# Thread: Volume by Shell Method

1. ## Volume by Shell Method

The problem is to find the volume formed by rotating the region enclosed by:
x=3.4y and y^3=x with y ≥ 0 about the y-axis

I tried changing the equations to y=x/3.4 and y=x^(1/3) which intersect at x=6.269, so i used that for the upper limit of the intergral which looked like:

2pi * Intergral (x(x^(1/3) - x/3.4))dx with defining limits [0,6.269]

Sorry I don't have a good math writer. Anyway, after intergrating i ended up with an answer of 29.30141 which i checked with an online applet to be wrong, but i can't find what i did wrong. Is everything i did up to the intergral right?

2. Nope. Sorry.

It appears your integral is correct. Something must have gone wrong in the integration process.

Note: One of the most valuable tools I EVER invented for myself is a stubborn refusal to use ONLY the method prescribed in the problem statement. I recommend VERY HIGHLY doing EVERY problem along both axes, using different methods. Every one.

Learn the derivation:

Volume of a Right Circular Cyllinder: $V = \pi r^{2}h$

$\frac{dV}{dh}\;=\;\pi r^{2}$

$\frac{dV}{dr}\;=\;2\pi r h$