Question

Use substitution and a table of integrals to find the area of the surface generated by revolving the curve \(y=e^{x}, 0 \leq x \leq 3\), about the \(x\) -axis. (Round the answer to two decimal places.)

Short answer

The surface area is approximately 61.81 square units.

Step-by-step solution

Surface Area Formula

The formula for the surface area generated by revolving a curve \( y = f(x) \) around the x-axis from \( x = a \) to \( x = b \) is given by: \[ \text{Surface Area} = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} \, dx \] For this problem, \( f(x) = e^x \), so we need to determine \( f'(x) \) and then evaluate the integral.

Finding the Derivative

Calculate the derivative of \( y = e^x \). The derivative is \( f'(x) = \frac{d}{dx}(e^x) = e^x \). Substitute \( f(x) \) and \( f'(x) \) into the formula: \[ \text{Surface Area} = 2\pi \int_{0}^{3} e^x \sqrt{1 + (e^x)^2} \, dx \].

Simplify the Expression Under the Square Root

The expression under the square root becomes: \( 1 + (e^x)^2 = 1 + e^{2x} \). Thus, the surface area integral is: \[ \text{Surface Area} = 2\pi \int_{0}^{3} e^x \sqrt{1 + e^{2x}} \, dx \].

Apply Substitution

Use substitution to simplify the integral. Let \( u = e^x \), hence \( du = e^x \, dx \). When \( x = 0, u = e^0 = 1 \), and when \( x = 3, u = e^3 \). Thus, the integral becomes: \[ 2\pi \int_{1}^{e^3} \sqrt{1 + u^2} \, du \].

Use Table of Integrals

Find the integral of \( \sqrt{1 + u^2} \) from a table of integrals, which is typically given as \[ \int \sqrt{1 + u^2} \, du = \frac{1}{2}(u \sqrt{1 + u^2} + \ln|u + \sqrt{1 + u^2}|) + C \]. Evaluate this from \( u = 1 \) to \( u = e^3 \).

Evaluate the Definite Integral

Plug the bounds into the evaluated integral:\[ \left[ \frac{1}{2}\left(u \sqrt{1 + u^2} + \ln|u + \sqrt{1 + u^2}|\right) \right]_{1}^{e^3} \]Calculate each part:At \( u = e^3 \), it becomes: \( \frac{1}{2}(e^3 \sqrt{1 + (e^3)^2} + \ln(e^3 + \sqrt{1 + (e^3)^2})) \) At \( u = 1 \), it becomes: \( \frac{1}{2}(1 \times \sqrt{1 + 1^2} + \ln(1 + \sqrt{1 + 1^2})) \)

Calculate the Meanings

At \( u = e^3 \):\( \approx 10.979 \) At \( u = 1 \):\( \approx 1.147 \) The definite integral evaluates to \( \approx 10.979 - 1.147 = 9.832 \).

Compute the Surface Area

Multiply the result from step 7 by \( 2\pi \) to find the surface area:\[ \text{Surface Area} = 2\pi \times 9.832 \approx 61.813 \]. Round to two decimal places: \( 61.81 \).

Key concepts

curve revolution

The concept of curve revolution deals with generating a three-dimensional surface by rotating a two-dimensional curve around an axis. Simply put, imagine taking a piece of a curve, like a semi-circle, and spinning it around the x-axis. This creates a 3D shape such as a sphere or a cylinder. In mathematics, especially calculus, finding the surface area of these revolution-created shapes is a common problem.

To find this surface area, we use a specific integral formula designed for surfaces of revolution. For example, when a curve defined by the function \(y = f(x)\) is rotated around the x-axis between \(x = a\) and \(x = b\), the formula becomes:

• Surface Area = \(2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} \, dx\)

This formula accounts for the path of the curve as it spins, considering both the distance it travels and changes in direction. This step is crucial in calculus for expanding the understanding of how curves interact with space in a rotational manner.

substitution method

The substitution method is a powerful tool in integration that helps simplify complex integrals, making them easier to solve.

Think of it like changing variables to clean up a messy math problem. In our exercise, the substitution method was used to convert the integral of \( e^x \sqrt{1 + e^{2x}} \) into a simpler form by substituting \( u = e^x \). This change of variables makes the integration process more manageable.

• Initial substitution: \(u = e^x\)
• Differential change: \(du = e^x dx\)

Substituting these values transforms the complex expression, allowing for a more straightforward calculation. This method is fundamental in calculus as it breaks down otherwise challenging problems into simpler parts, often turning an intimidating calculus problem into a puzzle that is much easier to finish.

table of integrals

The table of integrals is essentially a mathematician's best friend when it comes to solving integral problems quickly and efficiently. It is a collection of common integral formulas derived from different function types.

These tables provide quick solutions that can be used directly to solve integrals without doing the integration manually each time. In our exercise, we used the table of integrals to find the integral of \(\sqrt{1 + u^2}\):

• Integral formula used: \(\int \sqrt{1 + u^2} \, du = \frac{1}{2}(u \sqrt{1 + u^2} + \ln|u + \sqrt{1 + u^2}|) + C\)

Using a table of integrals saves time and reduces errors in calculations, especially during exams or time-sensitive situations. When you're solving problems involving integration regularly, becoming familiar with these tables can significantly boost your problem-solving speed.

integration techniques

Integration techniques are strategies used to solve integrals, either definite or indefinite, that appear in calculus problems. Various approaches exist, each suited to different types of functions and their complexities.

For instance, in our exercise, we applied the substitution method and then used a formula from a table of integrals, streamlining the integration process. Here are a few common integration techniques:

• Substitution: Changing variables to simplify an expression.
• Integration by parts: Breaking an integral into parts that are simpler to solve.
• Partial fractions: Decomposing complex rational expressions into simpler fractions for easier integration.

Each technique has its unique application, and sometimes, a combination of techniques is the most efficient way to solve a problem. Mastering these techniques is essential for tackling a wide variety of calculus problems, especially those involving curves, surfaces, and complex algebraic expressions.