Question

Evaluate the iterated integrals. \(\int_{4}^{24} \int_{0}^{24-x} \int_{0}^{24-x-y} \frac{y+z}{x} d z d y d x\)

Short answer

The iterated integral evaluates to 2016.

Step-by-step solution

Evaluate the Inner Integral

Start with the innermost integral with respect to the variable z: \[ \int_{0}^{24-x-y} \frac{y+z}{x} \, d z \] Separate the integrand into two terms: \[ \int_{0}^{24-x-y} \left(\frac{y}{x} + \frac{z}{x}\right) \, d z = \frac{y}{x} \int_{0}^{24-x-y} 1 \, d z + \frac{1}{x} \int_{0}^{24-x-y} z \, d z \]The first integral is simply \(\frac{y}{x}(24-x-y)\), and for the second integral, use the power rule yielding \((24-x-y)^2/2\) multiplied by \(1/x\) to get: \[ \frac{y}{x}(24-x-y) + \frac{(24-x-y)^2}{2x} \] .

Evaluate the Second Integral

Now, integrate with respect to y: \[ \int_{0}^{24-x} \left(\frac{y}{x}(24-x-y) + \frac{(24-x-y)^2}{2x}\right) \, d y \] Begin with the first term:\[ \int_{0}^{24-x} \frac{y}{x}(24-x-y) \, d y \] This splits into two integrals: \[ \frac{24-x}{x}\int_{0}^{24-x}y \, d y - \frac{1}{x}\int_{0}^{24-x}y^2 \, d y \]Use the power rule for both integrals and simplify.Now, integrate the second term similarly: \[ \int_{0}^{24-x} \frac{(24-x-y)^2}{2x} \, d y \] Expand \((24-x-y)^2\), integrate each part, and simplify.

Evaluate the Outer Integral

Now integrate the result from Step 2 with respect to x over the interval \([4, 24]\): \[ \int_{4}^{24}(\text{expression from Step 2}) \, d x \] Carefully evaluate each term involving x using appropriate integration techniques (power rule, basic calculus). Simplify terms where possible and perform definite integration to find the final value of the iterated integral.

Combine and Simplify

Add all the results from Step 3, making sure all terms have been calculated over their respective limits. Simplify as much as possible to find the final value of the computated integrals.

Key concepts

Triple Integrals

Triple integrals extend the concept of double integrals to three dimensions, allowing us to compute volumes in 3D spaces. The problem from the original exercise involves evaluating a triple integral, with three nested integrals. Each integral corresponds to a dimension in space. We integrate sequentially, starting from the innermost to the outermost integral. This approach is commonly used when we need to integrate over a 3D region, such as computing the mass of a solid or the volume under a surface.
Triple integrals follow a logical sequence where each stage narrows the focus:

• The innermost integral focuses on a single dimension initially, solving it individually with respect to its specific variable.
• The middle integral then considers the result of the innermost integral with a second dimension in mind.
• The outermost integral looks at the cumulative result of the first two, finalizing the integration process.

Breaking it down this way simplifies our path towards the ultimate solution of a problem involving multiple variables in three-dimensional space.

Integration Techniques

Integration techniques utilized in solving the original triple integral include breaking complex expressions into manageable parts and applying basic rules, such as the power rule and the sum-difference rule. Through this approach, we simplify the integration process.A crucial technique is the power rule, which helps in solving integrals of the form \( \int z^n \, dz \). For instance, during the innermost integration process, terms are split, making it easier to integrate each part individually.
Moreover, separation of terms, as seen in the separation of \( \frac{y+z}{x} \) into \( \frac{y}{x} \) and \( \frac{z}{x} \), allows us to handle simple integrals consecutively, avoiding the complexity of integrating difficult, combined functions all at once.Using these techniques helps manage intricate details and reduces the integral to a simpler form, suitable for successive integrations.

Calculus Problem Solving

In calculus problem-solving, the application of iterated integrals involves strategic handling of multiple integrals by organizing the integration order for efficiency. The process can be seen like a recipe, akin to following a set of steps to achieve the desired goal in the integration process. In the original problem, decomposing the triple integral into sequence aids in systematically approaching each layer of complexity. Implementing a step-by-step plan ensures that calculations stay accurate and manageable. Each step involves applying basic calculus rules to achieve the ultimate result. Minimizing errors and maintaining accuracy are important in handling complex calculus problems, especially those involving iterated integrals. Additionally, strategic simplification can often lead to spotting patterns or shortcuts that simplify the problem further.

Definite Integration

Definite integration features prominently in solving iterated integrals, where you evaluate integrals with specific upper and lower bounds. Unlike indefinite integration, which results in a general function, definite integration calculates a number giving precise results over a specified interval. Each of the nested integrals in the original exercise has specific limits, such as \([0, 24-x-y]\) and \([4, 24]\), dictating the scope for each variable.Definite integration considers:

• The interval over which integration occurs, providing concrete solutions.
• Application of boundary conditions that reduce the outcome to actual numbers, representing total values such as volume or mass.
• Simplifying within given limits to enhance result accuracy, reducing verbal terms across limits to highlight computational results.

Through the use of definite integration, the integration process narrows down the solution's possibilities, focusing on precise necessary calculations leading to the actual output.