Recent content by bpuk

umm the sqrt came from the spherical formula cos you have to sqrt the side you have as in x^2+y^2+z^2=R^2. and yesh i understand to integrate the eqn but in terms of the next step what is that lil r in reference to is that the main radius i know or the one of the smaller cut. as in if i know...

and going on from this i found the integral of this is like [.5*ln(x+sqrt(r^2+x^2)r^2 + .5*x*sqrt(r^2+x^2)] which seems a bit crazy.

using that \pi\int (radius^2- z^2)dz i found that intergrating it came out to be 1/2 ln (z+sqrt(r^2+x^2)r^2 + 1/2*x*sqrt(r^2+x^2)

hmm, but yeah that's what i thought but how do we then know the radius of each "slice" if we continued to work through the problem? and were the Z limits ok?

or is it the int[sqrt(radius^2 - z^2)] between Radius and radius - height??

so basically i will be working in terms of finding the integral (from the R to R-h) of the area of a cross section being pi*r^2??

i don't know how to edit but i know in order to do find the volume i will have to model a cone within th experiment obtian the limits and then derive a function but i am for the life of me unable to comprehend on how to start it

ok so i have a homework question that goes around the lines of a perfect sphere is used as a measureing bowl with radius R is given, this has a hole in the top in which it is filled with water (volume of this discounted as it is relatively small), we are asked to find at what height (relative to...