Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid generated by revolving about the \(x\)-axis the region bounded by the line \(y=6 x\) and the parabola \(y=6 x^{2}\).

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{6\pi}{5}\).

Step 1: Sketch the Region and Identify the Bounds

To find the volume of the solid of revolution, first sketch the region bounded by the two curves: the line \(y = 6x\) and the parabola \(y = 6x^2\). The bounds are determined where these two curves intersect. Set the equations equal: \(6x = 6x^2\), which simplifies to \(x(6 - 6x) = 0\). This yields \(x = 0\) and \(x = 1\). The region of interest is thus between \(x=0\) and \(x=1\).

Step 2: Set Up the Volume Integral

Utilize the disk method to set up the integral for the volume of the solid obtained by revolving the region around the x-axis. The radius of the disks is the difference in function values: \(R(x) = 6x - 6x^2\). The volume \(V\) is given by the integral:\[ V = \pi \int_0^1 (6x - 6x^2)^2 \, dx \]

Step 3: Expand and Simplify the Integrand

Expand the integrand \((6x - 6x^2)^2\):\[(6x - 6x^2)^2 = (36x^2 - 72x^3 + 36x^4)\].Now the integral becomes:\[ V = \pi \int_0^1 (36x^2 - 72x^3 + 36x^4) \, dx \]

Step 4: Evaluate the Integral

Integrate term by term:\[ \pi \left[ \int_0^1 36x^2 \, dx - \int_0^1 72x^3 \, dx + \int_0^1 36x^4 \, dx \right] \]Solving each:- \(\int_0^1 36x^2 \, dx = 36 \cdot \frac{x^3}{3} \bigg|_0^1 = 12\)- \(\int_0^1 72x^3 \, dx = 72 \cdot \frac{x^4}{4} \bigg|_0^1 = 18\)- \(\int_0^1 36x^4 \, dx = 36 \cdot \frac{x^5}{5} \bigg|_0^1 = \frac{36}{5}\)Putting these together, we get:\[ V = \pi (12 - 18 + \frac{36}{5}) = \pi \left(\frac{60}{5} - \frac{90}{5} + \frac{36}{5}\right) = \pi \left(\frac{6}{5}\right) \]

Step 5: Final Calculation

Calculate the final volume by plugging in the simplified integral result:\[ V = \pi \left(\frac{6}{5}\right) = \frac{6\pi}{5} \]

Key Concepts

Disk Method

The Disk Method is an intuitive approach in Integral Calculus used to find the volume of a solid of revolution. This is especially useful when a region is revolved around the x-axis, creating a "stack" of disks or circles. Think about it as adding up the volumes of infinitely many slices.To understand this, imagine stacking a bunch of thin, flat cylinders (or disks) vertically. By taking each of these disks and calculating their contributions to the overall volume, you can determine the total volume of the solid. Each disk's "radius" comes from the value of the function at a given point, since that determines how far it reaches out from the axis you're revolving around.

• The radius of the disk, if the solid is revolved around the x-axis, is given by the y-value of the function at that point.
• The thickness of each disk is an infinitesimally small change in x, denoted as dx.

Using the formula for the area of a circle, \(\pi r^2\), the volume of a single disk becomes \(\pi f(x)^2 dx\). The total volume is then the integral of these small volumes, evaluated over the range bound by the intersection points of the curves.

Integral Calculus

Integral Calculus, a cornerstone of calculus, is essential for understanding how quantities accumulate. In problems of volume of solids of revolution, like the one described here, integrals allow us to add up infinitely many infinitesimally small pieces to get the whole. In this case, we're adding up the volumes of infinitesimally thin disks to get the total volume of the solid. The process involves setting up an integral, which you can think of as a sum, but a very particular kind of sum called an "integral." Once your integral is set up—accounting for the function's shape over its interval—you evaluate it to find the total volume.

• To find the integral, you must identify the bounds, which are the points of intersection between the curves.
• You must apply integration rules to solve the integral, such as breaking it into simpler parts if necessary.

Integration can initially seem daunting due to its complexity, but it is incredibly powerful for calculating areas, volumes, and other quantities that accumulate continuously.

Intersection of Curves

Finding the intersection of curves is crucial in problems involving the volume of solids of revolution. This tells you where one curve meets another, essentially defining the bounds of the region you are looking to revolve.In this exercise, you dealt with two functions: a line, \(y = 6x\), and a parabola, \(y = 6x^2\). To find their intersection points, you equate the two functions:\[ 6x = 6x^2 \] This equation simplifies to \(x(6 - 6x) = 0\), resulting in the solutions \(x = 0\) and \(x = 1\). These are the x-values where the two curves intersect, setting the boundaries for the region of integration.

• Intersection points define the limits of your integration, as these are where the curves start and stop bounding your area.
• Accurately finding these points ensures that the calculated volume correctly represents the space between the curves.

Understanding how and why these intersections are found helps to set accurate boundaries for your integral, making the entire process of finding volumes more comprehensible.