I have to find the volume of a solid bound by $y=x$, $y=0$, $x=2$, and $x=4$, rotated by $x=1$. I'm doing fine with these problems when the regions are "right next" to the line upon which they rotate, but when they're apart I get very confused. What am I missing here? Any help (with this particular problem and or in general) would be greatly appreciated.
Thank you all!
Chris
On
You can use the washer method if you integrate with respect to $y$.
In cases where there isn't any kind of gap between the region and the axis of rotation, what we get is a solid, uh, thing. (Hence, I guess, the phrase solid of rotation, which sounds better.) So if we wanted a cylinder of radius $r$, we could set up the integral
$$\int_a^b \pi r^2 dy$$
which would give us the volume of a cylinder $b - a$ units high.
Of course, you could fit another, smaller cylinder inside; let's say it has radius $s$. Its integral is in essence the same: $\int_a^b \pi s^2 dy$. But what if we wanted not a cylinder but, I don't know, the volume of a cored pineapple? If the pineapple itself has radius $r$ and hole down the middle of radius $s$, then just calculating the first integral overestimates its volume — by exactly the amount given by the second integral. Therefore, we subtract one from the other. We can then combine these integrals to get
$$\int_a^b \pi r^2 - \pi s^2 dy$$
The idea with your question is similar, only each radius is a bit smaller; because we have the axis of rotation at $x=1$, it's closer to the region. If there weren't anything taken out of the middle, we'd have something like a cylinder of radius $3$; however, there is something taken out of the middle, and it has radius $1$. It's a skinnier pineapple, and its core is similarly skinnier — but that means the distance between the core and the edge is the same.
Anyway, your question requires two integrals via the washer method, one for $y \in [0,2]$ and one for $y \in [2, 4]$. With the above in mind, we can set up the first part of the integral as:
$$\int_0^2 \pi (4 - 1)^2 - \pi (2 - 1)^2 dy$$
(Of course, this can be simplified a lot.) What's left over is the diagonal part, which is, admittedly, slightly trickier. What do you get for that?
Because the region in question is a trapezoid, the shell method may be a bit simpler to contend with, as we can cover the entire revolved region "all in one go". The "slices" for this method are always parallel to the rotation axis, so the radii are "horizontal" and the "walls" of the shells are "vertical" here. The "thickness" of the shells is $ \ dx \ $ , so we will be integrating in the $ x-$ direction. The interval is then $ [ 2, 4 ] , $ the "radius arm" for a shell is $ \ r = (x-1) \ , $ and the "height" of each shell is $ \ h(x) = x - 0 = x \ . $ Here is a picture of the situation:
This is where a picture is valuable. If the trapezoid were in contact with the $ y-$ axis, the "radius arm" would simply have a length $ \ x \ $ over to the $ \ x-$ coordinate of the slice we were integrating. Since we are revolving the trapezoid around $ \ x = 1 \ , $ however, our reference is shifted to the right by one unit, so we must subtract 1 off of the length of the radius arm, making it $ \ r = x - 1 \ $ . (If the rotation axis were $ \ x = -1 \ $ , the radius arm would be one unit longer, thus $ \ r = x + 1 \ .) $
The infinitesimal volume of a cylindrical shell is $ \ dV = 2 \pi \cdot r \cdot h \ $ d(radius) , so that will be $ \ dV = 2 \pi \cdot (x-1) \cdot x \ \ dx \ $ here. The volume integral is then
$$ \int_2^4 \ \ 2 \pi \ (x-1) \ \cdot \ x \ \ dx \ . $$
By way of comparison, here are the two sets of "disk radii" dmk is describing for the disk method (a little more work to set up, but the integrals are still easy...).