Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis.
$xy = 1, x = 0, y = 1, y = 3$, rotated about the x-axis.
So I plan to do this via integration using horizontal rectangles.
I think the correct integral I need is this: radius = $y - 1$ height = $\frac{1}{y}$
$$ \int_1^3 2\pi(y-1)(\frac{1}{y}) \, dy$$
$$= 2\pi \int_1^3 (y-1)(y^{-1})$$
$$ 2\pi \int_1^3 (y^0 - y^{-1})$$
Does this look right?
On
You are almost correct.
Look specifically at your definition of radius : $r = y-1$
It appears that you are blending your concepts of how you define your radius with how you define the bounds of integration.
To get an intuition for where you are going wrong, simply draw a rough sketch of the 3-D shape you are creating, superimposed on the Cartesian coordinate system. From there, draw the outer most "shell" you could imagine on that shape, with thickness $\Delta y$ and measure the radius to the outer edge of that shell as well as the inside of the shell. I think you will quickly see where you have gone wrong!
Additionally, here is some reading on the derivation of the "Shell Method", that was helpful to me: http://mathonline.wikidot.com/calculating-volumes-cylindrical-shell-method
On
It doesn't look right.
A cylindrical shell around the $x$-axis of radius $y$ has circumference $2\pi y$, height $x={1\over y}$, and thickness $dy$. It follows that the volume of this shell is given by $${\rm dvol}=2\pi y\>{1\over y}\>dy=2\pi\>dy\ .$$ The volume $V$ of the solid under consideration therefore comes to $$V=\int_1^3 2\pi\>dy=4\pi\ .$$
$$\int_1^3 2\pi(y-1)(\frac{1}{y}) \, dy$$ is not the correct integral.
The problem with it is in the $r=(y-1)$ part.
Note that the radius is the distance from the x-axis which is $y=0$ to $y$ which is simply $r=y$