Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

The part of the surface \(z=\sqrt{4-y^{2}}\) that is directly above the square in the \(x y\)-plane with vertices \((1,0),(2,0),(2,1)\), and \((1,1)\)

Step-by-Step Solution

Answer

The surface directly above the square is defined by the region where \(1 \leq x \leq 2\), \(0 \leq y \leq 1\), and \(z = \sqrt{4-y^2}\).

Step 1: Understand the surface equation

The surface is defined by the equation \(z = \sqrt{4 - y^2}\). This is a part of a cylindrical surface with its radius depending on \(y\). Here, \(z\) is positive real values for \(4-y^2 \geq 0\), or equivalently for \(-2 \leq y \leq 2\).

Step 2: Identify the region above the square

The square has vertices \((1,0), (2,0), (2,1), (1,1)\). This defines a rectangular region in the \(xy\)-plane: \(1 \leq x \leq 2\) and \(0 \leq y \leq 1\).

Step 3: Determine the region for \(z\)

Given that \(z = \sqrt{4 - y^2}\) must be evaluated above the square, we focus on the region \(0 \leq y \leq 1\). In this range, \(4 - y^2\) remains positive, so \(z\) will have real values.

Step 4: Calculate \(z\) at the boundaries of \(y\)

At the boundaries: when \(y = 0\), \(z = \sqrt{4 - 0^2} = 2\). When \(y = 1\), \(z = \sqrt{4 - 1^2} = \sqrt{3}\). This shows that \(z\) ranges from \(\sqrt{3}\) to \(2\) over the square.

Step 5: Conclusion

The portion of the surface directly above the square is defined as \(z = \sqrt{4-y^2}\) with \(1 \leq x \leq 2\) and \(0 \leq y \leq 1\). The "height" \(z\) above this region will range between \(\sqrt{3}\) and \(2\).

Key Concepts

Cylindrical Surface

When dealing with multivariable calculus and 3D coordinate geometry, understanding cylindrical surfaces is essential. A cylindrical surface is a 3-dimensional shape that is formed by moving a line (the generator) parallel to a fixed line (the axis) while following a curve. The simplest examples are cylinders formed from circles, but cylinders can form from any other curves.
Our exercise involves the surface given by the equation \(z = \sqrt{4 - y^2}\). This represents a cylindrical surface because it remains consistent for all values of \(x\). In this case, \(z\) depends only on \(y\). This forms a semi-circular shape when plotted, and any cross-section along the \(x\)-axis would show a half-circle lying along the \(y\)-axis.

• The range of \(z\) is determined by \(y\) such that \(z\) can only be real when \(4 - y^2 \geq 0\).
• The resulting surface is part of a larger cylindrical shape that extends infinitely parallel to the \(x\)-axis.

Surface Integration

Surface integration is a vital tool in multivariable calculus, allowing us to compute the integral of a function over a surface in space. This technique becomes particularly useful when you need to find areas or evaluate physical quantities like mass or flux across a surface.
In this exercise, the surface \(z = \sqrt{4 - y^2}\) above a rectangular region highlights how surface integration works. To do surface integration, we set up a double integral over a defined region in the \(xy\)-plane, which projects to our desired surface.

• We choose limits based on the region: here, \(x\) ranges from 1 to 2 and \(y\) ranges from 0 to 1.
• The integration then considers each slice of the surface as \(z\)'s height at each \((x, y)\) coordinate within this range.

For this example, the height \(z\) contributes directly to what is integrated over the region, offering insights into different parts of the surface when approached this way.

3D Coordinate Geometry

3D coordinate geometry is the backbone for studying surfaces in multivariable calculus. With the fundamental concepts of the Cartesian coordinate system, we describe loci of points in three-dimensional space, analyze shapes, and solve practical problems.
In this exercise, the geometry underlies the placement and structure of the surface: the surface \(z = \sqrt{4 - y^2}\) describes a region in space. It helps to visualize this equation by looking at how the height \(z\) changes with \(y\), expanded over a consistent \(x\) region:

• The surface exists at positions \((x, y, z)\) where \(1 \leq x \leq 2\) and \(0 \leq y \leq 1\).
• This forms a visual boundary, encapsulating part of a cylindrical shape aligned along the \(x\)-axis.

The concept integrates geometry with algebraic equations to model real-world surfaces, prompting deeper understanding of spatial relationships in multivariable functions.