Berechnen Sie unter Verwendung des totalen Differentials die Oberfl?chen?nderung. \(\Delta O \approx d O\) eines Zylinders mit Boden und Deckel, dessen Radius \(r=10 \mathrm{~cm}\) um \(5 \%\) vergrö\betaert und dessen H?he \(h=25 \mathrm{~cm}\) gleichreitig um \(2 \%\) werkleinert wurde und vergleichen Sie diesen N?herungswert mit dem exakten Wert \(\triangle O\) exakt.

To calculate the surface area of a cylinder with both top and bottom bases, we use a specific formula. The cylinder's surface consists of two circular bases and a rectangular side that wraps around the cylinder.
The formula for the total surface area is: \[ O = 2\pi r(r + h) \]
Here, \( r \) is the radius and \( h \) is the height.
First, we calculate the area of one circular base, which is \( \pi r^{2} \). Since there are two bases, we multiply by 2.
Next, we calculate the area of the side, which is a rectangle. The height is \( h \) and the length is the circumference of the base, \( 2\pi r \).
This means the area of the side is \( 2\pi r h \). Adding these together gives the total surface area formula.
Understanding this formula is key to solving various problems related to cylinder surface area calculation.

Differential calculus helps us find how a function changes as its input changes. This is vital for approximating small changes in surface area.
The total differential of the surface area \( dO \) considers the partial derivatives with respect to both radius \( r \) and height \( h \).

Using the total differential formula: \[ dO = \frac { \partial O }{ \partial r } dr + \frac { \partial O }{ \partial h } dh \]
Here, \( \frac { \partial O }{ \partial r } \) and \( \frac { \partial O }{ \partial h } \) are the partial derivatives, representing the rate of change of surface area with respect to radius and height, respectively.
This approach approximates the change in surface area due to small changes in dimensions, providing a practical tool for real-world applications.

In mathematics, we often compare approximate values to exact ones to understand accuracy.
In our problem, we used the total differential to approximate the change in surface area \( dO \). We found it to be: \[ dO \approx 110 \text {cm}^2\]
However, we also calculated the exact change in the surface area \( \Delta O \), found directly from the differences in initial and final areas, resulting in the same: \[ \Delta O \approx 110 \text{cm}^2\]
This comparison shows that, for small changes, the approximate value given by differential calculus can be very close to the exact value. It simplifies complex calculations while maintaining a high degree of accuracy.
Understanding this concept ensures we apply the right method based on the desired precision.