Problem 4 - Thomas Calculus

Question

Evaluate the cylindrical coordinate integrals. \(\int_{0}^{\pi} \int_{0}^{\theta / \pi} \int_{-\sqrt{4-r^{2}}}^{3 \sqrt{4-r^{2}}} z d z r d r d \theta\)

Step-by-Step Solution

Answer

The value of the integral is \(\frac{157\pi}{60}\).

Step 1: Understand the Limits of Integration

The problem asks us to evaluate a triple integral in cylindrical coordinates with the integration order \(z, r, heta\). The limits given are: \(z\) from \(-\sqrt{4-r^2}\) to \(3\sqrt{4-r^2}\), \(r\) from \(0\) to \(\frac{\theta}{\pi}\), and \(\theta\) from \(0\) to \(\pi\). The variable \(z\) integration depends on \(r\), and \(r\) depends on \(\theta\).

Step 2: Integrate with respect to \(z\)

The inner integral is with respect to \(z\): \[ \int_{-\sqrt{4-r^2}}^{3\sqrt{4-r^2}} z \, dz \] Evaluating this integral, we get: \[ \left[ \frac{z^2}{2} \right]_{-\sqrt{4-r^2}}^{3\sqrt{4-r^2}} = \frac{(3\sqrt{4-r^2})^2}{2} - \frac{(-\sqrt{4-r^2})^2}{2}\] Simplifying, we obtain: \[ \frac{9(4-r^2)}{2} - \frac{4-r^2}{2} = \frac{36-9r^2 - 4 + r^2}{2} = \frac{32-8r^2}{2} = 16-4r^2 \]

Step 3: Integrate with respect to \(r\)

Next, integrate with respect to \(r\): \[ \int_0^{\frac{\theta}{\pi}} (16 - 4r^2) r \, dr \] Firstly, distribute \(r\) and rewrite the integral as: \[ 16\int_0^{\frac{\theta}{\pi}} r \, dr - 4\int_0^{\frac{\theta}{\pi}} r^3 dr \] Evaluating these, we get: \[ 16\left[ \frac{r^2}{2} \right]_0^{\frac{\theta}{\pi}} - 4\left[ \frac{r^4}{4} \right]_0^{\frac{\theta}{\pi}} \]\[ = 16\left( \frac{\theta^2}{2\pi^2} \right) - \frac{\theta^4}{4\pi^4} \]Further simplification gives: \[ = \frac{8\theta^2}{\pi^2} - \frac{\theta^4}{4\pi^4} \]

Step 4: Integrate with respect to \(\theta\)

Finally, integrate with respect to \(\theta\): \[ \int_0^{\pi} \left( \frac{8\theta^2}{\pi^2} - \frac{\theta^4}{4\pi^4} \right) d\theta \] Evaluate each part separately:For \(\frac{8\theta^2}{\pi^2}\): \[ \frac{8}{\pi^2} \int_0^{\pi} \theta^2 d\theta = \frac{8}{\pi^2} \left[ \frac{\theta^3}{3} \right]_0^{\pi} = \frac{8}{\pi^2} \cdot \frac{\pi^3}{3} = \frac{8\pi}{3} \]For \(\frac{-\theta^4}{4\pi^4}\): \[ \frac{-1}{4\pi^4} \int_0^{\pi} \theta^4 d\theta = \frac{-1}{4\pi^4} \left[ \frac{\theta^5}{5} \right]_0^{\pi} = \frac{-1}{4\pi^4} \cdot \frac{\pi^5}{5} = \frac{-\pi}{20} \]Combine the results: \[ \frac{8\pi}{3} - \frac{\pi}{20} = \frac{160\pi}{60} - \frac{3\pi}{60} = \frac{157\pi}{60} \]

Step 5: Conclusion: Calculate the Value of the Integral

After evaluating the integrals at each step, the final result for the given triple integral is \(\frac{157\pi}{60}\).

Key Concepts

Triple Integral

A triple integral allows us to evaluate a function over a three-dimensional space. In this exercise, we are integrating in cylindrical coordinates, which is often useful when dealing with problems that exhibit certain types of symmetry, like rotational symmetry around an axis.

Cylindrical coordinates are an extension of polar coordinates (used for two dimensions) into three dimensions. These coordinates use three variables:

• \( r \): the radial distance from the vertical z-axis.
• \( \theta \): the angle measured counter-clockwise from the positive x-axis.
• \( z \): the height above the xy-plane.

In the given integral \[ \int_{0}^{\pi} \int_{0}^{\theta / \pi} \int_{-\sqrt{4-r^{2}}}^{3\sqrt{4-r^{2}}} z \, dz \, r \, dr \, d\theta \] we are integrating a function \( f(r, \theta, z) = z \) through a volume where \( z \), \( r \), and \( \theta \) each have prescribed limits of integration. This allows us to evaluate how the function behaves in the defined space.

Limits of Integration

Understanding the limits of integration in a triple integral is crucial as it defines the region over which you're integrating. Here, the limits are given for each of the variables:

• \( z \): from \(-\sqrt{4-r^2}\) to \(3\sqrt{4-r^2}\)
• \( r \): from \(0\) to \(\theta/\pi\)
• \( \theta \): from \(0\) to \(\pi\)

These limits define a complex region in the three-dimensional space where the function \( z \) is limited by \( r \), and \( r \) is further limited by \( \theta \).

The limits of integration can sometimes depend on one another, as seen here with \( z \) depending on \( r \), and \( r \) depending on \( \theta \). This nesting is quite common in cylindrical coordinates as it reflects the geometric relationship within the system you are modeling.

Integration Order

The order of integration in triple integrals significantly affects how you set up and solve the problem. In this case, the order is \( z, r, \theta \). This means:

• First, integrate with respect to \( z \), while keeping \( r \) and \( \theta \) constant.
• Next, integrate with respect to \( r \), with \( \theta \) constant, but after the first integral.
• Finally, integrate with respect to \( \theta \).

Each step needs to be done sequentially, as the result of one integral becomes the function for the next.

Changing the order of integration could complicate the calculation or may even make it impossible without reevaluating the limits. That's why choosing the most efficient order, as given, is key to simplifying the task.

Mathematical Simplification

Mathematical simplification refers to reducing an expression to its simplest form. During integration, simplification can help avoid complicated expressions that could lead to calculation errors or difficulty.

• After integrating with respect to \( z \), the result \( 16 - 4r^2 \) is simplified by evaluating definite integrals and algebraically simplifying expressions.
• While integrating with respect to \( r \), you rewrite \((16 - 4r^2)r\) as two separate integrals to simplify their evaluation.
• Finally, combining results from independent parts and simplifying using common denominators helps in getting to the final answer: \(\frac{157\pi}{60}\).

Simplification often involves basic algebraic techniques like factoring, combining like terms, and recognizing patterns. All these make step-by-step computation easier and more manageable.