Problem 34 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the washer method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=|x|, y=2-x^{2}$$

Step-by-Step Solution

Answer

Question: Find the volume of the solid generated when revolving the region defined by the functions \(y=|x|\) and \(y=2-x^2\) around the x-axis using the washer method. Answer: \(\frac{2\pi}{3}\) cubic units.

Step 1: Sketch the curves and find the intersection points

First, plot the two given functions, \(y=|x|\) and \(y=2-x^2\). Observe where they intersect. In this case, they intersect at two points. To find those points, we can set the functions equal to each other: $$|x| = 2 - x^2$$ Since \(|x| = x\) when \(x \ge 0\) and \(|x|=-x\) when \(x<0\), we can split this into two cases: Case 1: \(x \ge 0\) \(x = 2 - x^2 \Rightarrow x^2 + x - 2 = 0\) Factoring gives \((x+2)(x-1)=0\), so we get \(x=1\) Case 2: \(x < 0\) \(-x = 2 - x^2 \Rightarrow x^2 - x - 2 = 0\) Factoring gives \((x+1)(x-2)=0\), so we get \(x=-1\) So, the region \(R\) is bounded by the curves between \(x=-1\) and \(x=1\).

Step 2: Set up an integral expression for the volume using the washer method formula

Using the washer method, we can express the volume of the solid generated when revolving the region \(R\) around the x-axis as follows: $$V = \pi \int_a^b (R^2 - r^2) dx$$ Here, \(R\) is the outer radius (from the x-axis to the curve \(y = 2 - x^2\)), and \(r\) is the inner radius (from the x-axis to the curve \(y = |x|\)). For \(-1\le x\le 0\), the outer radius is given by: $$R = 2 - x^2$$ and the inner radius is given by: $$r = -x$$ For \(0\le x\le 1\), the outer radius remains the same: $$R = 2 - x^2$$ and the inner radius now becomes: $$r = x$$ We will have to split the integral into two intervals and sum the volumes to get the total volume.

Step 3: Evaluate the integral to find the volume

Evaluate the integral for both intervals and sum them: $$V=\pi \left[\int_{-1}^0\left((2-x^2)^2-(-x)^2\right)dx + \int_0^1\left((2-x^2)^2-(x)^2\right)dx\right]$$ Now we can compute the integrals: $$V=\pi\left[\left.\left(\frac{-2}{3}x^3+x^5-\frac{1}{3}x^3\right)\right|_{-1}^{0}+\left.\left(\frac{-2}{3}x^3+x^5-\frac{1}{3}x^3\right)\right|_{0}^{1}\right]$$ Evaluating the integrals and adding them: $$V=\pi \left[\left(-\frac{-2}{3}+\frac{1}{3}\right)+\left(\frac{-2}{3}+\frac{1}{3}\right)\right]=\pi\left[\frac{1}{3}+\frac{1}{3}\right]=\frac{2\pi}{3}$$ So the volume of the solid generated when revolving the region \(R\) around the x-axis is \(\frac{2\pi}{3}\) cubic units.

Key Concepts

Volume of Solids of Revolution

When you revolve a region around an axis, a 3D solid is formed. This is a fascinating concept in calculus because it allows us to calculate the volume of such complex solids using integration. The washer method is particularly useful here. It helps us handle situations where there is a hollow region, like a donut:

• Outer Radius (R): Distance from the axis of rotation to the outer curve. In our example, this is given by the curve \(y = 2 - x^2\).
• Inner Radius (r): Distance from the axis of rotation to the inner curve. For this exercise, it is \(y = |x|\), which changes form based on whether \(x\) is positive or negative.

The washer method formula is:\[ V = \pi \int_a^b (R^2 - r^2) \, dx \] This formula calculates the volume by subtracting the volume of the inner solid (hole) from the outer solid. The rotation about the x-axis involves computing the integral over the given bounds.

Intersection of Curves

The intersection points of curves are crucial in defining the boundaries of a region. To solve our problem, we first need to determine where the two curves \(y = |x|\) and \(y = 2 - x^2\) intersect. This helps establish the limits for our integration:

• Finding Intersection Points: Set the equations equal to each other: \(|x| = 2 - x^2\). Solve this equation separately for two cases: \(x \ge 0\) and \(x < 0\).
• Simplification and Solution: For \(x \ge 0\), solve \(x^2 + x - 2 = 0\) to find \(x = 1\). For \(x < 0\), solve \(x^2 - x - 2 = 0\), leading to \(x = -1\).
• Integration Limits: The intersection points \(x = -1\) and \(x = 1\) serve as boundaries for using the washer method.

Understanding intersections is key in determining the "active" interval over which integration is possible.

Integral Calculus

Integral calculus allows us to find areas and volumes of regions defined by curves. In the washer method, integrals compute the volume of solids by accumulating infinitely small disk areas:

• Integral Setup: For this problem, we split the integral into two parts: from \(-1\) to \(0\) and from \(0\) to \(1\). Each part corresponds to one of the curves forming the inner radii.
• Solving the Integral: Calculate each integral separately:\[ \pi \int_{-1}^{0} ((2-x^2)^2 - (-x)^2) \, dx + \pi \int_{0}^{1} ((2-x^2)^2 - x^2) \, dx \]
• Full Evaluation: Plug the evaluated results back to sum the total volume of the solid. It calculates to \(\frac{2\pi}{3}\).

Integral calculus provides a powerful method to transition from 2D to 3D problems, making these calculations possible.