Problem 71 - Calculus Single Variable

Question

In each of Exercises 69-76, calculate the volume of the solid obtained when the region \(\mathcal{R}\) is rotated about the given line \(\ell\) \(\mathcal{R}\) is the region between the curve \(y=\cos (x)\) and the \(x\) -axis, \(\pi / 2 \leq x \leq 3 \pi / 2 ; \ell\) is the line \(x=3 \pi\).

Step-by-Step Solution

Answer

The volume is \( 6\pi^2 + 2\pi \).

Step 1: Understand the problem

We need to find the volume of a solid formed by rotating the region between the curve \( y = \cos(x) \) and the \( x \)-axis over the interval \( \pi/2 \leq x \leq 3\pi/2 \) around the line \( x = 3\pi \).

Step 2: Define the method

Since the rotation axis \( x = 3\pi \) is parallel to the \( y \)-axis, we will use the method of cylindrical shells. The formula for the volume \( V \) using cylindrical shells is given by \[ V = 2\pi \int_{a}^{b} (r(x))(h(x)) \, dx \]where \( r(x) \) is the distance from the axis of rotation, and \( h(x) \) is the height of the shell.

Step 3: Determine the functions

For rotation around \( x = 3\pi \), the distance (radius) from any point \( x \) to the line is \( |x - 3\pi| \). Since we consider only the section \( \pi/2 \leq x \leq 3\pi/2 \), this distance becomes \( 3\pi - x \). The height \( h(x) \) of the shell at point \( x \) is simply the value of the function \( y = \cos(x) \).

Step 4: Compute the integral

Substitute \( r(x) = 3\pi - x \) and \( h(x) = \cos(x) \) into the cylindrical shell formula.\[V = 2 \pi \int_{\pi/2}^{3\pi/2} (3\pi - x)\cos(x) \, dx\]Split the integral:\[V = 2\pi \left[ 3\pi \int_{\pi/2}^{3\pi/2} \cos(x) \, dx - \int_{\pi/2}^{3\pi/2} x \cos(x) \, dx \right]\]The first integral, \( \int \cos(x) \, dx = \sin(x) \), evaluated as:\[3\pi \left[ \sin(x) \right]_{\pi/2}^{3\pi/2} = 3\pi \left[ 0 - 0 \right] = 0\]For the second integral \( \int x \cos(x) \, dx \), we use integration by parts. Let \( u = x \), \( dv = \cos(x) \, dx \) then \( du = dx \), \( v = \sin(x) \):\[\int x \cos(x) \, dx = x \sin(x)\bigg|_{\pi/2}^{3\pi/2} - \int \sin(x) \, dx \bigg|_{\pi/2}^{3\pi/2}\]Calculating the limits,\[x \sin(x)\bigg|_{\pi/2}^{3\pi/2} = \left( 3\pi/2 \cdot -1 - \pi/2 \cdot 1 \right) = -3\pi/2 - \pi/2 = -2\pi\]For the second part, \( \int \sin(x) \, dx = -\cos(x) \), so:\[\left[-\cos(x)\right]_{\pi/2}^{3\pi/2} = \left[-(-1) - (-0)\right] = 1\]Thus the second integral evaluates as:\[-2\pi - 1 = -2\pi - 1\]Finally,\[V = 2\pi \left( 0 + 2\pi + 1 \right) = 6\pi^2 + 2\pi\]

Step 5: Conclusion

The volume of the solid obtained by rotating the region about the line \( x = 3\pi \) is \( 6\pi^2 + 2\pi \).

Key Concepts

Cylindrical Shell Method

To find the volume of a solid formed by rotating a region around a line, we often use the cylindrical shell method. This technique is particularly useful when the axis of rotation is parallel to the axis of one of the region's dimensions, in this case, the y-axis. Imagine the region being rotated as consisting of infinitely thin cylindrical shells. The main idea is to calculate the volume of each shell and sum them up using integration. For each shell, defined over the range of x, we consider:

• The radius, which measures the distance from the axis of rotation to the shell. In our scenario with the rotation line at \(x = 3\pi\), the radius is \(3\pi - x\).
• The height, represented by the function value \(h(x)\), which in our case is \(\cos(x)\).

The formula for such a solid using the cylindrical shell method is: \[ V = 2\pi \int_{a}^{b} (r(x))(h(x)) \, dx \] By visualizing the region as a stack of these shells, the approach becomes an intuitive way to compute the volume of complex shapes through rotation.

Integration by Parts

Integration by parts is a technique from calculus that helps integrate products of functions. It is based on the product rule for differentiation and is particularly helpful when dealing with integrals involving products of polynomial and trigonometric or exponential functions.The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] where \(u\) and \(dv\) are continuously differentiable functions.In our scenario, to solve the integral \(\int x \cos(x) \, dx\), we used:

• \(u = x\), making \(du = dx\).
• \(dv = \cos(x) \, dx\), thus \(v = \sin(x)\).

Applying integration by parts gives: \[ x \sin(x) - \int \sin(x) \, dx \] Calculating further allows the integral to be solved in terms of elementary functions. Using integration by parts efficiently breaks down complex integrals into simpler, more manageable pieces.

Trigonometric Functions

Trigonometric functions are foundational mathematical functions based on angles, most commonly seen as sine, cosine, and tangent functions. They are periodic and oscillate within the range -1 to 1. For regions involving trigonometric functions such as \(y = \cos(x)\), their period and symmetry properties often aid in simplifying integration processes.In our exercise, \(y = \cos(x)\) is the height of the cylindrical shells. Since this function oscillates between -1 and 1, over [\(\pi/2, 3\pi/2\)], it covers a full wave: descending from 0, hitting -1, and returning to 0.Key properties used:

• \(\cos(x) = 0\) in this region at \(x = \pi\).
• Sine derivative: \(\frac{d}{dx}\cos(x) = -\sin(x)\).

These properties help determine the function's contribution to the shape and nature of the bounding region in the problem.

Definite Integrals

Definite integrals extend the basic concept of integration by evaluating the integral's total accumulation over an interval. They help not only in finding areas under curves but also in computing volumes when combined with methods like the cylindrical shell.The process requires evaluating the integral expression between upper and lower limits. This exercise involves the integral: \[ \int_{\pi/2}^{3\pi/2}(3\pi - x)\cos(x) \, dx \] It is split into simpler parts, each solved within the limits. For example, \(\int \cos(x) \, dx = \sin(x)\) is evaluated between \(\pi/2\) and \(3\pi/2\), which simplifies the calculation to zero.Important aspects include:

• The need to correctly calculate limits for each component of the integral.
• Understanding how trigonometric identities affect the area or volume calculations.

Definite integrals offer precision in computation, enabling the accurate determination of physical properties, such as volume, of three-dimensional objects formed by rotations.