Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-20, sketch the solid \(S\). Then write an iterated integral for $$ \iiint_{S} f(x, y, z) d V $$ $$ \begin{gathered} S=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 3 \\ \left.0 \leq z \leq \frac{1}{6}(12-3 x-2 y)\right\\} \end{gathered} $$

Step-by-Step Solution

Answer

Integrate \( \int_{0}^{1} \int_{0}^{3} \int_{0}^{\frac{1}{6}(12-3x-2y)} f(x, y, z) \, dz \, dy \, dx \).

Step 1: Understand the Region S

The region \( S \) is defined by three inequalities: \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 3 \), and \( 0 \leq z \leq \frac{1}{6}(12-3x-2y) \). This suggests that \( S \) is a volume under the plane \( z = \frac{1}{6}(12-3x-2y) \) above the rectangle formed by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 3 \).

Step 2: Sketch the Solid S

To sketch \( S \), consider the x-y plane bounded by the rectangle. For each point \( (x,y) \) in this rectangle, \( z \) ranges from 0 to the plane defined by \( z = \frac{1}{6}(12 - 3x - 2y) \). The sloping plane intersects the z-axis at \( z = 2 \) when both \( x \) and \( y \) are zero. The solid appears as a portion of space beneath this descending plane.

Step 3: Set up the Iterated Integral

The iterated integral for \( \iiint_{S} f(x, y, z) dV \) is constructed using the bounds described:\[iint \int f(x, y, z) dz \, dy \, dx\]The integration with respect to \( z \) is the innermost, with limits from 0 to \( \frac{1}{6}(12-3x-2y) \).\[iint \left(\int_{0}^{\frac{1}{6}(12-3x-2y)} f(x, y, z) dz \right) dy \, dx\]The integration with respect to \( y \) next, iterating from 0 to 3.\[\int_{0}^{1} \left( \int_{0}^{3} \left( \int_{0}^{\frac{1}{6}(12-3x-2y)} f(x, y, z) dz \right) dy \right) dx\]

Step 4: Write the Full Iterated Integral

Now, we can write down the full iterated integral as follows:\[\int_{0}^{1} \int_{0}^{3} \int_{0}^{\frac{1}{6}(12-3x-2y)} f(x, y, z) \, dz \, dy \, dx\]This represents the volume integral of the function \( f(x, y, z) \) over the solid region \( S \).

Key Concepts

Understanding Triple Integrals

In calculus, a triple integral allows you to integrate a function over a three-dimensional region. This is particularly useful for finding volumes under surfaces or in complex three-dimensional shapes. The concept of triple integrals extends the idea of single and double integrals to three dimensions. A triple integral is defined over a volume, which is usually bounded by surfaces or planes. In terms of notation, a triple integral is written as \[ \iiint_{V} f(x, y, z) \, dV \]where \( V \) represents the volume over which you are integrating, and \( f(x, y, z) \) is the integrand function. When setting up a triple integral, it is crucial to correctly determine the limits of integration for each variable—these describe the bounds of the volume for which the integral is evaluated. Triple integrals can be visually understood as the accumulation of infinitesimal volumes, effectively summing up properties over the specified region.

Determining Limits of Integration

Limits of integration in a triple integral determine the scope of the region over which you're integrating. Setting correct limits is essential as they define the boundaries of the solid or volume. In the given exercise:

• The limits for \( x \) are from \( 0 \) to \( 1 \), establishing the range along the x-axis.
• For \( y \), the limits extend from \( 0 \) to \( 3 \), setting the scope along the y-axis.
• The limits for \( z \) are particularly interesting, as they vary with both \( x \) and \( y \): from \( 0 \) to \( \frac{1}{6}(12-3x-2y) \). This specifies a plane that descends based on values of \( x \) and \( y \).

These inequalities define a volume in three-dimensional space, beneath a slanted plane, enclosed by x-y-z limits. Understanding these boundaries is key to solving any triple integral problem, as they essentially instruct you on how to "slice" the integration process.

Finding Volume Under a Surface

One common application of triple integrals is to find the volume under a surface. In this context, the triple integral sums up infinitesimal volumes defined by the function and the given bounds.Consider the exercise, where the region \( S \) is beneath a surface defined by the plane \( z = \frac{1}{6}(12-3x-2y) \). To find the volume:- Visualize the shape described by the boundaries and how it intersects with the defined plane.- The volume can be expressed as an integral of \( 1 \) (for simply finding volume) over the region \( S \).This region is what you're integrating over to find the total volume. It involves evaluating the integral over the specified limits for \( x \), \( y \), and \( z \), which collectively bound the region dictating the "height" in the \( z \)-dimension beneath the slanted surface.

Applications in Calculus 3

Triple integrals are a vital part of Calculus 3, usually introduced alongside other multivariable calculus topics like vector fields and surface integrals. Understanding how to work with triple integrals is crucial for dealing with problems involving three-dimensional shapes and higher-dimensional calculus concepts. In Calculus 3:

• You learn how to set up and evaluate triple integrals effectively, often involving transformations to polar, cylindrical, or spherical coordinates to simplify calculations.
• Such skills are important for evaluating physical and geometric problems, including those related to mass, volume, and other properties of a material or system in three-dimensional space.
• Grasping these integrations aids significantly when tackling real-world applications in engineering, physics, and higher mathematics."

Triple integrals form a foundational tool in assessing and analyzing complex systems across multiple fields due to their capacity to handle volumetric data and represent physical properties spread over a region.