91Ó°ÊÓ

The volume of a cylinder is a crucial concept when exploring three-dimensional geometry. In simple terms, the volume of a cylinder indicates how much space is enclosed within a cylinder. To compute this, you use the formula:
\[ V = \pi r^2 h \]
where:

• \(V\) is the volume
• \(r\) is the radius of the base
• \(h\) is the height of the cylinder

You can visualize this formula as stacking circular disks (represented by \(\pi r^2\)) on top of each other to fill up the height \(h\). In our exercise, we're given that the volume equals \(12\pi\). We use this information to determine another variable, such as the height, as a function of the radius. By rearranging the formula, the height \(h\) is expressed as \(h(r) = \frac{12}{r^2}\). This allows us to understand how changes in the radius influence the height, given the volume remains constant.

Calculating the surface area of a cylinder involves understanding both the lateral area and the area of the circular bases. The total surface area \(A\) is a combination of the curved (side) surface and the top and bottom surfaces:
\[ A = 2\pi rh + 2\pi r^2 \]
This formula can be broken down as:

• \(2\pi rh\): Curved surface area surrounding the cylinder, like the label on a can.
• \(2\pi r^2\): Combined area of the two circular ends (top and bottom).

When we express the surface area as a function of the radius, knowing the height as \(h(r) = \frac{12}{r^2}\), we substitute \(h\) into our formula:
\[ A = 2\pi r\left(\frac{12}{r^2} + r\right) \]
Simplifying, we find:
\[ A(r) = \frac{24\pi}{r} + 2\pi r^2 \]
This illustrates how \(r\) affects the overall surface area, offering insight into optimizing shapes for specific purposes like packaging.

Understanding the domain of a function is a key aspect of mathematics, especially in problems involving real-world constraints. The domain represents all possible input values for which the function is defined.
For our given problem, the domain is primarily restricted by the condition that the radius \(r\) of the cylinder must be greater than zero (\(r > 0\)). This is because:

• The radius cannot be zero or negative as cylinders with those radii wouldn't exist physically.
• A radius of zero yields a height or surface area that doesn't logically occur in real-world applications.

Thus, both functions derived from the exercise — height as \(h(r) = \frac{12}{r^2}\) and surface area as \(A(r) = \frac{24\pi}{r} + 2\pi r^2\) — share the domain \(r > 0\). This constraint reflects the practical limitations on how we define these mathematical functions.