Evaluating The Triple Integral Of Xyz Over A Specified Region

In this article, we delve into the step-by-step evaluation of the triple integral ∫₀¹ ∫₀¹ ∫√(x²+y²)¹ xyz dz dy dx. Triple integrals extend the concept of double integrals to three dimensions, allowing us to calculate the volume, mass, or other properties of a three-dimensional region. The given integral represents the integral of the function f(x, y, z) = xyz over a specific volume defined by the limits of integration. Understanding how to set up and evaluate these integrals is a fundamental skill in multivariable calculus, with applications in physics, engineering, and various other scientific fields. The process involves integrating the function successively with respect to each variable, carefully managing the limits of integration at each step.

Understanding Triple Integrals

Before we dive into the specifics of this problem, it's crucial to grasp the basic concept of triple integrals. Triple integrals are used to calculate the integral of a function over a three-dimensional region. They are the three-dimensional analogue of double integrals, which are used for integration over two-dimensional regions. In essence, a triple integral can help determine properties such as volume, mass, or the average value of a function over a three-dimensional space. To truly understand how triple integrals work, it's essential to visualize the region of integration and the function being integrated.

Triple integrals are represented mathematically as ∭_V f(x, y, z) dV, where f(x, y, z) is the function being integrated, and V is the volume of the region of integration. The dV represents the differential volume element, which can be expressed in various coordinate systems such as Cartesian (dx dy dz), cylindrical (r dr dθ dz), or spherical (ρ² sin φ dρ dθ dφ). The choice of coordinate system often depends on the symmetry and shape of the region of integration. For instance, regions with circular or cylindrical symmetry are often easier to handle in cylindrical coordinates, while regions with spherical symmetry are better suited for spherical coordinates. Understanding these coordinate systems and when to use them is crucial for simplifying the integral and making it easier to evaluate. The limits of integration define the boundaries of the region V, and these limits can be constants or functions of other variables, depending on the shape of the region.

The order of integration matters and can sometimes significantly affect the complexity of the calculation. While it is often possible to change the order of integration, careful consideration must be given to how the limits of integration will change accordingly. In some cases, changing the order can make the integral significantly easier or harder to evaluate. Visualizing the region of integration can be incredibly helpful in determining the best order of integration and the correct limits for each variable. This visualization can often be done through sketching the region or using computer software to create a 3D representation.

In the context of physics and engineering, triple integrals have numerous applications. They can be used to calculate the mass of an object with variable density, the center of mass, moments of inertia, and the flux of a vector field through a surface. In fluid dynamics, triple integrals can be used to determine the flow rate of a fluid through a given volume. In electromagnetism, they can be used to calculate the electric or magnetic field generated by a charge or current distribution. The versatility of triple integrals makes them an indispensable tool in many quantitative disciplines, highlighting the importance of mastering the techniques for setting up and evaluating these integrals.

Step 1: Setting up the Integral

Our first step involves understanding the given triple integral: ∫₀¹ ∫₀¹ ∫√(x²+y²)¹ xyz dz dy dx. This integral is set up in Cartesian coordinates, with the integration order being dz dy dx. This means we will first integrate with respect to z, then y, and finally x. The limits of integration define the region over which we are integrating. The outermost integral has limits from 0 to 1 for x, the middle integral has limits from 0 to 1 for y, and the innermost integral has limits from √(x² + y²) to 1 for z. These limits define a volume in the first octant bounded by the planes x = 0, x = 1, y = 0, y = 1, and the surfaces z = √(x² + y²) and z = 1.

To better understand this region, it's helpful to visualize the surfaces defined by the limits of integration. The surface z = √(x² + y²) represents a cone opening upwards from the origin. The surface z = 1 is a horizontal plane parallel to the xy-plane. The planes x = 0, x = 1, y = 0, and y = 1 define a square region in the xy-plane. The volume of integration is thus the region bounded below by the cone and above by the plane z = 1, and laterally by the square region in the xy-plane. This three-dimensional understanding of the integration region is crucial for evaluating the integral correctly. It helps in anticipating any simplifications that might arise and in checking the reasonableness of the final result.

The integrand, xyz, is a relatively simple function, but the complexity of the integral comes from the non-constant limits of integration, particularly the lower limit z = √(x² + y²). This limit suggests that converting to cylindrical coordinates might simplify the integral, as the expression x² + y² becomes r² in cylindrical coordinates. However, for this example, we will proceed with the Cartesian setup first to demonstrate the process and then discuss how cylindrical coordinates could offer an alternative approach.

