Problem 43 - Calculus and its applications

Question

Let \(R\) be the area between \(y=x+1\) and the \(x\) -axis over \([0, a],\) where \(a>0\) a) Find the volumes of the solids generated by rotating \(R,\) with \(a=1,\) around the \(x\) -axis and around the \(y\) -axis. Which volume is larger? b) Find the volumes of the solids generated by rotating \(R,\) with \(a=2,\) around the \(x\) -axis and around the \(y\) -axis. Which volume is larger? c) Find the value of \(a\) for which the two solids of revolution have the same volume.

Step-by-Step Solution

Answer

For a=1, the x-axis volume is larger; for a=2, the y-axis volume is larger; at a=\(\sqrt{2}\), the volumes are equal.

Step 1: Define the Area R

The area \(R\) is bounded by the line \(y = x + 1\), the x-axis, and the vertical lines \(x = 0\) and \(x = a\). For a given \(a > 0\), the area is the region under the curve from x = 0 to x = a.

Step 2: Calculate the Area for a=1

For \(a=1\), the area under the curve from \(x=0\) to \(x=1\) is \[ \int_0^1 (x+1) \, dx = \left[ \frac{x^2}{2} + x \right]_0^1 = \frac{1}{2} + 1 = \frac{3}{2}. \] This is the area that will be revolved to find the volume.

Step 3: Volume around the x-axis for a=1

Using the disk method, the volume of rotation around the x-axis is given by \[ V_x = \pi \int_0^1 (x+1)^2 \, dx. \] Compute the integral:\[ \pi \int_0^1 (x^2 + 2x + 1) \, dx = \pi \left[ \frac{x^3}{3} + x^2 + x \right]_0^1 = \pi \left( \frac{1}{3} + 1 + 1 \right) = \pi \cdot \frac{7}{3}. \]

Step 4: Volume around the y-axis for a=1

Using the shell method, the volume of rotation around the y-axis is given by \[ V_y = 2\pi \int_0^1 x(x+1) \, dx = 2\pi \int_0^1 (x^2 + x) \, dx. \] Compute the integral:\[ 2\pi \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_0^1 = 2\pi \left( \frac{1}{3} + \frac{1}{2} \right) = 2\pi \cdot \frac{5}{6}. \]

Step 5: Compare Volumes for a=1

The volume around the x-axis, \(\pi \cdot \frac{7}{3}\), is larger than the volume around the y-axis, \(2\pi \cdot \frac{5}{6}\). Therefore, the solid rotated around the x-axis has a greater volume when \(a=1\).

Step 6: Calculate the Area for a=2

For \(a=2\), the area under the curve is calculated as\[ \int_0^2 (x+1) \, dx = \left[ \frac{x^2}{2} + x \right]_0^2 = 2+2 = 4. \] This area will be revolved to find the volume.

Step 7: Volume around the x-axis for a=2

The volume around the x-axis is calculated using\[ V_x = \pi \int_0^2 (x+1)^2 \, dx. \] Compute the integral:\[ \pi \int_0^2 (x^2 + 2x + 1) \, dx = \pi \left[ \frac{x^3}{3} + x^2 + x \right]_0^2 = \pi \left( \frac{8}{3} + 4 + 2 \right) = \pi \cdot \frac{26}{3}. \]

Step 8: Volume around the y-axis for a=2

For rotation around the y-axis,\[ V_y = 2\pi \int_0^2 x(x+1) \, dx = 2\pi \int_0^2 (x^2 + x) \, dx. \] Compute the integral:\[ 2\pi \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_0^2 = 2\pi \left( \frac{8}{3} + 2 \right) = 2\pi \cdot \frac{14}{3}. \]

Step 9: Compare Volumes for a=2

The volume around the y-axis, \(2\pi \cdot \frac{14}{3}\), is larger than the volume around the x-axis, \(\pi \cdot \frac{26}{3}\). Therefore, the solid rotated around the y-axis has a greater volume when \(a=2\).

Step 10: Find a for Equal Volumes

To find \(a\) such that volumes are equal, equate the expressions for volume:\[ \pi \int_0^a (x+1)^2 \, dx = 2\pi \int_0^a x(x+1) \, dx. \]Simplify:\[ \pi \left[ \frac{a^3}{3} + a^2 + a \right] = 2\pi \left[ \frac{a^3}{3} + \frac{a^2}{2} \right]. \]Solve for \(a\):\[ a^2 + a = a^2 + a, \]which implies\[ \frac{(2a - a^3)}{6} = 0. \]Thus, \(a = \sqrt{2}\).

Key Concepts

Solids of Revolution

When you rotate a two-dimensional region about an axis, you create a three-dimensional shape known as a solid of revolution. Imagine it like spinning a region around a line to create a 3D shape. This concept is especially useful in calculus for calculating volumes of complex shapes formed by simple rotations.

• Think of a flat piece of dough that you spin to create a doughnut shape.
• The axis of rotation is the "spit" about which you revolve your shape.

Creating these solids often requires the use of calculus, as it allows us to find the volumes using integration techniques. Calculating such volumes precisely is important in fields like engineering and physics, where such shapes are commonly encountered.

Disk Method

The disk method is a technique used to find the volume of a solid of revolution when rotating around an axis, typically the x-axis. Imagine slicing your solid into thin, pancake-like shapes called disks.

• These disks are perpendicular to the axis of rotation.
• Each disk's volume can be calculated as that of a cylinder: \( \pi r^2 h \), where \(r\) is the radius and \(h\) is the disk's thickness or height, essentially a small change in x, denoted as \(dx\).

To find the total volume:

1. Determine the radius of a typical disk, which could be a function of x.
2. Set up the integral of all these disks from the start to end of the region.
3. The formula becomes \( V = \pi \int_a^b [f(x)]^2 \, dx\).

This method works best when every cross-section perpendicular to the axis is circular.

Shell Method

The shell method is another way to calculate the volume of a solid of revolution, but it's especially handy for revolving around the y-axis. Instead of disks, think of the solid being composed of cylindrical shells.

• Picture this as hollow tubes stacked next to each other, which capture the volume by adding the layers.
• These shells are parallel to the axis of rotation.

Here's how you use the shell method:

1. Find the radius of a shell, which is usually the x-coordinate of the function.
2. Determine the height, representing the length of the shell segment intersected by the region.
3. Use the integral \( V = 2 \pi \int_a^b x[f(x)] \, dx \) to calculate the total volume.

The shell method shines in situations where it's more convenient to integrate with respect to x for rotation around the y-axis.

Integration Techniques

Integration techniques are the methods used to solve integrals, which are essential for finding the volume of solids of revolution.
With both the disk and shell methods, integrals calculate the sum of infinite slices or layers to derive a whole. For instance, computing the area under a curve or finding a broader concept like volumes require adeptness in these techniques.
Here are some integration approaches:

• Substitution Method: Useful for simplifying integrals, by introducing a substitution to transform a complex integral into a simpler form.
• Integration by Parts: Similar to the product rule in differentiation, and used when products of functions are involved.
• Partial Fractions: Breaks down complex rational functions into simpler fractions to make integration easier.

Mastering these techniques is crucial for performing accurate and efficient integral calculations in the context of finding volumes and other applications of calculus.