1. ## Help please!! - Volume using method of cylindrical shells

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y = 6sqrt(x) and y = 6x^2 about the y axis.
I keep getting ~9.425 which is wrong, but I'm not sure why yet!
Here is how I come to my answer time and time again...
S = integral sign loL
V = S 2pi(x(6sqrt(x) - 6x^2))
V = 2pi S (6x - 6x^3)
V = 2pi [3x^2 - (3/2)x^4] from 0-1
V = 9.4248

2. Hello, sgares!

Your algebra is off . . .

Use the method of cylindrical shells to find the volume of the solid obtained
by rotating the region bounded by $\displaystyle y \:= \:6\sqrt{x}\text{ and }y \:= \:6x^2$ about the y-axis.
Your set-up is correct . . .

$\displaystyle V \;=\;2\pi\int^1_0x\left(6x^{\frac{1}{2}} - 6x^2\right)\,dx$

But this is: .$\displaystyle V \;=\;12\pi\int^1_0\left(x^{\frac{3}{2}} - x^3\right)\,dx$

. . . Try it now . . .

3. Your set up seems correct. I think you just have an arithmetic error towards the end.

$\displaystyle 2{\pi}\int_{0}^{1}\left[x(6\sqrt{x}-6x^{2})\right]dx=\frac{9{\pi}}{5}\approx{5.65}$

4. Haha wow! Embarrassing... thank goodness I didn't go to the engineering tutor center for this! This site seriously is the internets. Bookmarked for life.

Thanks so much btw! )<