38. Consider The Solid Region That Lies Under The Surface Z = X Vy And Above The Rectangle R= [0, 2]

Understanding the Solid Region Under the Surface \( Z = XY \) Above the Rectangle \( R = [0, 2] \)

38. Consider The Solid Region That Lies Under The Surface \( Z = XY \) And Above The Rectangle \( R = [0, 2] \). This problem involves analyzing a three-dimensional region bounded by a surface and a rectangular base. Such regions are fundamental in multivariable calculus, especially in setting up and evaluating triple integrals for volume calculations, moments, and other physical or geometric quantities. To comprehend this region thoroughly, we need to explore the geometric setup, the limits of integration, and the methods to evaluate integrals over this region.

Geometric Interpretation of the Surface and Region

The Surface \( Z = XY \)

The surface described by \( Z = XY \) is a hyperbolic paraboloid, a common saddle-shaped surface in three-dimensional space. Its key characteristics include:

• It passes through the origin \( (0, 0, 0) \).
• The surface is symmetric with respect to the axes, with positive \( Z \) values where \( X \) and \( Y \) have the same sign, and negative \( Z \) where they have opposite signs.
• The surface exhibits saddle behavior, curving upward in one direction and downward in the perpendicular direction.

The Base Rectangle \( R = [0, 2] \times [0, 2] \)

The rectangle \( R \) in the \( XY \)-plane spans from \( X = 0 \) to \( X = 2 \) and \( Y = 0 \) to \( Y = 2 \). Visualizing this rectangle is crucial:

• It occupies the first quadrant, where both \( X \) and \( Y \) are non-negative.
• Within this domain, \( Z = XY \) is always non-negative, since the product of two non-negative numbers is non-negative.

Defining the Solid Region

Boundaries of the Region

The solid region is bounded below by the rectangle \( R \) in the \( XY \)-plane and above by the surface \( Z = XY \). Specifically:

• Base boundary: The 2D rectangle \( R \) in the \( XY \)-plane.
• Upper boundary: The surface \( Z = XY \), which extends upward from the \( XY \)-plane.

Visualizing the Region

Imagine the rectangle \( R \) lying flat on the \( XY \)-plane. Above each point \( (x, y) \) in \( R \), the surface extends upward to the height \( Z = XY \). The three-dimensional volume enclosed between the rectangle and the surface forms the solid region of interest.

Mathematical Description of the Region

Limits of Integration

To set up an integral over this volume, we need to define the limits for \( X \), \( Y \), and \( Z \). Since the base is fixed and the surface is explicitly given, the natural choice is:

• \( x \) varies from 0 to 2.
• For each fixed \( x \), \( y \) varies from 0 to 2.
• \( Z \) varies from 0 (the \( XY \)-plane) up to \( XY \).

Describing the Volume Element

The differential volume element in Cartesian coordinates is \( dV = dx\,dy\,dz \). To integrate over the solid, we can set up the integral as:

V = \int{x=0}^{2} \int{y=0}^{2} \int_{z=0}^{XY} dz\,dy\,dx

Evaluating the Volume of the Solid Region

Setting Up the Triple Integral

Given the bounds, the volume \( V \) can be computed as:

V = \int{0}^{2} \int{0}^{2} \int_{0}^{xy} dz\,dy\,dx

Integrating with respect to \( z \) first:

V = \int{0}^{2} \int{0}^{2} xy\,dy\,dx

Reducing to a Double Integral

The problem simplifies to calculating a double integral:

V = \int{0}^{2} \left( \int{0}^{2} xy\,dy \right) dx

Computing the Inner Integral

For a fixed \( x \), evaluate:

\int{0}^{2} xy\,dy = x \int{0}^{2} y\,dy = x \left[ \frac{y^{2}}{2} \right]_0^2 = x \times \frac{4}{2} = 2x

Computing the Outer Integral

Now, integrate with respect to \( x \):

V = \int{0}^{2} 2x\,dx = 2 \int{0}^{2} x\,dx = 2 \left[ \frac{x^{2}}{2} \right]_0^2 = 2 \times \frac{4}{2} = 4

Interpretation of the Result

The volume of the solid region bounded above by \( Z = XY \) over the rectangle \( R = [0, 2] \times [0, 2] \) is 4 cubic units. This calculation provides a quantitative measure of the space enclosed between the surface and the base in the specified domain.

Applications and Further Implications

Applications in Physics and Engineering

• Calculating mass or charge distribution over a region with density proportional to \( XY \).
• Determining the volume for fluid or material flow in a bounded region.

Extensions to Surface Integrals and Moments

Beyond volume calculation, similar setups allow for computing surface integrals, moments of inertia, and center of mass by integrating different functions over the same region.

Visual Aids and Graphical Representation

Utilizing 3D graphing tools can enhance understanding of the region. Visualizing the surface \( Z = XY \) over the rectangle reveals the saddle shape and how the volume accumulates as \( x \) and \( y \) vary from 0 to 2.

• Plot the surface \( Z = XY \) over the domain.
• Shade the region under the surface within the rectangle.
• Observe how the height \( Z \) increases with \( x \) and \( y \).

Summary and Key Takeaways

• The solid region is bounded below by the rectangle \( R = [0, 2] \times [0, 2] \) and above by the surface \( Z = XY \).
• Its volume can be computed via a straightforward double integral after integrating out \( Z \).
• The calculated volume is 4 cubic units, illustrating the application of multivariable calculus techniques.
• This analysis serves as a foundation for more complex volume and surface integral calculations in applied mathematics and physics.

Conclusion

Analyzing the solid region under the surface \( Z = XY \) above the rectangle \( R = [0, 2] \) demonstrates fundamental techniques in multivariable calculus. Setting up the correct limits, understanding the geometric interpretations, and performing the integration systematically allows us to evaluate the volume of complex 3D regions efficiently. Such skills are essential in various scientific and engineering fields where spatial analysis and integral calculus are pivotal.