91Ó°ÊÓ

The cylindrical shells method is a powerful technique to calculate the volume of solids of revolution. It is especially useful when revolving a region around an axis that is not parallel to its predominant orientation. For a region revolved around the y-axis, cylindrical shells visualize this region as a collection of thin-walled cylinders. These shells have a small thickness, denoted as \( dx \), which simplifies the integration process.

The radius of a shell is defined as the distance from the y-axis. In this example, that radius is simply \( x \). The height of each shell corresponds to the function value, \( f(x) \), at that x-coordinate. Here, the height is given by \( \sin x \).

Key steps in using cylindrical shells:

• Determine the shell's radius from the axis of rotation.
• Calculate the height of the shell from the function given.
• Combine these in the formula \( V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \),

which leads to an integral that represents the volume of these cylinders summed over the desired interval.

Integration by parts is a method applied when the integral of a product of two functions is involved. It comes from the product rule for differentiation and is particularly useful when dealing with two functions that are both non-trivial to integrate directly.

The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]

Here, we need to choose \( u \) and \( dv \) wisely to simplify the integral. In this exercise, we set \( u = x \), making \( du = dx \), and \( dv = \sin x \, dx \), leading to \( v = -\cos x \).

Substituting these into the integration by parts formula, we get:
\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx \]
While solving this, we find:

• The first term \(-x \cos x\) is evaluated,
• Then solve the simpler integral \( \int \cos x \, dx = \sin x \).

These steps effectively reduce a complex integral into simpler parts, which can then be solved to find the integral over an interval.

Definite integrals provide a way to evaluate an integral over a specific interval \([a, b]\). Calculating a definite integral involves finding the antiderivative and then systematically applying the boundaries of the interval to this antiderivative.

After employing integration by parts for our function \( x \sin x \), we have obtained the antiderivative:
\[-x \cos x + \sin x\]

When computing a definite integral, we first evaluate the antiderivative at the upper limit \( b = \pi \) and subtract the evaluation at the lower limit \( a = 0 \):
\[ \left[-x \cos x + \sin x\right]_{0}^{\pi} \]

Breaking it down further:

• Substituting \( x = \pi \), it simplifies to \( \pi \).
• Substituting \( x = 0 \), results in 0.

Thus, the final evaluation gives the area under the curve between these bounds, which is used for computing the volume.