Before proceeding with the integration, it is always a good practice to check if the integrand is continuous over the region of integration. In this case, f(x, y, z) = xyz is continuous everywhere, so we do not need to worry about any singularities or discontinuities within the integration region. This ensures that the integral is well-defined and that we can proceed with the standard integration techniques. Setting up the integral correctly is often the most critical step in evaluating triple integrals, as it lays the foundation for the subsequent calculations. A clear understanding of the integration region and the integrand helps in choosing the appropriate coordinate system and the order of integration, which can significantly impact the ease of evaluating the integral.

Step 2: Integrating with Respect to z

The first step in evaluating the triple integral ∫₀¹ ∫₀¹ ∫√(x²+y²)¹ xyz dz dy dx is to integrate with respect to z, treating x and y as constants. This means we are evaluating the inner integral ∫√(x²+y²)¹ xyz dz. The integrand is xyz, and we are integrating with respect to z, so we apply the power rule for integration, which states that ∫z^n dz = (z^(n+1))/(n+1) + C, where C is the constant of integration. In this case, the power of z is 1, so we have ∫xyz dz = xy ∫z dz = xy(z²/2) + C. Since we are evaluating a definite integral, we do not need to include the constant of integration, as it will cancel out when we evaluate the limits.

Now, we need to evaluate the antiderivative at the limits of integration for z, which are z = √(x² + y²) and z = 1. Substituting these limits into the antiderivative, we get:

xy[(1)²/2 - (√(x² + y²))²/2] = xy[1/2 - (x² + y²)/2] = (xy/2)(1 - x² - y²)

This result represents the value of the inner integral after integrating with respect to z. It is now a function of x and y, which we will integrate in the subsequent steps. It's crucial to handle the limits of integration carefully at this stage to ensure the correctness of the result. A common mistake is to incorrectly substitute the limits or to forget to apply the fundamental theorem of calculus. The expression we obtained, (xy/2)(1 - x² - y²), is a polynomial in x and y, which is continuous over the region of integration in the xy-plane. This continuity is important because it allows us to proceed with the next steps of integration without worrying about any singularities or discontinuities.

This step demonstrates the fundamental principle of iterated integration, where we treat the other variables as constants while integrating with respect to one variable. This approach allows us to break down a multi-dimensional integral into a series of single-variable integrals, which are often easier to evaluate. The result of this step, (xy/2)(1 - x² - y²), will be the integrand for the next integration step with respect to y. It is a good practice to simplify this expression as much as possible before proceeding to the next integration to minimize the complexity of the calculations.

Step 3: Integrating with Respect to y

Having completed the integration with respect to z, we now move on to integrating with respect to y. We are faced with the integral ∫₀¹ (xy/2)(1 - x² - y²) dy. It’s beneficial to first simplify the integrand to make the integration process smoother. Distributing the terms, we get:

∫₀¹ (xy/2 - x³y/2 - xy³/2) dy

Now, we can integrate term by term with respect to y. Applying the power rule for integration, which states ∫y^n dy = (y^(n+1))/(n+1) + C, we get:

(x/2)∫₀¹ y dy - (x³/2)∫₀¹ y dy - (x/2)∫₀¹ y³ dy

Integrating each term, we have:

(x/2)(y²/2)|₀¹ - (x³/2)(y²/2)|₀¹ - (x/2)(y⁴/4)|₀¹

Now, we evaluate the antiderivatives at the limits of integration, y = 0 and y = 1:

(x/2)((1)²/2 - (0)²/2) - (x³/2)((1)²/2 - (0)²/2) - (x/2)((1)⁴/4 - (0)⁴/4)

This simplifies to:

(x/2)(1/2) - (x³/2)(1/2) - (x/2)(1/4) = x/4 - x³/4 - x/8

Combining like terms, we get:

x/4 - x/8 - x³/4 = x/8 - x³/4

This result represents the value of the integral after integrating with respect to y. It is now a function of x, which we will integrate in the final step. It’s crucial to double-check the calculations at this stage to ensure that the antiderivatives and the limits of integration have been applied correctly. Any errors made in this step will propagate through the rest of the calculation, so accuracy is paramount. The expression we obtained, x/8 - x³/4, is a polynomial in x, which is continuous over the interval of integration. This continuity ensures that we can proceed with the final integration step without any concerns about singularities or discontinuities.

