Problem 1 - Thomas Calculus

Question

Evaluate the cylindrical coordinate integrals. \(\int_{0}^{2 \pi} \int_{0}^{1} \int_{r}^{\sqrt{2-r^{2}}} d z r d r d \theta\)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{4\pi}{3} (\sqrt{2} - 1) \).

Step 1: Set up the integral

The given integral is evaluated in cylindrical coordinates. The limits for \( \theta \) are from 0 to \( 2\pi \), for \( r \) are from 0 to 1, and for \( z \) are from \( r \) to \( \sqrt{2-r^2} \). The integrand is simply 1 (implied inside the structure), so we are integrating regarding the region defined by these limits.

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

Compute the first integral by integrating with respect to \( z \): \[ \int_{r}^{\sqrt{2-r^{2}}} d z = [z]_{r}^{\sqrt{2-r^{2}}} = (\sqrt{2-r^{2}} - r). \] Substitute these limits to complete the integral.

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

The integral now becomes \[ \int_{0}^{1} (\sqrt{2-r^{2}} - r) \cdot r \, d r. \] Distribute \( r \) to get \[ \int_{0}^{1} (r\sqrt{2-r^{2}} - r^{2}) \, d r. \] This requires us to evaluate two separate integrals.

Step 4: Evaluate \( \int_{0}^{1} r \sqrt{2-r^{2}} \, d r \)

We use the substitution \( u = 2 - r^2 \), \( du = -2r \, dr \). So, \( r \, dr = -\frac{1}{2} \, du \) and the limits change from \( r = 0 \) to \( u = 2 \), and \( r = 1 \) to \( u = 1 \). The integral becomes \[ \int_{2}^{1} \sqrt{u} \left( -\frac{1}{2} \right) \, du = \frac{1}{2} \int_{1}^{2} u^{1/2} \, du. \] Evaluate it to yield \[ \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{1}^{2} = \frac{1}{3} [(2^{3/2} - 1^{3/2})]. \] Calculate to find \( \frac{1}{3} (2\sqrt{2} - 1) \).

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

Integrate \( r^2 \) to find: \[ \int_{0}^{1} r^2 \, d r = \left[ \frac{r^3}{3} \right]_{0}^{1} = \frac{1}{3} - 0 = \frac{1}{3}. \]

Step 6: Combine results of integrals over \( r \)

Combining the results from Step 4 and 5, we have the integral over \( r \) as: \( \frac{1}{3}(2\sqrt{2} - 1) - \frac{1}{3} \). Simplify to get: \( \frac{1}{3}(2\sqrt{2} - 2) = \frac{2}{3}(\sqrt{2} - 1) \).

Step 7: Integrate with respect to \( \theta \)

Perform the final integration with respect to \( \theta \): \[ \int_{0}^{2\pi} \frac{2}{3} (\sqrt{2} - 1) \, d\theta. \] This results in \[ \frac{2}{3} (\sqrt{2} - 1) \int_{0}^{2\pi} \, d\theta = \frac{2}{3} (\sqrt{2} - 1) \cdot 2\pi = \frac{4\pi}{3} (\sqrt{2} - 1). \]

Key Concepts

Triple Integral

When we talk about a triple integral, we're exploring the integration of a function over a three-dimensional region. This process allows us to find volumes, as well as other properties like mass, when density is part of the integrand. A triple integral involves integrating three times, one for each dimension of the space. In this particular exercise, we have a function that, technically, is simply 1. This implies that we are purely calculating the volume of the region determined by our bounds in cylindrical coordinates.

The structure of the triple integral is set as:

• First, integrate with respect to one variable while treating the others as constants.
• Next, integrate the resulting expression with respect to the second variable.
• Finally, integrate the last result with regard to the third variable.

This sequential operation enables us to deal with each independent direction in the three-dimensional space. For this specific problem, we sequentially integrate over the coordinates \( z \), \( r \), and \( \theta \). Understanding the order and the limits of integration is key to effectively solving a triple integral.

Cylindrical Coordinates

Cylindrical coordinates are a crucial way to simplify the calculation of integrals over volumes, particularly when dealing with symmetrical problems related to cylinders or circular shapes. This coordinate system uses three values to describe points in three-dimensional space:

• \( r \) — the radius, or the shortest distance from the point to the \( z \)-axis.
• \( \theta \) — the angle measured from the positive \( x \)-axis, projected on the \( XY \)-plane.
• \( z \) — similar to the Cartesian system, it still represents the vertical height.

In our integral, the setup is determined by these coordinates. The limits of \( \theta \) cover the full circle from \( 0 \) to \( 2\pi \), so we're examining the full rotational angle. \( r \) varies from 0 to 1, indicating that our region of interest has a maximum radius of 1. Lastly, \( z \) is bound between \( r \) and \( \sqrt{2-r^2} \), defining the height of the region in relation to both the radius and the cylindrical symmetry around its axis. This system simplifies integration for problems involving rotation and symmetry, making it especially useful for the current scenario.

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables, similar to the substitution method in calculus. This approach is particularly useful when dealing with functions that involve compositions, such as nested square root terms.

• Identify a substitution that simplifies the integral, often a term that is a composite function or has nested operations.
• Express the differential of substitution \( du \) in terms of the original variables.
• Change the limits of the integral if it's a definite integral, according to the new variable \( u \).
• Integrate the function with respect to \( u \), then substitute back if necessary to find the answer in the original variable.

In step 4 of our example, this method is employed when we let \( u = 2 - r^2 \), which turns the integral into a more manageable form. Bringing the integral from \( r \) into \( u \) space allows us to untangle the complexity arising from the nested square root, making the calculation possible in a straightforward manner.