1. ## calculus disk mehod

the region boud by y=x^(1/2) ; y=0 when x=0 ad x=4 is revoved about the x-axis. find the value of x in the interval [0,4] that divides the solid into two parts of equal volume.

please help me thank you i appreciate it!!! i just need a hint on how to set up the problem..like the integral. i can do the rest

2. Hi

I do not want to tell rubbish (and actually I cannot find again the relevant page on Wikipedia )

So I try to find the formula by myself and here is my result

The volume of a revolution solid around x-axis with a boundary given by y=f(x) is (to be confirmed)

$V = \pi \int_{x_0}^{x_1} f^2(x) dx$

I checked this formula on simple solids and it seems to work properly

3. okay thanks i got x equals the square root of 8 as my answer. does anyone know if that's correct?

4. the region boud by y=x^(1/2) ; y=0 when x=0 ad x=4 is revoved about the x-axis. find the value of x in the interval [0,4] that divides the solid into two parts of equal volume.

please help me thank you i appreciate it!!! i just need a hint on how to set up the problem..like the integral. i can do the rest
hi
well acc. to me also same formula should be applied

u appply it and see if u r gettng the answer
i suppose it should be proceeded as
2*V1(from 0 to x1)= V2(from 0 to 4)

from here you will get the value of x that divides solid into two parts

okay thanks i got x equals the square root of 8 as my answer. does anyone know if that's correct?
ya thats the same answer i m gettng
cross check with the answer given

5. thanks so much