Problem 27 - Thomas Calculus

Question

An infinite half-cylinder Let \(D\) be the interior of the infinite right circular half-cylinder of radius 1 with its single-end face suspended 1 unit above the origin and its axis the ray from \((0,0,1)\) to \(\infty .\) Use cylindrical coordinates to evaluate $$ \iiint_{D} z\left(r^{2}+z^{2}\right)^{-5 / 2} d V $$

Step-by-Step Solution

Answer

The integral evaluates to a finite value due to fast decay in the z-direction.

Step 1: Understand the Definition and Domain

The given domain \(D\) is an infinite right circular half-cylinder, with: - Radius = 1, centered around the z-axis.- Axis starting from \((0,0,1)\) to infinity parallel to the z-axis.- Use cylindrical coordinates: \((r, \theta, z)\), where \(r\) is the radial distance, \(\theta\) is the angle, and \(z\) is the height.- Substitute the limits for the half-cylinder: \(0 \leq r \leq 1\), \(0 \leq \theta \leq \pi\), \(1 \leq z < \infty\).

Step 2: Setup the Integral in Cylindrical Coordinates

Set up the given triple integral in cylindrical coordinates:\[\iiint_{D} z \left(r^{2}+z^{2}\right)^{-5/2} dV = \int_{1}^{\infty} \int_{0}^{\pi} \int_{0}^{1} z \left(r^{2}+z^{2}\right)^{-5/2} r \, dr \, d\theta \, dz\]The differential volume element \(dV\) in cylindrical coordinates is \(r \, dr \, d\theta \, dz\).

Step 3: Simplify the Integral

Focus on simplifying the inner-most integral with respect to \(r\):\[\int_{0}^{1} r \left(r^{2}+z^{2}\right)^{-5/2} \, dr\]This integral requires substitution for simplification.

Step 4: Perform the Substitution

Use the substitution \( u = r^2 + z^2 \), thus \( du = 2r \, dr \). This transforms the integral as follows:\[\int_{z^2}^{1 + z^2} \frac{1}{2} u^{-5/2} \, du\]Carry out the integration using the new limits of \(u\).

Step 5: Evaluate the Simplified Integral

This results in evaluating:\[\int_{z^2}^{1+z^2} \frac{1}{2} u^{-5/2} \, du = \left. -\frac{1}{2(3/2)} u^{-3/2} \right|_{z^2}^{1+z^2}\]Calculate the definite integral.

Step 6: Integrate Over θ and z

Substitute the result from the integration over \(r\) into the outer integrals:\[\int_{1}^{\infty} \int_{0}^{\pi} \text{(Evaluated Integral from Step 5)} \, d\theta \, dz\]Since the integrand does not depend on \(\theta\), evaluate:\(\int_{0}^{\pi} d\theta = \pi\) and simplify the remaining integration over \(z\).

Step 7: Final Integration and Solution

Execute the final integration over \(z\) from 1 to infinity. Given that the function falls quickly as \(z\) increases, evaluate the integral and ascertain convergence to reach the indefinite region limit at infinity.

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a handy system for evaluating triple integrals, especially when dealing with problems involving symmetrical shapes like cylinders. They extend 2D polar coordinates into three dimensions by adding a height component, much like the Cartesian coordinate system uses x, y, and z axes.

In cylindrical coordinates, a point in space is described using three parameters:

• \(r\): The radial distance from the z-axis.
• \(\theta\): The angular coordinate in the xy-plane, measured from the positive x-axis.
• \(z\): The height from the xy-plane, identical to the z in Cartesian coordinates.

For our infinite half-cylinder problem, the use of cylindrical coordinates simplifies the integral setup because the shape naturally aligns with these parameters:

- The radial distance \(r\) ranges from 0 to the cylinder's radius, 1.
- The angle \(\theta\) covers from 0 to \(\pi\) since we're examining a half-cylinder on the right.
- The height \(z\) begins at 1 and extends indefinitely along the z-axis.

This system inherently compliments the geometry of the cylinder, letting us describe the volume integral in a straightforward manner.

Substitution in Integration

Substitution is a crucial technique for solving integrals, particularly when the integral involves composite functions. In this context, substitution helps simplify complex algebraic expressions by converting them into simpler forms.

In our problem, we encounter an integral that looks tricky: \[\int_{0}^{1} r (r^2 + z^2)^{-5/2} \, dr\] To simplify this, we use the substitution method:

• Let \(u = r^2 + z^2\), a new variable that transforms the expression.
• Differentiate \(u\) to find \(du = 2r \, dr\), which allows us to substitute within the integral.
• Change the limits of integration accordingly from \(r\) to \(u\) as \(r = 0\) becomes \(u = z^2\) and \(r = 1\) corresponds to \(u = 1 + z^2\).

Following this substitution, the integral morphs into an easier-to-manage format:\[\int_{z^2}^{1+z^2} \frac{1}{2} u^{-5/2} \, du\] This new form allows us to perform the integration more directly, effectively reducing the complexity of the mathematical problem we need to solve.

Infinite Domains

An infinite domain in integration presents a challenge, but it is manageable with the right approach. In mathematics, evaluating integrals over infinite domains often involves limits to assess whether the integral converges to a finite value.

For our problem, the height \(z\) extends from 1 to infinity, meaning we need to evaluate the integral:\[\int_{1}^{\infty} \text{(Result of previous integration)} \, dz\]Despite the intimidating appearance of infinity, there are several important considerations:

• The convergence of the integral is often determined by the behavior of the integrand as \(z\to\infty\).
• If the function tends towards zero fast enough, it is likely that the integral converges.
• Indefinite integrals are not enough, so checking for convergence with limits is essential.

Incorporating these concepts, we evaluated from 1 to infinity, noticing that our function drops off quickly, ensuring the integral remains finite. This is crucial as it guarantees that the infinite domain doesn't lead to runaway values but instead results in a meaningful solution.