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...
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...
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...
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.
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...