Problem 24 - Thomas Calculus

Question

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) Parabolic band The portion of the paraboloid \(z=x^{2}+y^{2}\) between the planes \(z=1\) and \(z=4\)

Step-by-Step Solution

Answer

The surface area is \(\frac{\pi}{12} (17\sqrt{17} - 5\sqrt{5})\).

Step 1: Set Up the Parametrization

To parametrize the paraboloid, we can use cylindrical coordinates. Take \(x = r \cos\theta\), \(y = r \sin\theta\), and \(z = r^2\). For the section between \(z = 1\) and \(z = 4\), we have \(1 \leq r^2 \leq 4\). So, \(1 \leq r \leq 2\), with \(\theta\) ranging from \(0\) to \(2\pi\).

Step 2: Find the Surface Element

The surface element for the parametrized surface can be found using the cross product of the partial derivatives of the position vector \(\mathbf{r}(r, \theta) = (r \cos\theta, r \sin\theta, r^2)\). Compute \(\frac{\partial \mathbf{r}}{\partial r}\) and \(\frac{\partial \mathbf{r}}{\partial \theta}\), and take their cross product to get the normal vector.

Step 3: Calculate the Cross Product

\(\frac{\partial \mathbf{r}}{\partial r} = (\cos\theta, \sin\theta, 2r)\) and \(\frac{\partial \mathbf{r}}{\partial \theta} = (-r \sin\theta, r \cos\theta, 0)\). The cross product \(\frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta}\) yields \((-2r^2 \cos\theta, -2r^2 \sin\theta, r)\).

Step 4: Find the Magnitude of the Normal Vector

The magnitude of the normal vector \(|\mathbf{N}|\) is given by \(\sqrt{(-2r^2 \cos\theta)^2 + (-2r^2 \sin\theta)^2 + r^2}\). Simplifying, this results in \(\sqrt{4r^4 + r^2}\), which equals \(r \sqrt{4r^2 + 1}\).

Step 5: Set Up the Double Integral

The area \(A\) is given by the double integral \(A = \int_{0}^{2\pi} \int_{1}^{2} r \sqrt{4r^2 + 1} \, dr \, d\theta\), integrating \(r \sqrt{4r^2 + 1}\) over \(r\) from \(1\) to \(2\) and \(\theta\) from \(0\) to \(2\pi\).

Step 6: Perform the Integration

First, integrate with respect to \(r\): \(\int_{1}^{2} r \sqrt{4r^2 + 1} \, dr\). Use substitution, where \(u = 4r^2 + 1\), \(du = 8r \, dr\), giving \(dr = \frac{du}{8r}\). Adjust limits: when \(r = 1\), \(u = 5\), and when \(r = 2\), \(u = 17\). Perform the integration: \(\frac{1}{8} \int_{5}^{17} \sqrt{u} \, du\). The antiderivative of \(\sqrt{u}\) is \(\frac{2}{3} u^{3/2}\). Evaluate from 5 to 17: \(\frac{1}{8} \left[ \frac{2}{3} (17^{3/2} - 5^{3/2}) \right]\).

Step 7: Final Calculation

Compute \(17^{3/2}\) and \(5^{3/2}\), evaluate the expression, then multiply with \(2\pi\) after solving for \(u\). Thus, area \(A = \frac{\pi}{12} \left((17\sqrt{17}) - (5\sqrt{5}) \right)\).

Key Concepts

Parametrization

When dealing with complex surfaces such as paraboloids, it's often helpful to express the surface using parametrization. This method involves describing the surface with a set of equations that rely on parameters, rather than Cartesian coordinates. Parametrization simplifies the integration process, especially when evaluating surface areas. In the case of the paraboloid defined by the equation \(z = x^2 + y^2\) between the planes \(z = 1\) and \(z = 4\), we use a parametrization that incorporates cylindrical coordinates. These coordinates are well-suited for circular or cylindrical shapes and involve parameters like radius \(r\) and angle \(\theta\). By setting \(x = r \cos\theta\), \(y = r \sin\theta\), and \(z = r^2\), we transform the paraboloid into a more manageable form for integration.
The range for \(r\) and \(\theta\) depends on the sections of the paraboloid we are considering. Here, \(r\) ranges from 1 to 2, which corresponds to the values \(r^2\) covering the range of \(z\) from 1 to 4, and \(\theta\) spans from 0 to \(2\pi\). This transformation enables us to explore the paraboloid using polar-like coordinates while staying true to its original shape.

Double Integral

The concept of a double integral is pivotal when calculating surface areas of three-dimensional objects like paraboloids. A double integral sums up areas over a two-dimensional region, providing us with the total surface area when applied over the entire surface. In our paraboloid example, we evaluate the double integral over the region defined by \(r\) and \(\theta\), specifically integrating over \(r\) from 1 to 2 and \(\theta\) from 0 to \(2\pi\).

• To set up the double integral, we need the expression for the surface element. The integrand, in this case, involves the magnitude of the cross product of partial derivatives of the parametrization.
• Integrating with respect to \(r\) and then \(\theta\) allows us to account for the entire surface within the specified bounds.

The result from solving this double integral gives us the surface area of the specified portion of the paraboloid.

Cylindrical Coordinates

Cylindrical coordinates serve as an extension of polar coordinates into three dimensions. These coordinates are particularly suitable for surfaces that exhibit rotational symmetry or any problems involving circular boundaries, such as our paraboloid. In the cylindrical coordinate system:

• The variables \(r\), \(\theta\), and \(z\) are used, where \(r\) represents the radial distance from the z-axis.
• \(\theta\) is the angular coordinate, typically measured from a reference direction.
• \(z\) represents the height or vertical position, akin to the z-axis value in Cartesian coordinates.

Applying cylindrical coordinates to the given paraboloid problem helps in simplifying complex expressions and computations. The paraboloid \(z = x^2 + y^2\) is easily expressed as \(z = r^2\) in cylindrical coordinates. Thus, transformations become straightforward, leading to a more efficient calculation process.

Cross Product

The cross product is a mathematical operation that is central to finding the surface element in surface area integrals. It involves taking two vectors associated with the surface: the partial derivatives of the parametrization with respect to the chosen parameters. These vectors are typically tangents to the surface. In our paraboloid example, we compute the cross product of the derivatives \(\frac{\partial \mathbf{r}}{\partial r}\) and \(\frac{\partial \mathbf{r}}{\partial \theta}\).
This operation results in a new vector - the normal vector. The significance of this vector lies in its perpendicularity to the surface, which allows it to represent the infinitesimal area element necessary for integration. The computed normal vector for our surface \(\mathbf{N} = (-2r^2 \cos\theta, -2r^2 \sin\theta, r) \) encapsulates this direction, playing a crucial role in defining the integrand for our double integral.

Magnitude of a Vector

Understanding the magnitude of a vector is essential when evaluating surface areas, particularly when working with the surface element. After finding the normal vector using the cross product, its magnitude plays a key role in determining the size of the infinitesimal surface element. For our paraboloid, the magnitude of the vector \( \mathbf{N} = (-2r^2 \cos\theta, -2r^2 \sin\theta, r) \) is calculated.

• The magnitude \(|\mathbf{N}|\) simplifies the square root of the sum of squares of its components, calculated as \(\sqrt{(-2r^2 \cos\theta)^2 + (-2r^2 \sin\theta)^2 + r^2}\).
• Simplifying this expression gives \(r \sqrt{4r^2 + 1}\), highlighting the importance of handling vector components appropriately during integration.

This magnitude is the factor that gets integrated over the specified region, contributing to the final surface area calculation.