91Ó°ÊÓ

A double integral extends the concept of integration to functions of two variables, over a specified region in the plane. Imagine it as stacking an infinite number of tiny 3D blocks, summing their volumes to find the total volume under a surface. For our problem, the region of interest is defined as the rectangle with boundary conditions given by the surface equation.
The expression \( V = \int_{0}^{1} \int_{0}^{2} (4 - y^2) \, dy \, dx \) exemplifies this perfectly. Here, we first integrate with respect to \( y \), then with respect to \( x \). This sequential process considers a small area in the rectangle, computes the contribution of the column of blocks directly beneath the surface defined by \( z = 4 - y^2 \), and repeats across the entire area.

Surface integration involves integrating a function over a surface or area. It plays a crucial role in finding the volume under a curve or a more complex surface. In our scenario, using the surface integration method, we need to account for how the function \( z = 4 - y^2 \) changes across the entire rectangle.
This method involves:

• Identifying the surface function, which in this context is \( 4 - y^2 \).
• Determining the area over which to integrate, represented by the rectangle \( R: 0 \leq x \leq 1, \; 0 \leq y \leq 2 \).

Through this approach, we convert the problem of finding volume into manageable segments, each represented by the limits of integration as specified in our setup.

To compute the volume using these integrals, we rebuild the volume piece by piece. We start by calculating the area under the curve for a single slice of the region. This means evaluating the integral of \( 4 - y^2 \) with respect to \( y \), and subsequently, with respect to \( x \).
Performing this involves:

• Calculating \( \int_{0}^{2} (4 - y^2) \, dy \), resulting in \( \frac{16}{3} \).
• Integrating the constant \( \frac{16}{3} \) across the entire range of \( x \): \( \int_{0}^{1} \frac{16}{3} \, dx \).

Combining these steps, the method effectively ‘sweeps’ through all areas of the region under the surface, accumulating the total volume. Thus, the computed volume under the curve \( z = 4 - y^2 \) over the specified rectangle is \( \frac{16}{3} \).