Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the mass of the surface \(z=1-\left(x^{2}+y^{2}\right) / 2\) over \(0 \leq x \leq 1,0 \leq y \leq 1\), if \(\delta(x, y, z)=k x y\).

Step-by-Step Solution

Answer

Integrate \( kxy \sqrt{1 + x^2 + y^2} \) over \( x \) and \( y \) from 0 to 1.

Step 1: Define the Surface and Density Function

The surface is given by the equation \( z = 1 - \frac{x^2 + y^2}{2} \). We observe that this is a paraboloid. The density function of the surface is given by \( \delta(x, y, z) = kxy \). We want to find the mass of the surface as defined by these parameters.

Step 2: Establish the Limits for Integration

The surface is defined over the region \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \). These will be our limits of integration for \( x \) and \( y \), in the double integral for computing the surface integral.

Step 3: Determine the Surface Element Area

To integrate over a surface, we need to express the surface element area \( dS \). For a surface \( z = f(x, y) \), the area element \( dS \) is given by \( \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dx \, dy \).

Step 4: Calculate Partial Derivatives

Calculate the partial derivatives \( \frac{\partial f}{\partial x} = -x \) and \( \frac{\partial f}{\partial y} = -y \). Thus, \( dS = \sqrt{1 + x^2 + y^2} \, dx \, dy \).

Step 5: Setup the Surface Integral for Mass

The mass \( M \) of the surface is the integral of the density function over the surface: \[ M = \int_{0}^{1} \int_{0}^{1} kxy \sqrt{1 + x^2 + y^2} \, dx \, dy \]

Step 6: Evaluate the Inner Integral

Evaluate the integral with respect to \( y \): \[ \int_{0}^{1} kxy \sqrt{1 + x^2 + y^2} \, dy \]

Step 7: Evaluate the Outer Integral

Substitute the result from the inner integral and integrate with respect to \( x \): \[ \int_{0}^{1} (\text{Inner Integral Result}) \, dx \] The exact evaluation requires the substitution and solving of these integrals, which may involve more complex integration techniques.

Step 8: Final Calculation

Perform the definite integration to obtain the mass \( M \). This could involve numerical methods or symbolic computation for the exact value.

Key Concepts

Surface Mass Calculation

When dealing with calculus surface integrals, calculating the mass of a surface like a paraboloid involves a few critical steps. The surface mass is essentially the density distributed over the entire surface area. To determine it, you need to

• Identify the surface described mathematically, such as our paraboloid.
• Recognize the density function, like \(\delta(x, y, z) = kxy\), which indicates how mass is distributed across the surface.
• Set up the limits of integration over the defined region, here \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\).
• Calculate the surface area element \(dS\), integrating the density over this element.

The result from integrating the density \(kxy\) over the surface provides the total mass. Solving integrals with correct limits ensures an accurate calculation of surface mass.

Paraboloid Surface

A paraboloid surface like \(z = 1 - \frac{x^2 + y^2}{2}\) is a common geometric shape in calculus. It can be visualized as a bowl-shaped surface that opens downwards or upwards. Understanding the geometric nature is essential:

• The equation describes how every point \((x, y)\) on the plane corresponds to a height \(z\).
• For this surface, heights decrease as \(x\) and \(y\) increase, forming the typical bowl shape.
• The region \(0 \leq x \leq 1, 0 \leq y \leq 1\) is the bounded section for our integration.

Grasping the structure of a paraboloid aids in visualizing the problem and setting correct integration boundaries. Recognizing such surfaces is pivotal when dealing with multivariable calculus, as it affects how you compute the surface element \(dS\).

Density Function Integration

Integrating a density function over a surface is crucial for determining surface mass. Here, the density is represented by \(\delta(x, y, z) = kxy\). This function tells us:

• How dense or thick the surface is at each point along the \(x\) and \(y\) coordinates.
• The variable density reflects changes in mass distribution over the surface area.
• Critical to use it properly in the surface integral \( M = \int \int kxy \sqrt{1 + x^2 + y^2} \, dx \, dy \).

Integrating this function over the specified region captures the total mass by adding up all the infinitesimal contributions of mass from each point on the surface. It's important to frame it within the right bounds and integrate over the correct area element \(dS\).

Partial Derivatives in Calculus

Partial derivatives are derivatives where we focus on how a function changes with respect to one variable, keeping others constant. They are essential in understanding the small changes in surfaces:

• For the paraboloid \(z = 1 - \frac{x^2 + y^2}{2}\), compute \(\frac{\partial f}{\partial x} = -x\) and \(\frac{\partial f}{\partial y} = -y\).
• These derivatives highlight the rate of change of the surface along the \(x\) and \(y\) axes.
• Used to derive the surface area element \(dS = \sqrt{1 + x^2 + y^2} \, dx \, dy\).

Partial derivatives are not only a tool for linear approximation but also help in identifying critical points and understanding how a surface curves. They fundamentally underpin the concepts needed to accurately set up and solve integrals involving surfaces in multivariable calculus.