Problem 47 - Calculus Early Transcendentals

Question

Use a line integral to find the area of the surface. The surface that extends upward from the parabola \(y=x^{2}(0 \leq x \leq 2)\) in the \(x y\) -plane to the plane \(z=3 x\)

Step-by-Step Solution

Answer

The area of the surface is approximately 34.55.

Step 1: Understand the Problem

We are asked to find the area of the surface that is bounded by the parabolic curve in the xy-plane, extending upwards to the plane given by \( z = 3x \). The parabola given is \( y = x^2 \) for \( 0 \leq x \leq 2 \).

Step 2: Set Up the Parametric Equations

Set the parametric equations for the curve in the xy-plane. The curve can be defined parametrically as: \( \mathbf{r}(t) = \langle t, t^2, 0 \rangle \) where \( t \) ranges from 0 to 2. This represents the parabola.

Step 3: Parametrize the Surface

To describe the surface extending to the plane \( z = 3x \), modify the parameterization to include the z-component. Thus, the parameterization of the surface becomes \( \mathbf{R}(t, z) = \langle t, t^2, z \rangle \) subject to \( 0 \leq z \leq 3t \).

Step 4: Calculate the Surface Element

For the surface area, calculate the partial derivatives \( \mathbf{R}_t = \langle 1, 2t, 0 \rangle \) and \( \mathbf{R}_z = \langle 0, 0, 1 \rangle \). The cross product of these gives the normal vector \( \mathbf{R}_t \times \mathbf{R}_z = \langle 2t, -1, 0 \rangle \). The magnitude of this vector is \( \sqrt{(2t)^2 + (-1)^2 + 0^2} = \sqrt{4t^2 + 1} \).

Step 5: Set Up and Evaluate the Integral

The area of the surface is obtained by integrating the magnitude of the cross product over the specified region \( 0 \leq t \leq 2 \) and \( 0 \leq z \leq 3t \): \[ \int_{0}^{2} \int_{0}^{3t} \sqrt{4t^2 + 1} \, dz \, dt \]. First, integrate with respect to \( z \): \[ \int_{0}^{2} 3t\sqrt{4t^2+1} \, dt \].

Step 6: Solve the Final Integral

Solve the integral \( \int_{0}^{2} 3t\sqrt{4t^2+1} \, dt \). Use the substitution \( u = 4t^2 + 1 \), \( du = 8t \, dt \), yielding \[ \frac{3}{8} \int_{1}^{17} \sqrt{u} \, du \].Evaluating this gives the final area: \[ \frac{3}{12} \left[ \frac{2}{3}u^{3/2} \right]_{1}^{17} = \frac{1}{4} ( 2 \times 17^{3/2} - 2 \times 1^{3/2}) \].

Step 7: Final Computation

Compute the final result:\[ \frac{1}{4} (2 \cdot 17^{3/2} - 2) \approx \frac{1}{4} (2 \cdot 70.0967 - 2) = \frac{1}{4} \cdot 138.1934 \approx 34.5484 \].

Key Concepts

Parametric Equations

Parametric equations allow us to define a set of points on a curve or a surface using one or more parameters, rather than using explicit Cartesian coordinates. For instance, a curve in the plane can be represented by a pair of equations, each a function of the same parameter. In our exercise, the given parabola in the xy-plane, described as \( y = x^2 \), is expressed through parametric equations:

• \( x(t) = t \)
• \( y(t) = t^2 \)
• \( z(t) = 0 \)

Here, \( t \) serves as the parameter ranging from 0 to 2. Such representation is especially useful when dealing with complex curves and surfaces, as it simplifies the process of performing calculations like derivatives and integrals.

Surface Area

Finding the surface area of a parametrically defined surface involves integrating over the surface. In our example, the surface extends upwards from the parabola \( y = x^2 \) to the plane \( z = 3x \). To find the area, the surface was adjusted to include the z-component of the parametrization, resulting in\( \mathbf{R}(t, z) = \langle t, t^2, z \rangle \).

The surface element required for integration is determined by the cross product of the partial derivatives of this parameterization. This approach provides a straightforward way of computing the area of complex, non-flat surfaces, as long as the surface can be adequately parametrized in terms of one or more parameters.

Cross Product

The cross product is a mathematical operation used in vector calculus to find a vector perpendicular to two given vectors in three dimensions. In the context of finding the surface area of a parametrically defined surface, it's used to determine the normal vector to the surface.

For our parameterized surface \( \mathbf{R}(t, z) = \langle t, t^2, z \rangle \), we compute two partial derivatives:

• \( \mathbf{R}_t = \langle 1, 2t, 0 \rangle \)
• \( \mathbf{R}_z = \langle 0, 0, 1 \rangle \)

The cross product of these vectors, \( \mathbf{R}_t \times \mathbf{R}_z = \langle 2t, -1, 0 \rangle \), gives us a normal vector. Its magnitude, \( \sqrt{4t^2 + 1} \), is what we integrate over the surface to find the area. This technique effectively bridges the gap between geometry and calculus for calculating surface areas in 3D space.

Substitution Method

The substitution method is a powerful technique in calculus used to simplify integrals. It involves changing variables to transform an integral into a more manageable form. This technique was instrumental in solving the integral \( \int_{0}^{2} 3t\sqrt{4t^2+1} \, dt \) in our exercise.

By setting \( u = 4t^2 + 1 \) and \( du = 8t \, dt \), the problem simplifies, allowing us to replace the original variable with the new one. This transformation enables us to integrate using simpler limits and functions. The integral becomes \( \frac{3}{8} \int_{1}^{17} \sqrt{u} \, du \), which is much easier to evaluate.

Using substitutions in integrals aligns closely with recognizing patterns within calculus, making it a versatile and essential tool for finding solutions to otherwise complex problems.