Question

Evaluate the triple integral. $$ \int_{0}^{1} \int_{0}^{x} \int_{0}^{x+y}(x+y+z) d z d y d x $$

Short answer

The triple integral evaluates to \( \frac{7}{8} \).

Step-by-step solution

Understand the Limits of Integration

The given triple integral is in the form \( \int_{0}^{1} \int_{0}^{x} \int_{0}^{x+y} (x+y+z) \, dz \, dy \, dx \). The limits of integration for \( z \) are from \( 0 \) to \( x+y \), for \( y \) from \( 0 \) to \( x \), and for \( x \) from \( 0 \) to \( 1 \). This setup means we integrate \( z \) first, then \( y \), and finally \( x \).

Integrate Over Variable z

Perform the innermost integration with respect to \( z \): \[ \int_{0}^{x+y} (x+y+z) \, dz = \left[ (x+y)z + \frac{1}{2}z^2 \right]_{0}^{x+y} \]\[ = (x+y)(x+y) + \frac{1}{2}(x+y)^2 - 0 \]\[ = (x+y)^2 + \frac{1}{2}(x+y)^2 = \frac{3}{2}(x+y)^2 \].

Integrate Over Variable y

Substitute the result from Step 2 into the next integral and integrate with respect to \( y \): \[ \int_{0}^{x} \frac{3}{2}(x+y)^2 \, dy \].Use substitution \( u = x+y \), so \( du = dy \) and when \( y = 0 \), \( u = x \) and when \( y = x \), \( u = 2x \).After substitution: \[ \int_{x}^{2x} \frac{3}{2}u^2 \, du = \frac{3}{2} \left[ \frac{u^3}{3} \right]_{x}^{2x} \]\[ = \frac{3}{2} \left( \frac{(2x)^3}{3} - \frac{x^3}{3} \right) \]\[ = \frac{3}{2} \left( \frac{8x^3}{3} - \frac{x^3}{3} \right) = \frac{3}{2} \cdot \frac{7x^3}{3} = \frac{7x^3}{2} \].

Integrate Over Variable x

Substitute the result from Step 3 into the final integration and integrate with respect to \( x \): \[ \int_{0}^{1} \frac{7x^3}{2} \, dx \]\[ = \frac{7}{2} \left[ \frac{x^4}{4} \right]_{0}^{1} \]\[ = \frac{7}{2} \cdot \frac{1}{4} = \frac{7}{8} \].

Conclude the Calculation

The evaluation of the triple integral gives the final result. Each integration step systematically reduces the dimensions of the problem until only a simple polynomial integral remains.

Key concepts

calculus

Calculus is a branch of mathematics that deals with continuous change. It involves concepts such as differentiation and integration, which help us understand and solve problems related to rates of change and areas under curves. In this triple integral problem, calculus is used to assess how a sum of variables can change over a specific region.

There are two primary components:

• Differentiation, which involves finding the rate at which a quantity changes.
• Integration, which focuses on accumulating quantities, such as areas under curves or volumes within boundaries.

For this exercise, we delve into integration – specifically, a triple integral – which extends the idea of accumulation to three dimensions. By calculating \(\int_{0}^{1} \int_{0}^{x} \int_{0}^{x+y}(x+y+z)\, d z \, d y \, d x\), we find a volume under a surface defined by the function \(x+y+z\) over the specified bounds.

Understanding calculus as applied here helps us model and solve real-world problems in which multiple variables interact over spatial regions.

integration techniques

Integration techniques are strategies used to simplify integrals, especially when dealing with complex or multiple integrals. For a triple integral like the one in our exercise, applying proper techniques effectively simplifies calculations.

Here's how we approached each integration:

• **Iterative integration**: This involves integrating one variable at a time, working from the innermost to the outermost bounds.
• **Substitution**: Used to simplify the integration process, especially for challenging integrals. In Step 3, we substitute \(u = x+y\) to streamline the integral calculation.

By using these techniques, we broke down a potentially cumbersome triple integral into manageable single-variable integrals. Each layer we peel back brings clarity and simplifies the computation, ultimately leading us to our solution, \(\frac{7}{8}\).

These integration techniques are essential in solving problems methodically, ensuring accuracy and efficiency.

iterated integrals

Iterated integrals represent a structured approach to solving multi-variable integrals, such as double or triple integrals. Rather than tackling the entire function all at once, an iterated approach handles variables successively.

Here's a breakdown:

• **Integration Order**: The sequence in which we integrate matters, as seen in \(\int_{0}^{1} \int_{0}^{x} \int_{0}^{x+y}(x+y+z)\, d z \, d y \, d x\). We start with \(z\), progress to \(y\), and finish with \(x\). The order aligns with the provided limits.
• **Limit of Integration**: Each integral’s limits depend on the preceding variable’s bounds. For example, integrating over \(z\) from 0 to \(x+y\) directly affects the next steps.

This process makes it possible to evaluate integrals in a systematic and manageable manner. Iterated integrals facilitate the decomposition of multidimensional problems into simpler tasks.

Employing iterated integrals transforms a daunting triple integral task into a series of straightforward operations, illustrating the power and utility of this technique in calculus.