Problem 54 - Thomas Calculus

Question

Find the area of the surfaces. The portion of the plane \(y+z=4\) that lies above the region cut from the first quadrant of the \(x\) -plane by the parabola \(x=4-z^{2}\)

Step-by-Step Solution

Answer

The area of the surface is \(\frac{16\sqrt{3}}{3}\).

Step 1: Understand the Problem Setup

The plane given by the equation \(y + z = 4\) needs to be analyzed over a specific region of the \(xz\)-plane. We are focused on the first quadrant region (where both \(x\) and \(z\) are non-negative) cut by the parabola \(x = 4 - z^2\).

Step 2: Identify Limits of Integration

Since the parabola \(x = 4 - z^2\) bounds the region, determine integration limits for \(z\) first. For the first quadrant, \(z\) ranges from \(0\) to \(2\), because when \(z = 2\), \(x = 4 - 2^2 = 0\). \(x\) ranges within \(0\) and \(4 - z^2\) for a given \(z\).

Step 3: Parameterize the Surface

Write the plane equation \(y = 4 - z\) in terms of parameters \(x\) and \(z\), since \(y + z = 4\) implies \(y = 4 - z\). Consider a parametrization: \(\mathbf{r}(x, z) = (x, 4 - z, z)\).

Step 4: Calculate the Surface Area Element

Express the partial derivatives of the parametrized surface with respect to \(x\) and \(z\): \(\frac{\partial \mathbf{r}}{\partial x} = (1, 0, 0)\) and \(\frac{\partial \mathbf{r}}{\partial z} = (0, -1, 1)\). Compute their cross product to obtain the vector determining the surface area element: \(\mathbf{n} = (1, 1, 1)\).

Step 5: Integrate to Find Surface Area

Compute the surface area by integrating over the region. The magnitude of the cross product vector is \(\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\). Therefore, the surface area is \(\int_{0}^{2} \int_{0}^{4-z^2} \sqrt{3} \; dx \, dz\).

Step 6: Evaluate the Integral

First integrate with respect to \(x\): \(\int_{0}^{4-z^2} \sqrt{3} \, dx = \sqrt{3} (4-z^2)\). Then integrate with respect to \(z\): \(\int_{0}^{2} \sqrt{3} (4-z^2) \, dz = \sqrt{3} \left[4z - \frac{1}{3}z^3 \right]_{0}^{2}\).

Step 7: Calculate the Final Result

Evaluate the definite integral: \(\sqrt{3} \left[(4\cdot2) - \frac{1}{3}(2)^3\right] = \sqrt{3} \left[8 - \frac{8}{3}\right] = \sqrt{3} \left[\frac{24}{3} - \frac{8}{3}\right] = \sqrt{3} \frac{16}{3}\). Thus, the area is \(\frac{16\sqrt{3}}{3}\).

Key Concepts

Parametrization

In mathematics, "parametrization" refers to expressing the coordinates of points that make up a surface or curve in terms of one or more parameters. This method is particularly useful when working with surfaces in three-dimensional space and it helps to describe complex shapes.

For the given exercise, we have a plane described by the equation \(y + z = 4\). We need to express this surface in terms of two parameters, commonly \(x\) and \(z\), which are independent variables. By re-writing the plane equation in the format \(y = 4 - z\), we consider the parametrization:

• \(\mathbf{r}(x, z) = (x, 4 - z, z)\)

Using this parametrization, each point on the plane is uniquely determined by the coordinates \(x\) and \(z\). This is significant as it simplifies the process of determining other properties of the surface, such as its area.

The choice of parameters depends on the problem's constraints, in this case, focusing on the \(xz\)-plane for appropriate representation.

Limits of Integration

``Limits of integration" define the range over which a variable in a definite integral can vary. These limits are essential to accurately determine the extent of the region over which you are integrating.

For this exercise, identify the region bounded within the first quadrant by the parabola \(x = 4 - z^2\). Since we are within the first quadrant, both \(x\) and \(z\) must be non-negative.

• The \(z\) variable varies from \(0\) to \(2\), as substituting \(z = 2\) into the parabola equation yields \(x = 0\).
• For each fixed \(z\), \(x\) ranges from \(0\) to \(4 - z^2\).

These limits ensure that every part of the required region is covered in the integration, helping to correctly determine the surface area.

Definite Integral

The concept of a "definite integral" is central in calculus for finding the accumulated quantity, such as area or volume, especially when dealing with continuously varying factors.

In this problem, the surface area of a plane segment must be calculated over a specific area. The surface area is given as a double integral:

• Integrate \(\int_{0}^{2} \int_{0}^{4-z^2} \sqrt{3} \, dx \, dz\)

First, integrate with respect to \(x\), keeping \(z\) constant, giving \( \sqrt{3} (4-z^{2})\). Then, integrate this result with respect to \(z\), calculating at the limits 0 to 2. This multi-step integration is key for evaluating the exact area of the surface.

The process demonstrates how definite integrals can represent complex areas through the summation of infinitesimal parts.

Cross Product

The "cross product" is a vector operation that finds a vector perpendicular to two given vectors in 3D space. It's vital for calculating surface elements in surface area problems.

In this task, compute the cross product between the partial derivatives of the parametrized surface to get a normal vector:

• \(\frac{\partial \mathbf{r}}{\partial x} = (1, 0, 0)\)
• \(\frac{\partial \mathbf{r}}{\partial z} = (0, -1, 1)\)

The cross product of these vectors is \(\mathbf{n} = (1, 1, 1)\), determining the orientation and unit element of the surface.

The magnitude of this vector, \(\sqrt{3}\), scales the integral, representing the surface area element properly. The cross product thus proves instrumental in translating the geometric problem into a solvable analytical form.