# Cylindrical shell

1. Sep 10, 2006

### mbrmbrg

Use the method of cylindrical shells to find the volume of the solid generated by revolving aabout the indicated axis the region bounded by the given curves.

x=y, x+2y=3, y=0; about the x axis.

(I don't know how to paste a graph into the post. But I graphed my curves.)

$$V=\int_{a}^{b}2\pi\L(x)dx$$

But since this rotates around the x axis instead of the y axis, replace all x's with y's.
I took the 2pi out of the integral b/c it's a constant. After that I'm very unsure of what's going on.

y stayed y.

L(x), the height of the cylinder, is now the width of the cylinder, and is defined as the length of the horizontal line, which is the distance between the two curves.
So L(x)= (-2y+3) - y = -3y+3.

My interval is the largest distance beteween my two curves, which is the x-axis on the interval [0,3]

I shoved that into the equation, factored out a 3 and removed it from the integrand, leaving me with
$$6\pi\int_{0}^{3}(y-3y)dx\\=(6\pi)[\frac{y^2}{2}-\frac{y^3}{3}]^3_0\\=6\pi\frac{-27}{6}\\=-27\pi$$

But the book says the answer is pi.

Last edited: Sep 10, 2006
2. Sep 11, 2006

### Galileo

You don't need the book to tell it is wrong. Volume should always be positive! This gives you a hint on where it might have gone wrong. If you look at the integrand (you should integrate wrt dy) you see it's y-y^2 (ignoring the typo's). Which is negative when y is greater than 1.
So recheck your interation limits and remember you are integrating wrt y.

3. Sep 11, 2006

### mbrmbrg

OK, so I switched my limits to lower limit=0 and upper limit=1 (b/c the curves intersect at y=1).
But when I set L(x)=the horizontal line between the two curves=-2y+3-y=3-3y...

right. It works. Thank you!!