Calculating Volume Of A Cylinder's Surface Segment

Hey guys! Today, we're diving deep into a super interesting problem involving 3D geometry and calculus: figuring out the volume spanned by a surface peeled from a cylinder. It sounds complex, right? But don't sweat it! We'll break it down step-by-step. We're talking about a specific section of a cylinder defined by $x^2+y^2=1$ in our good old $xyz$-space, and we're focusing on the part where $z$ goes from 0 to 2. Imagine a sheet of paper, super thin, just wrapping around this curved surface. What's the volume this 'paper' encloses or spans? Let's get our hands dirty and explore this!

Understanding the Setup: The Cylinder and the Surface

First things first, let's really get a grip on the cylinder we're dealing with. The equation $x^2+y^2=1$ describes a cylinder whose base is a circle of radius 1 centered at the origin in the $xy$-plane, and it extends infinitely along the $z$-axis. Think of it like a perfectly round pipe. Now, we're not looking at the entire infinite cylinder. We're slicing it. The condition 0 leq z leq 2 tells us we're only interested in the portion of the cylinder between the $xy$-plane (where $z=0$) and a plane parallel to it, 2 units up (where $z=2$). This gives us a finite cylinder, like a can of soda, but hollow since we're interested in the surface.

Now, about that sheet of paper. It has negligible thickness, which is a classic calculus simplification. This means we can treat it as a purely 2D surface embedded in 3D space. This 'paper' is wrapped around the cylindrical surface $x^2+y^2=1$ within the bounds 0 leq z leq 2. The crucial part is understanding what 'volume spanned' means in this context. Since the paper has no thickness, it doesn't have a volume in the traditional sense. Instead, it's acting as a boundary. The problem is asking for the volume of the region enclosed by this cylindrical surface and potentially other surfaces implied by the peeling process. Without further constraints on how the paper is 'peeled' or what other boundaries are involved, there could be ambiguity. However, in problems like this, it often implies the volume inside the cylinder bounded by the given $z$ limits. If the paper is simply wrapped around the cylindrical surface $S$, and no other boundaries are specified, the volume spanned would naturally refer to the volume enclosed by this cylindrical surface segment. This is the volume of the solid cylinder with radius 1 and height 2.

Let's make sure we're on the same page, guys. The surface $S$ is the lateral surface of a cylinder. Its equation is $x^2+y^2=1$ for 0 leq z leq 2. If we were asked for the surface area of this paper, that would be different. But we're after volume. This usually means the space contained within certain boundaries. If the paper is just the cylindrical surface itself, the volume it 'spans' is the volume of the solid cylinder defined by x^2+y^2 leq 1 and 0 leq z leq 2. Let's assume this standard interpretation unless more information suggests otherwise. So, the task boils down to calculating the volume of this solid cylinder.

Calculating the Volume: Cylindrical Coordinates to the Rescue!

Alright, now that we've clarified what we're calculating – the volume of the solid cylinder defined by x^2+y^2 leq 1 and 0 leq z leq 2 – let's figure out how to calculate it. For problems involving cylinders, spheres, or cones, cylindrical coordinates are often our best friends. They simplify the equations and the integration process significantly. In cylindrical coordinates $(r, heta, z)$, our familiar Cartesian coordinates $(x, y, z)$ are related by:

x = r leq ext{cos}( heta) y = r leq ext{sin}( heta) $z = z$

The Jacobian determinant for the volume element in cylindrical coordinates is dV = r leq dr leq d heta leq dz. This little '$r$' factor is super important – don't forget it!

Now, let's translate our cylinder's boundaries into these new coordinates:

1. The radius $r$: The condition x^2+y^2 leq 1 becomes r^2 leq 1. Since $r$ represents a distance, it must be non-negative. So, 0 leq r leq 1.
2. The angle $ heta$: To cover the entire circular base of the cylinder, $ heta$ needs to sweep through a full circle. This means 0 leq heta leq 2leq pi.
3. The height $z$: The problem explicitly gives us the bounds for $z$ as 0 leq z leq 2.

So, our volume integral in cylindrical coordinates is:

V = iiint_E dV

where $E$ is the region defined by 0 leq r leq 1, 0 leq heta leq 2leq pi, and 0 leq z leq 2. Substituting the volume element dV = r leq dr leq d heta leq dz, we get:

V = int_0^2 int_0^{2pi} int_0^1 r leq dr leq d heta leq dz

This is an iterated integral, and we can solve it by integrating with respect to one variable at a time, starting from the inside.

First, integrate with respect to $r$:

int_0^1 r leq dr = {raw{r^2}}{2}|_{raw{0}}^1 = {1^2}{2} - {0^2}{2} = {1}{2}

Now, substitute this result back into the integral and integrate with respect to $ heta$:

int_0^{2pi} {{1}{2}} d heta = {1}{2} int_0^{2pi} d heta = {1}{2} [ heta]_{raw{0}}^{2pi} = {1}{2} (2leq pi - 0) = pi

Finally, integrate with respect to $z$:

int_0^2 pi leq dz = pi int_0^2 dz = pi [z]_{raw{0}}^2 = pi (2 - 0) = 2leq pi

So, the volume is V = 2leq pi. Pretty neat, huh? This is the standard formula for the volume of a cylinder: V = {pi r^2 h}. In our case, $r=1$ and $h=2$, so V = {pi (1^2) (2)} = 2leq pi. The calculus approach confirms the geometric formula!

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