Problem 23 - Thomas Calculus

Question

Evaluate the cylindrical coordinate integrals in Exercises \(23-28\) $$ \int_{0}^{2 \pi} \int_{0}^{1} \int_{r}^{\sqrt{2-r^{2}}} d z r d r d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \( \pi \sqrt{2} - \pi \).

Step 1: Understand the Integral Limits

We have the integral \[ \int_{0}^{2 \pi} \int_{0}^{1} \int_{r}^{\sqrt{2-r^{2}}} d z \ r \ d r \ d \theta \] in cylindrical coordinates with limits for \(z\) from \(r\) to \(\sqrt{2-r^2}\), \(r\) from 0 to 1, and \(\theta\) from 0 to \(2\pi\). The order of integration is \(z\), \(r\), \(\theta\).

Step 2: Integrate with respect to z

First, integrate with respect to \(z\):\[ \int_{r}^{\sqrt{2-r^2}} d z = \left[ z \right]_{r}^{\sqrt{2-r^2}} = \sqrt{2 - r^2} - r.\]

Step 3: Setup for Integration with respect to r

Now substitute the \(z\) integral result into the radial integral:\[ \int_{0}^{1} r(\sqrt{2-r^2} - r) d r.\] Here, the expression inside the integral simplifies to a function of \(r\) that we must now integrate.

Step 4: Breakdown the inner integral

Separate the terms for easier integration:\[ \int_{0}^{1} (r\sqrt{2-r^2} - r^2) d r.\] This gives two integrals:\[ \int_{0}^{1} r\sqrt{2-r^2} \, d r - \int_{0}^{1} r^2 \, d r.\]

Step 5: Evaluate the integral \( \int_{0}^{1} r\sqrt{2-r^2} \, d r \)

Let \[ u = 2 - r^2 \Rightarrow du = -2r \, dr \Rightarrow -\frac{1}{2}du = r \, dr. \]Change the limits for \(u\) from 2 when \(r=0\) to 1 when \(r=1\):\[ -\frac{1}{2} \int_{2}^{1} \sqrt{u} \, du = \frac{1}{2} \int_{1}^{2} \sqrt{u} \, du \]Solve the integral:\[ \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} = \frac{1}{2} \left( \frac{2}{3} (2^{3/2}) - \frac{2}{3} \right). \]

Step 6: Evaluate the integral \( \int_{0}^{1} r^2 \, d r \)

This is a straightforward integral:\[ \int_{0}^{1} r^2 \, d r = \left[ \frac{r^3}{3} \right]_{0}^{1} = \frac{1}{3}. \]

Step 7: Evaluate and Simplify

Substitute the evaluated integrals back and simplify:\[ \int_{0}^{1} r\sqrt{2-r^2} \, d r - \int_{0}^{1} r^2 \, d r = \left( \frac{1}{2} \cdot \frac{2}{3} (2\sqrt{2} - 1) \right) - \frac{1}{3} \] This computes to a real number after performing the calculations.

Step 8: Integrate with respect to \( \theta \)

Our expressions simplify to a constant after integrating over \(r\). Now, integrate with respect to \(\theta\):\[ \int_{0}^{2\pi} d \theta . \]Knowing the above simplifies to a constant result \(C\):\[ C\cdot \theta \bigg|_0^{2\pi} = C \cdot 2\pi. \]

Step 9: Compute the Final Result

Insert the computed value of \(C\) from the radial integral, multiplying by \(2\pi\) to get the total value of the complete integral.

Key Concepts

Multiple Integrals: Capturing Dimensions

When dealing with several variables and dimensions in calculus, multiple integrals allow us to explore spaces that are beyond the traditional one-dimensional integrals. In the example provided, we are using a triple integral. This means we're integrating over three different variables: \(z\), \(r\), and \(\theta\).
This approach lets us evaluate computations over three-dimensional regions, encompassing volumes.
Using cylindrical coordinates—ideal for shapes like cylinders or circular formations—simplifies the computation by aligning with natural symmetry.
The order in which these integrals are computed matters, and it's crucial to follow it precisely, as skipping or altering the order can lead to different, potentially incorrect results.

• Triple integrals expand integration concepts into three dimensions.
• Cylindrical coordinates effectively handle circular and cylindrical shapes.
• The order of integration must be followed specifically.

Integration Limits: Foundation of Accuracy

Every integral is confined within specific limits which determine the bounds of integration. These limits play a significant role in the evaluation and accuracy of the solution, as they delineate where the integration occurs.
In the exercise you examined, the limits are

• \(\theta: 0 \text{ to } 2\pi\)
• \(r: 0 \text{ to } 1\)
• \(z: r \text{ to } \sqrt{2-r^2}\)

These values are carefully selected to fit the shape and region of study, which is standard in conversion from one coordinate system to another, ensuring the integral assesses the intended region.
Integration limits directly impact the outcome, so selecting the appropriate bounds is integral to reaching the correct solution. Correct understanding and application of these limits determine the bounds of where and how the function is evaluated within the space.

Coordinate Transformation: The Shift Makes it Simple

Transforming the coordinates simplifies otherwise challenging integrals, particularly in cases involving non-linear shapes. This coordinate transformation lets us switch to a more suitable system, like cylindrical coordinates, where the problem is more naturally expressed, and solution strategies are clearer.
Switching to cylindrical coordinates utilizes \(r\) (radius), \(\theta\) (angle), and \(z\) (height), making this system highly effective for problems involving circular or cylindrical shapes.

• Why Transform: Aligns the problem with coordinate symmetry.
• How it Works: Converts rectangular coordinates \((x, y, z)\) to cylindrical \((r, \theta, z)\).
• Benefits: Reduces complexity and exploits rotational symmetry.

This step is not just about conversion—it's about simplifying the integral to make it more manageable, ensuring the computation effectively captures the physical or geometric properties of the space being studied.

Integral Calculus: Bringing it All Together

Integral calculus allows us to accumulate quantities that are distributed continuously over a region or space. It is fundamentally about understanding the "sum" over continuous dimensions.
In our specific exercise, integral calculus helps in finding the volume under a curved surface within the bounds specified by the cylindrical coordinates.
Here,

• We first integrate over \(z\), capturing height variation.
• Next, we integrate over the radial dimension \(r\), handling distance from the origin.
• Ultimately, we handle the angular portion \(\theta\).

Each step necessitates careful application of calculus principles to ensure that we sum the infinitesimally small changes effectively over the defined region.
This method is powerful and widely applicable, enabling us to solve real-world problems involving volumes, areas, masses, and other quantities in multi-dimensional spaces.