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I am studying Calculus II and I had a question about finding the surfaces of revolution.

Why don't you add the top and bottom circle when you find the surface? If one is trying to find the surface area of a solid obtained by rotating an arc, shouldn't the top and bottom circles be added too?

For example, the surface area of a cylinder calculated by the area formula says it is $2\pi rh$ which is only the sides. I thought it would be $2\pi rh+2\pi r^2$. Same with cones and with rotating arcs.

Why is this so?

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    $\begingroup$ The surface is the set of points coinciding with the points covered by the rotating curve. This does not usually include the "top" and "bottom". $\endgroup$ – David Mitra Feb 24 '16 at 13:53
  • $\begingroup$ Because the top and the bottom do not belong to the surface of revolution. $\endgroup$ – Alex M. Feb 24 '16 at 13:54
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To expand on David Mitra and Alex M's comments:

Consider a particular case: The curve is $x = 1$, for $y \in [0, 1]$, and we are rotating it about the $y$-axis. Since this is a 3D phenomenon, we must consider our curve as a set of points $C$ in $\Bbb R^3$, so we need to also note that $z = 0$. Therefore $C = \{(1,y, 0)\mid 0\le y \le 1\}$. Note that every point in $C$ is a distance of $1$ from the $y$-axis.

What points lie in the surface of rotation? For a point $(x, y, z)$ to lie in the surface, it must be the rotation of some point in $C$. If $(1,y, 0)$ is a point of $C$, then the rotations of this particular point form a circle about the $y$-axis of radius $1$. That is, the rotations form the set $\{(x, y, z) \mid x^2 + z^2 = 1\}$. The surface of rotation is the union of theses sets for every point of $C$, which is the set $\{(x,y,z)\mid x^2 + z^2 = 1 \text{ and } 0\le y \le 1\}$. There is not "top" or "bottom" to this surface because the top and bottom are not the rotations of points in the original curve.

If instead I defined $C$ to be the curve $\{(x,0),(x,1),(1,y)\mid 0\le x \le 1, 0\le y \le 1\}$, which forms 3 sides of a square, then the rotations of the top and bottom of the square would sweep out the tops and bottoms of the cylinder, and we would get a closed cylinder as the surface of rotation. But without the top and bottom in the original curve, we don't get a top and bottom in the cylinder.

Your problem is likely that you are thinking of a solid of rotation instead of a surface of rotation. In a solid of rotation, we start with a surface (most commonly, a planar area) and rotate it. The original surface has a closed boundary, so when that boundary is rotated, it results in a closed surface that is the boundary of the solid of rotation. But for a surface of rotation, there is no requirement that the original curve is closed, and so the surface of rotation does not need to be closed, either.

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