91Ó°ÊÓ

A double integral extends the concept of an integral to functions of two variables, usually within a specific region in the xy-plane. In the context of finding volumes, a double integral enables us to calculate the volume under a surface by integrating over a two-dimensional region. The double integral can be represented by:

• \[ \int \int_{R} f(x, y) \, dx \, dy \]

This integral computes the accumulation of infinitesimally small slices of the volume under the curve defined by \( f(x, y) \).
For instance, if you have a function \( f(x, y) = x^n + y^n + xy \) over a region \( R = [0, 1] \times [0, 1] \), you set up a double integral to find the volume:\[ V = \int_0^1 \int_0^1 (x^n + y^n + xy) \, dx \, dy. \]
Each of these components \( x^n \), \( y^n \), and \( xy \) is separately integrated over the region \( R \) to find their contributions to the total volume.

In multivariable calculus, calculating the volume under a surface refers to determining the space enclosed between the surface (like a curved ceiling) and a specific region in the base plane. This is particularly useful in applications such as physics and engineering, where understanding such volumes is crucial.
The concept uses a double integral to sum up all the infinitesimal volumes under the surface across the designated region. In our example, the function \( f_n(x, y) \) represents the height of the surface above point \( (x, y) \) in the region \( R \).
By integrating \( f_n(x, y) \) over \( R \), we find the volume of the solid state below the surface and above where \( x \) and \( y \) range from 0 to 1:

• The volume under the surface \( z = x^n + y^n + xy \) over \( R \) is calculated as \( V_n = \int_0^1 \int_0^1 (x^n + y^n + xy) \, dx \, dy \).

The solution step by step reveals how we systematically evaluate the integrals for each term to arrive at the final volume expression.

Understanding the limit of a sequence is a fundamental aspect of calculus. It involves analyzing what value a sequence approaches as its terms continue indefinitely. In this context, we are examining the sequence of volumes \( V_n \) as \( n \) increases without bound.
Mathematically, if a sequence \( a_n \) has a limit \( L \) as \( n \) approaches infinity, we denote it as \( \lim_{n \to \infty} a_n = L \).
In our problem, after computing the volume \( V_n = \frac{2}{n+1} + \frac{1}{4} \), we examine the limit of this expression as \( n \to \infty \):

• Noting that \( \frac{2}{n+1} \to 0 \) as \( n \to \infty \),
• The remaining term, \( \frac{1}{4} \), is constant.

Hence, the sequence of volumes converges to \( \frac{1}{4} \), meaning the total volume stabilizes at this value as \( n \to \infty \). Understanding these limits helps in predicting the behavior of functions across infinite steps or time.