Disk Method of revolution when there is a gap
The question is to find the vol of the solid formed by revolving the region bounded by $y=\sqrt x, x=1, x=4$ about the $y$ axis using the disk method.
I can easily see that this will be a $dy$ slice if we are using the disk method to revolve around a vertical axis.
My problem is not knowing what to do with that gap from $0$ to $1$. If the problem had said revolve around the line $x=1$ where the region touches the axis, then I can easily work this. How to I deal with this gap?
If there is a gap must I use a "washer" method?
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$\begingroup$The washer method is simply a generalization of the disk method; alternatively, you can conceptualize the washer method as the subtraction of one volume calculated via disks from another volume also calculated via disks.
When the integration is performed along the same axis as the rotation axis, then disks/washers are employed. When the integration is performed along the perpendicular axis as the rotation axis, then cylindrical shells are employed.
Your particular integral via disks/washers (the difference is largely semantic) is simply $$V = \int_{y=0}^1 \pi (4^2 - 1^2) \, dy + \int_{y=1}^2 \pi (4^2 - (y^2)^2) \, dy.$$ The first integral is just the volume of a thick washer of inner radius $1$ and outer radius $4$ and height $1$. Alternatively, we can write $$V = \int_{y=0}^2 \pi (4^2 - (y^2)^2) \, dy - \int_{y=0}^1 \pi ((1^2) - (y^2)^2) \, dy.$$ Note that this is the difference of two volumes, both of which are calculated via washers. You could also set it up as $$V = (4^2)(2)\pi - (1^2)(1)\pi - \int_{y=1}^2 \pi (y^2)^2 \, dy,$$ which computes the volume as comprising the difference of a large cylindrical block of radius $4$ and height $2$, minus a smaller interior block of radius $1$ and height $1$, minus a third volume which is computed via disks on the interval $y \in [1,2]$ whose radius at $y$ is $y^2$. The result is the same.
Yet another way to compute this volume is to use cylindrical shells: $$V = \int_{x=1}^4 2 \pi x \sqrt{x} \, dx.$$ This is in my opinion much more simple and elegant--in fact, more natural to visualize.
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