Problem 38 - Thomas Calculus in SI Units

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\) -axis. $$y=4-x^{2}, \quad y=2-x$$

Step-by-Step Solution

Answer

The volume is \( \frac{105\pi}{3} \) cubic units.

Step 1: Identify the Intersection Points

To find the region bounded by the curves, we first determine the points of intersection by setting the equations equal to each other: \(4 - x^2 = 2 - x\). Solving for \(x\), we get \(-x^2 + x + 2 = 0\). Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the intersection points to be \(x = -1\) and \(x = 2\).

Step 2: Determine the Functions in Relation to Each Other

The curves \(y = 4 - x^2\) (a downward-facing parabola) and \(y = 2 - x\) (a straight line) intersect at \(x = -1\) and \(x = 2\). In the interval \([-1, 2]\), the parabola lies above the line.

Step 3: Set Up the Volume Integral Using the Disk Method

For the volume \(V\) generated by revolving the region between \(x = -1\) and \(x = 2\) around the x-axis, we use the disk method. The volume can be calculated as: \[ V = \pi \int_{-1}^{2} \left((4 - x^2)^2 - (2 - x)^2 \right) \, dx \].

Step 4: Evaluate the Integral

First, simplify the expression:\((4 - x^2)^2 = 16 - 8x^2 + x^4\) and \((2 - x)^2 = 4 - 4x + x^2\).Substitute back:\[ V = \pi \int_{-1}^{2} \left((16 - 8x^2 + x^4) - (4 - 4x + x^2) \right) \, dx \].Simplify the integrand:\[ V = \pi \int_{-1}^{2} (12 - 4x^2 + x^4 + 4x - x^2) \, dx \],which further simplifies to\[ V = \pi \int_{-1}^{2} (12 + 4x - 5x^2 + x^4) \, dx \].

Step 5: Integrate and Calculate the Volume

Evaluate the integral:First, find the antiderivative:\[ \int (12 + 4x - 5x^2 + x^4) \, dx = 12x + 2x^2 - \frac{5}{3}x^3 + \frac{1}{5}x^5 + C \].Evaluate from \(-1\) to \(2\):\[ V = \pi \left[(12(2) + 2(2^2) - \frac{5}{3}(2)^3 + \frac{1}{5}(2)^5) - (12(-1) + 2(-1)^2 - \frac{5}{3}(-1)^3 + \frac{1}{5}(-1)^5)\right] \].

Step 6: Final Calculation

Calculate the values:For \(x = 2\):\[ 12(2) + 2(2^2) - \frac{5}{3}(8) + \frac{1}{5}(32) = 24 + 8 - \frac{40}{3} + \frac{32}{5} \].Calculate these:\(= 32 - 13.33 + 6.4 = 25.067\).For \(x = -1\):\[ 12(-1) + 2(1) - \frac{5}{3}(-1) + \frac{1}{5}(-1) = -12 + 2 + \frac{5}{3} - \frac{1}{5} \].Calculate these:\(= -10 + 1.667 - 0.2 = -8.533 \).Finally:\[ V = \pi (25.067 + 8.533) = 105\frac{\pi}{3} \] cubic units.

Key Concepts

Disk Method

The disk method is a technique used to find the volume of a solid of revolution. When we revolve a region around an axis, we can imagine slicing the solid into thin, flat circular disks. Each disk has a small thickness and a radius determined by the distance from the axis to the function describing the boundary of the region.

• To determine the volume using the disk method, we integrate these disks over a specific interval.
• For a region under a curve revolved around the x-axis, each disk has a radius equal to the vertical distance from the x-axis to the curve, which is given by the function value.
• The volume of an individual disk with thickness \(dx\) is \(\pi [f(x)]^2 dx\).

By combining all these tiny volumes through integral calculus, we sum up to get the volume of the entire solid.
By setting up an integral from the start to the end of the interval, we can find the total volume of the solid generated.

Intersection Points

Finding intersection points is crucial when determining the region bounded by curves that is being revolved. This indicates where one curve meets or crosses the other.

• In this exercise, we set the equations equal to identify these points: \(4 - x^2 = 2 - x\).
• Solving this equation reveals the points where the parabola and the line intersect.

When we solve the quadratic equation obtained, we find the solutions \(x = -1\) and \(x = 2\). These values are the x-coordinates where the two curves intersect, defining the interval over which integration will take place.
Knowing the intersection points helps to properly set up the integral, so you include the entire region needed to accurately compute the volume of the solid.

Integral Calculus

Integral calculus is the branch of mathematics focused on finding the total sum from the accumulation of quantities, which is essential for calculating areas, volumes, and other concepts involving continuous change.

• In the disk method, we use integrals to calculate the volume by adding up infinitesimally small disks.
• We set up a definite integral to evaluate the area or volume over a specified interval from the curve's intersections.

The integral expression to compute the volume of the solid is \(\pi \int_{-1}^{2} ((4 - x^2)^2 - (2 - x)^2) \, dx\).
Through simplifying this integrand, we rewrite it as \(\pi \int_{-1}^{2} (12 + 4x - 5x^2 + x^4) \, dx\) to make the integration straightforward.
Calculus then allows us to find the antiderivative, evaluate it at the bounds, and subtract the results to find the overall volume.

Parabolas and Lines

Understanding the geometric shapes of the curves involved, such as parabolas and lines, is helpful for visualizing the region of revolution and how they interact.

• The function \(y = 4 - x^2\) represents a downward-facing parabola, opening downwards because the term \(x^2\) is negative.
• The function \(y = 2 - x\) represents a straight line with a negative slope, indicating the line is sloping downwards.

In the context of this problem, the parabola \(y = 4 - x^2\) is above the line \(y = 2 - x\) between the intersection points.
This relationship tells us which function's value to use as the outer radius when setting up the integral for the disk method.
Knowing this is essential for correctly applying the disk method and forming the integral to calculate the volume, ensuring all calculations accurately represent the region under consideration.