Step 4: Integrating with Respect to x

In the final step, we integrate the result from the previous step with respect to x. We need to evaluate the integral ∫₀¹ (x/8 - x³/4) dx. This is a straightforward integral involving polynomials, so we can apply the power rule for integration, which states ∫x^n dx = (x^(n+1))/(n+1) + C. Applying this rule to our integral, we get:

∫₀¹ (x/8) dx - ∫₀¹ (x³/4) dx = (1/8)∫₀¹ x dx - (1/4)∫₀¹ x³ dx

Now, we integrate each term:

(1/8)(x²/2)|₀¹ - (1/4)(x⁴/4)|₀¹

Next, we evaluate the antiderivatives at the limits of integration, x = 0 and x = 1:

(1/8)((1)²/2 - (0)²/2) - (1/4)((1)⁴/4 - (0)⁴/4)

This simplifies to:

(1/8)(1/2) - (1/4)(1/4) = 1/16 - 1/16 = 0

Therefore, the final result of the triple integral ∫₀¹ ∫₀¹ ∫√(x²+y²)¹ xyz dz dy dx is 0. This means that the signed volume of the region, weighted by the function xyz, is zero. This result might seem surprising, but it is a valid outcome given the symmetry of the region and the nature of the integrand. The integrand xyz changes sign within the region of integration, and these sign changes can result in cancellations that lead to a zero integral.

Step 5: Conclusion and Alternative Approaches

In conclusion, we have successfully evaluated the triple integral ∫₀¹ ∫₀¹ ∫√(x²+y²)¹ xyz dz dy dx and found that it equals 0. This result underscores the importance of careful calculation and the potential for symmetry and cancellations to simplify integrals. We systematically integrated with respect to z, then y, and finally x, carefully applying the limits of integration at each step. While the Cartesian coordinate system worked for this problem, it is worth considering alternative approaches, such as using cylindrical coordinates, which can often simplify integrals involving cylindrical symmetry.

Alternative Approach: Cylindrical Coordinates

As mentioned earlier, cylindrical coordinates might offer a more streamlined approach for this particular integral due to the presence of the term √(x² + y²) in the limits of integration. In cylindrical coordinates, we have x = r cos θ, y = r sin θ, and z = z. The volume element dV in Cartesian coordinates (dz dy dx) transforms to r dz dr dθ in cylindrical coordinates. The expression x² + y² becomes r², so √(x² + y²) becomes r.

To convert the integral to cylindrical coordinates, we need to determine the new limits of integration. The original limits for x and y were 0 to 1. In the xy-plane, this represents a square. In polar coordinates (which are the basis for cylindrical coordinates), this square can be described in terms of r and θ. The limits for r will range from 0 to the line that connects (1,0) to (1,1) and then to (0,1), which means r goes from 0 to sec θ for 0 ≤ θ ≤ π/4 and from 0 to csc θ for π/4 ≤ θ ≤ π/2. The limits for θ are from 0 to π/2, covering the first quadrant. The limits for z change from √(x² + y²) to r and 1, remaining the same.

The integral in cylindrical coordinates becomes:

∫₀^(π/2) ∫? ∫ᵣ¹ (r cos θ)(r sin θ)z *r dz dr dθ

Here, the '?' represents the r limits which are sec θ from 0 ≤ θ ≤ π/4 and csc θ for π/4 ≤ θ ≤ π/2.

The integral becomes more complex to set up in cylindrical coordinates due to the r limits, though the innermost integral simplifies somewhat. For this particular problem, the Cartesian approach might be more straightforward due to the shape of the integration region.

Summary

Evaluating triple integrals requires a systematic approach and a solid understanding of multivariable calculus concepts. We have demonstrated a step-by-step evaluation of the integral ∫₀¹ ∫₀¹ ∫√(x²+y²)¹ xyz dz dy dx, highlighting the process of integrating with respect to each variable and applying the limits of integration. Additionally, we discussed the potential benefits and complexities of using cylindrical coordinates as an alternative approach. The key to mastering triple integrals lies in practice, careful calculation, and the ability to visualize the region of integration and the integrand.