on the way the ice skater pulled her arms in? But the equation tells me that always when the distance is shortened the tangential velocity increases.