Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(R\) be the region in the first quadrant below the curve \(y=x^{-2 / 3}\) and to the left of \(x=1\) (a) Show that the area of \(R\) is finite by finding its value. (b) Show that the volume of the solid generated by revolving \(R\) about the \(x\) -axis is infinite.

Step-by-Step Solution

Answer

(a) The area is 3 square units. (b) The volume is infinite.

Step 1: Understand the Region

First, identify the region in the first quadrant described in the problem. We have the function \( y = x^{-2/3} \), and we are considering the region below this curve and to the left of \( x = 1 \). This essentially means the region of interest is from \( x = 0 \) to \( x = 1 \) under the curve.

Step 2: Set Up the Integral for Area

The area \( A \) of the region can be expressed as an integral from \( 0 \) to \( 1 \) of the function \( y = x^{-2/3} \). Therefore, the integral is: \[ A = \int_{0}^{1} x^{-2/3} \, dx. \]

Step 3: Evaluate the Area Integral

Compute the integral \( \int_{0}^{1} x^{-2/3} \, dx \). The antiderivative of \( x^{-2/3} \) is \( \frac{x^{1/3}}{1/3} = 3x^{1/3} \). Evaluate this from 0 to 1: \[ A = \left[ 3x^{1/3} \right]_{0}^{1} = 3(1^{1/3}) - 3(0^{1/3}) = 3 \times 1 - 3 \times 0 = 3. \] Thus, the area is finite and equal to 3 square units.

Step 4: Set Up the Integral for Volume

To find the volume generated by revolving the region \( R \) around the x-axis, we use the disk method. The volume \( V \) is given by: \[ V = \pi \int_{0}^{1} (x^{-2/3})^2 \, dx. \] This simplifies to: \[ V = \pi \int_{0}^{1} x^{-4/3} \, dx. \]

Step 5: Evaluate the Volume Integral

Evaluate the integral \( \int_{0}^{1} x^{-4/3} \, dx \). The antiderivative of \( x^{-4/3} \) is \( \frac{x^{-1/3}}{-1/3} = -3x^{-1/3} \). Evaluate this from 0 to 1:\[ V = \pi \left[ -3x^{-1/3} \right]_{0}^{1} = \pi (-3(1) - (-3(\infty))) = -3\pi + \infty. \]The volume diverges to infinity because the expression \( x^{-1/3} \) approaches infinity as \( x \) approaches 0. Thus, the volume is infinite.

Key Concepts

Area Under a Curve

The area under a curve is a fundamental concept in calculus, representing the region between the graph of a function and the x-axis over a given interval. In the exercise, the curve is defined by the function \( y = x^{-2/3} \) from \( x = 0 \) to \( x = 1 \). This is essentially an integral problem where the area \( A \) is calculated using an integral:\[A = \int_{0}^{1} x^{-2/3} \, dx\]To find this measure, we evaluate the integral. The antiderivative of \( x^{-2/3} \) is \( 3x^{1/3} \). Calculating this from 0 to 1 yields the finite result:

• The area is \( 3 \) square units, indicating that the region is bounded, and the integral converges to a finite number.

The area under the curve often translates to a real-world quantity like distance traveled, signal strength over time, or any other quantity where you're summing over a continuum.

Volume of Revolution

When a region under a curve is revolved around an axis, it forms a three-dimensional solid, and the volume of this solid can be calculated using calculus. This is known as the volume of revolution. In the exercise, the region below \( y = x^{-2/3} \) from \( x = 0 \) to \( x = 1 \) is revolved about the x-axis.To determine the volume \( V \), we apply the disk method, which involves integrating the square of the function. The formula for this case is:\[V = \pi \int_{0}^{1} (x^{-2/3})^2 \, dx = \pi \int_{0}^{1} x^{-4/3} \, dx\]Calculating this integral results in:

• The antiderivative is \( -3x^{-1/3} \).
• Evaluated from 0 to 1, the integral diverges as \( x\) approaches zero.

This means the volume is infinite, demonstrating how certain shapes can extend indefinitely when revolved about an axis.

Improper Integrals

Improper integrals arise when we integrate functions over unbounded intervals or when the integrand becomes unbounded within the interval of integration. In the exercise, we deal with an improper integral when calculating the volume of revolution.The function \( x^{-4/3} \) becomes problematic as \( x \) approaches zero because \( 0^{-4/3} \) tends towards infinity. This is addressed by evaluating it as an improper integral:

• The integral \( \pi \int_{0}^{1} x^{-4/3} \, dx \) is divergent.
• The area under \( x^{-2/3} \) was finite, yet the volume calculation diverges due to the higher power of \( x \) leading to infinite limits at zero.

Improper integrals are crucial for understanding limits and convergence, particularly in areas like engineering and physics where quantities can approach infinity. Hence, they teach us not just about convergent sums, but also about the nature of mathematical infinity.