Recent content by Caldus

I am trying to solve a problem dealing with applying integrals to physics. Here's one that I am having trouble with: A trough is 4 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of x^6 from x = -1 to x =1. The trough is full of...

How can an arclength be negative? Is that even possible?

Yeah that was a typo. I actually used y^1/3 there. My final answer ended up being -5330.25 which makes no sense. :\

Oh wait I think I know what you mean now. You just complete the square or whatever and then cancel out the square with the square root. So now I can do the integral. Problem is that I get a negative answer which doesn't make sense for the length of an arc. :\ Guess I'm still doing it wrong...

I can't figure it out. What is the square root of it? Do you try to form a perfect square with it all or what?

So then how do I go about coming up with a numerical solution to it? Am I going to have to use limits?

What do you mean by expanding it out? And the calculator will not give me an answer since 0 is not in the domain of y^(-1/3)...

But isn't it just easier to do it the way I did (since it is given that y is between two values and not x)?

Can't seem to finish this problem: Find the length of the curve: x = 3y^{4/3}-((3/32)y^{2/3}) And -343 <= y <= 125. I used the formula: \int_{a}^{b}\ (1 + (x')^{2})^{1/2}dx \int_{-343}^{125}\ (1+(4y^{1/3}-(1/16)y^{-1/2})^{2})^{1/2}dx But how do you find this integral? Is this...

Anyone? These problems are really frustating me...

Wouldn't it be y = 4 and not y = 2 for the other one? I tried your answer and it did not work while I tried my answer using y = 4 and of course it says incorrect. I'm running out of ideas as to what I could have done wrong for both of these problems.

Yeah I suppose so. Grrr this stupid program. It must be a bug or something...

OK, got an idea for the last problem but still not getting the right answer. OK so if I place a sphere of radius 11 in the center of the x-y plane, then the top right quadrant curve of that sphere will equal (r^2 - x^2)^(1/2) correct? So then I can do the volume like so (using vertical...

I ended up getting 5.864306287 as the answer but the online homework program says it's still incorrect. I agree that the small radius is 1. How did you come up with 2 - x^2 as the big radius though? I don't get how to determine what the outer radius is in this case. Thanks for any help...

Thanks, I eventually figured it out. Still stuck on the other two problems...