TITLE: Calculating a Solid of Revolution with Horizontal and Vertical Slices
QUESTION [1 upvotes]: The first part of the problem my question is related to is as follows:
Take the region bounded by the curve $y = \sqrt{x}$ and $y = \frac{x}{2}$, rotate it around the line $x=-1$, and calculate the volume of the solid of revolution formed.
I was able to answer this relatively easily. The second part of the question asks for the volume to be calculated in two ways, by slicing both vertically and horizontally. I calculated the volume slicing horizontally, and my question is: is it possible to even do this slicing vertically? When the region is divided into horizontal slices, the "washer" slices of the solid of revolution approach a circle of zero height as the number of slices goes to infinity. When the region is divided up vertically however, as the number of slices goes to infinity the washer slices approach sort-of a line shape, if that makes sense.
I don't see how it is possible to calculate the volume slicing vertically, so if you could anyone shed some light on the topic that would be great, thanks.

REPLY [2 votes]: Perhaps it is asking you to do it by cylindrical shells. Take a vertical slice of the region we are rotating, sloppily from $x$ to $x+dx$. When we rotate it, we get a cylindrical shell of radius $1+x$, height $\sqrt{x}-\frac{x}{2}$, thickness $dx$, so approximate volume $2\pi (1+x)\left(\sqrt{x}-\frac{x}{2}\right)\,dx$.
"Add up" (integrate) from $x=0$ to $x=4$. Our volume is
$$\int_{x=0}^4 2\pi (1+x)\left(\sqrt{x}-\frac{x}{2}\right)\,dx.$$
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