Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals. \(\int_{1}^{4} \int_{z-1}^{2 z} \int_{0}^{y+2 z} d x d y d z\)

Step-by-Step Solution

Answer

The value of the iterated integral is \(\frac{763}{6}\).

Step 1: Understand the Boundaries and Variables

The iterated integral given is \(\int_{1}^{4} \int_{z-1}^{2z} \int_{0}^{y+2z} dx \, dy \, dz\). This integral represents a volume bounded in three dimensions. The variables \(x\), \(y\), and \(z\) are integrated in that order, so start with the inner integral in \(x\), followed by \(y\), and finally \(z\). The limits of integration depend on the variables associated with each integral.

Step 2: Evaluate the Inner Integral with Respect to x

The innermost integral is \(\int_{0}^{y+2z} dx\). Integrating \(dx\) from 0 to \(y + 2z\) gives the result \(x\) evaluated from 0 to \(y + 2z\). Thus, the result is \( (y + 2z) - 0 = y + 2z\).

Step 3: Evaluate the Middle Integral with Respect to y

Now consider the middle integral: \(\int_{z-1}^{2z} (y + 2z) \, dy\). Integrate \(y\) to get \( \frac{1}{2}y^2 + 2zy\), then substitute the limits of \(y\) from \(z-1\) to \(2z\). This results in: \[\left(\frac{1}{2}(2z)^2 + 2z(2z)\right) - \left(\frac{1}{2}(z-1)^2 + 2z(z-1)\right).\] Calculating this gives you: \[2z^2 + 4z^2 - \frac{1}{2}(z^2 - 2z + 1) - 2z^2 + 2z\] Simplifying gives \[5z^2 + z - \frac{1}{2}\].

Step 4: Evaluate the Outer Integral with Respect to z

Finally, evaluate the outer integral: \(\int_{1}^{4} \left(5z^2 + z - \frac{1}{2}\right) dz\). Integrate term-by-term:\[\int 5z^2 \, dz = \frac{5}{3}z^3, \quad \int z \, dz = \frac{1}{2}z^2, \quad \int \left(-\frac{1}{2}\right) dz = -\frac{1}{2}z\]Substitute the limits \(1\) to \(4\):\[\left(\frac{5}{3}(4)^3 + \frac{1}{2}(4)^2 - \frac{1}{2}(4)\right) - \left(\frac{5}{3}(1)^3 + \frac{1}{2}(1)^2 - \frac{1}{2}(1)\right),\]Calculate and simplify to get the final result: \(\frac{763}{6}\).

Key Concepts

Multivariable Calculus

Multivariable Calculus is a branch of mathematics that deals with functions of multiple variables. It extends the basic concepts of calculus, such as derivatives and integrals, into higher dimensions. In this context, we are dealing with iterated integrals, which is a key component of multivariable calculus. Iterated integrals allow us to evaluate complex integrals step-by-step by integrating one variable at a time.
Understanding how to work with multiple variables is crucial in solving problems involving three-dimensional spaces, like calculating volumes or other quantities that are dependent on several changing factors. It bridges the gap between single-variable calculus and the real-world applications where we need to compute changes in more than one dimension.
In iterated integrals, the order of integration matters and is usually determined by the limits of integration. By integrating one variable at a time, we can break down multi-dimensional problems into more manageable, solvable steps. This process is essential in solving problems involving volume integration, especially in physics and engineering.

Volume Integration

Volume Integration helps us to calculate the volume of 3D objects. It is based on extending the concept of area under a curve to three dimensions. In our exercise, we are using iterated integrals to determine a volume within a specified region.
The solution involves evaluating the integral \(\int_{1}^{4} \int_{z-1}^{2z} \int_{0}^{y+2z} dx \, dy \, dz\\). This process is essentially cutting the volume into infinitesimally small pieces and summing these pieces up to get the total volume.
Key points of volume integration include:

• Belonging to multivariable calculus, which means handling more than one variable or dimension.
• Using a systematic approach to solving complex 3D problems by evaluating each integral from the inside out.
• Each integral corresponds to adding up slices of the volume, integrating one dimension at a time until the whole region is accounted for.

Volume integration is widely used in engineering and physics when determining the mass of 3D bodies, calculating heat or fluid distribution, or analyzing electromagnetic fields.

Limit of Integration

Limits of Integration are the boundaries within which we calculate the integral. They indicate the range over which the function is to be integrated for each variable. In the provided problem, integration starts from the innermost limit in variable \(x\), progresses to \(y\), and ends with the outermost integration in \(z\).
The limits can be constant (definite values) or variable, depending on the context. For the solution given:

• The inner integral, \(x\), runs from 0 to \(y + 2z\).
• The middle integral, \(y\), is evaluated from \(z-1\) to \(2z\).
• The outer integral, \(z\), is solved from 1 to 4.

Understanding these limits is critical in solving iterated integrals correctly. They determine the portion of space over which you're summing up volume elements in this case. Limits help describe the shape and bounds of the region being integrated over, akin to setting the parameters for slicing a loaf of bread precisely.