1. ## volume of revolution

hello guys, just wanted to see if i derived this question correctly.

A bowl is formed by rotating about the y-axis the area contained between that part of the curve 2y=x^2 from x = 0 to x = 4, the line x=4 and the x-axis. calculate the maximium volume of water he bowl could hold.

i have derived the formula

Volume = $\displaystyle \int^0 $$\displaystyle \pi \displaystyle (x^2/2)^2 dx the upper limit is 0 and the lower limit should be 4. i cant seem to get that in latex form.. is this correct though 2. Originally Posted by sigma1 hello guys, just wanted to see if i derived this question correctly. A bowl is formed by rotating about the y-axis the area contained between that part of the curve 2y=x^2 from x = 0 to x = 4, the line x=4 and the x-axis. calculate the maximium volume of water he bowl could hold. i have derived the formula Volume = \displaystyle \int^0$$\displaystyle \pi$ $\displaystyle (x^2/2)^2$ dx

the upper limit is 0 and the lower limit should be 4. i cant seem to get that in latex form..
is this correct though
Well lets look at it like this. We have a parabola bounded by the vertical lines x=0, x=4 and the horizontal line y=0

So this would appear to be best for the method of cylindrical shells.

The interior of the bowl corresponds to the revolving region given by

$\displaystyle x^2/2<=y<=8$

The area element at position x has height

$\displaystyle 8-x^2/2$

and generates the clindrical shell of volume

$\displaystyle dV = 2\pi x(8-x^2/2)$

Edit- Was it just my imagination or didn't I see a moderator post in this thread? They had something different then me but I don't see why what I've done would be wrong. Although it has been a while since I did cylindrical shells. Can you post again?

3. disks w/r to y ...

$\displaystyle V = \pi \int_0^8 2y \, dy$