Problem 10 - University Calculus: Early Transcendentals

Question

The integrals we have seen so far suggest that there are preferred orders of integration for cylindrical coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integrals. $$\int_{0}^{2} \int_{r-2}^{\sqrt{4-r^{2}}} \int_{0}^{2 \pi}(r \sin \theta+1) r d \theta d z d r$$

Step-by-Step Solution

Answer

The integral evaluates to \( 8\pi \).

Step 1: Analyze the Integral and Limits

The given integral is in cylindrical coordinates: \( \int_{0}^{2} \int_{r-2}^{\sqrt{4-r^{2}}} \int_{0}^{2 \pi}(r \sin \theta+1) r d \theta d z d r \). The order of integration is \( d\theta \), \( dz \), \( dr \). The bounds are: \( r \) from 0 to 2, \( z \) from \( r-2 \) to \( \sqrt{4-r^{2}} \), and \( \theta \) from 0 to \( 2\pi \).

Step 2: Integrate with Respect to \( \theta \)

For the innermost integral \( \int_{0}^{2 \pi} (r \sin \theta + 1) r \,d\theta \), split it into two integrals: \( r^2 \int_{0}^{2\pi} \sin \theta \, d\theta + r\int_{0}^{2\pi} 1 \,d\theta \). The integral of \( \sin \theta \) over a full period is 0, so it's 0. The second integral is \( r(2\pi) \), so the result is \( 2\pi r \).

Step 3: Integrate with Respect to \( z \)

Now, evaluate the integral \( \int_{r-2}^{\sqrt{4-r^2}} 2\pi r \, dz \). Here, \( 2\pi r \) is a constant with respect to \( z \), so the integral is \( 2\pi r (\sqrt{4-r^2} - (r-2)) \). This simplifies to \( 2\pi r(\sqrt{4-r^2} - r + 2) \).

Step 4: Integrate with Respect to \( r \)

The outer integral is \( \int_{0}^{2} 2\pi r(\sqrt{4-r^2} - r + 2) \, dr \). Distribute and integrate each term: 1. \( 2\pi \int_{0}^{2} r\sqrt{4-r^2} \, dr \)2. \(-2\pi \int_{0}^{2} r^2 \, dr \)3. \( 4\pi \int_{0}^{2} r \, dr \).Evaluate each separately for a total result.

Step 5: Evaluate Each Integral Separately

1. Let \( u = 4 - r^2 \), \( du = -2r \,dr \), changing limits from 0 to 4, so it's \( -\pi \int_{4}^{0} \, u^{1/2} \, du = \pi \int_{0}^{4} \sqrt{u} \, du = \pi \left[ \frac{2}{3}u^{3/2} \right]_{0}^{4} = \frac{8\pi}{3} \).2. \( -2\pi \int_{0}^{2} r^2 \, dr = -2\pi \left[ \frac{r^3}{3} \right]_{0}^{2} = -\frac{16\pi}{3} \).3. \( 4\pi \int_{0}^{2} r \, dr = 4\pi \left[ \frac{r^2}{2} \right]_{0}^{2} = 8\pi \).Sum the results: \( \frac{8\pi}{3} - \frac{16\pi}{3} + 8\pi = \frac{24\pi}{3} = 8\pi \).

Step 6: Conclusion - Combine Results

Add all the integrals together: the total after simplifying all terms is \( 8\pi \). This is the evaluated value of the given integral over the given limits in cylindrical coordinates.

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height component, typically denoted by \( z \). They are particularly useful in problems involving symmetry around an axis, such as cylindrical shapes. In cylindrical coordinates, any point in space can be described using three parameters: \( r \), \( \theta \), and \( z \):

• \( r \) represents the radial distance from the z-axis.
• \( \theta \) is the angle in the xy-plane measured from the positive x-axis.
• \( z \) is the height above the xy-plane, similar to the Cartesian z-coordinate.

When converting an integral from Cartesian coordinates to cylindrical coordinates, the element of volume \( dx \ dy \ dz \) changes to \( r \ dr \ d\theta \ dz \). This is because the projection on the xy-plane forms a circle, and the \( r \) term accounts for the distance from the origin, which influences the volume element size. Thus, when setting up an integral in cylindrical coordinates, don’t forget to include an extra \( r \) term, often referred to as the Jacobian of the transformation.

Integration Techniques

Integration in cylindrical coordinates often involves tricky regions or boundaries that make the initial boundary reading crucial. Here, let's consider two main techniques:

• Substitution: This technique simplifies the integration by transforming the variable. In this example, when evaluating the integral with respect to \( r \), substitution was used to transform the expression \( r\sqrt{4-r^2} \). Letting \( u = 4 - r^2 \), reduced the problem to a simpler integral involving \( u^{1/2} \).


• Splitting the integral: When dealing with expressions such as \( (r \sin \theta + 1) r \), it's often practical to split it into two separate integrals. Each part can then be integrated over the same limits, which simplifies evaluation and recognition of symmetry, such as the integral over a full period of \( \sin \theta \) being zero.

By mastering these techniques, one can efficiently solve complex integrals, even when they initially appear daunting.

Order of Integration

The order of integration is essential when solving multiple integrals because it affects the complexity of the integration and whether the boundaries can be easily determined or not. Here’s why the preferred order is often chosen:

• In the original problem, the integral was \( \int d\theta \ dz \ dr \). Such an order reduces the expression to simpler forms step-by-step. First integrating with respect to \( \theta \), which usually leads to familiar trigonometric integrals, helps simplify the expression quickly.


• Changing the order to integrate with respect to \( z \) or \( r \) first can complicate the integration as the limits for \( z \) depend on \( r \). Therefore, the chosen order facilitates an easier step-wise simplification of the integration.


• Adjusting the order of integration might involve different integral limits but can be advantageous if it simplifies the calculations. Thus, understanding the geometrical shape and its symmetry can dictate the best integration order.

Experimenting with different orders of integration can sometimes lead to more straightforward calculations, though solving it in unconventional orders requires careful evaluation of changed limits.