Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The iterated integral is \( \int_{0}^{1} \int_{0}^{\sqrt{z}} \int_{0}^{y^2} f(x, y, z) \, dx \, dy \, dz \).

Step 1: Understand the Region

First, we need to interpret the region described by the inequalities: \( S = \{ (x, y, z) : 0 \leq x \leq y^2, 0 \leq y \leq \sqrt{z}, 0 \leq z \leq 1 \} \). This defines a 3-dimensional region in the space. The constraint \( 0 \leq x \leq y^2 \) means that for a given \( y \), \( x \) can range from 0 to \( y^2 \). Similarly, \( 0 \leq y \leq \sqrt{z} \) limits \( y \) to values between 0 and \( \sqrt{z} \). Finally, \( 0 \leq z \leq 1 \) sets the range for \( z \).

Step 2: Sketch the Solid

To sketch \( S \), start by considering the bounds for \( z: 0 \leq z \leq 1 \). For each \( z \), consider the inequality \( 0 \leq y \leq \sqrt{z} \), which forms a triangular region in the \( yz \)-plane if \( z \) is seen as a constant. Within these constraints, \( 0 \leq x \leq y^2 \) gives a parabola in the \( xy \)-plane. This configuration creates a solid that is constrained by parabolic sheets along the \( x \)-axis and right triangles expanding as \( z \) increases from 0 to 1.

Step 3: Write the Iterated Integral

Using the information from each bound, the iterated integral can be constructed. The integration order will be \( dz \), \( dy \), \( dx \) as tree-dimensional integrations are executed innermost to outermost. Thus, the iterated integral becomes: \[ \int_{0}^{1} \int_{0}^{\sqrt{z}} \int_{0}^{y^2} f(x, y, z) \, dx \, dy \, dz \] The limits reflect the constraints provided for \( S \).

Key Concepts

Iterated Integrals

Iterated integrals are a powerful tool in calculus used to compute the total amount of a function across a certain region. In the context of triple integrals, they enable the computation over a three-dimensional space.
The expression \[ \iiint_{S} f(x, y, z) \, dV \] indicates that you're summing up the values of a function \( f(x, y, z) \) across every point in the region \( S \). This involves nesting or iterating separate integrals for each of the variable dimensions: \( x \), \( y \), and \( z \).

• Inner integrals process the innermost variables first: Here, \( x \).
• The next integral handles the next variable \( y \), followed by
• the outermost integral which processes the remaining variable \( z \).

The limits for each of these integrals are crucial, as they describe how one variable affects others. They define the region where the function is evaluated, ensuring that we only sum over the appropriate portion of the space.

3D Region Sketching

Sketching a 3D region defined by inequalities helps visualize the area over which you're integrating. This is especially useful when dealing with triple integrals. In this example, the region \( S \) is defined by \((x, y, z) \) inequalities: \[ 0 \leq x \leq y^2, 0 \leq y \leq \sqrt{z}, 0 \leq z \leq 1. \] These constraints define a bounded space in 3D that can be understood in layers:

• Start with the bounds for \( z \), which is between 0 and 1. This gives a height limit for our region.


• For any fixed \( z \), \( y \) ranges from 0 to \( \sqrt{z} \). This results in a semi-circular region in the \( yz \)-plane.


• Within each of these \( yz \) regions, for any given \( y \), \( x \) varies from 0 to \( y^2 \). This means the width along the \( x \)-axis is defined by parabolas as \( y \) changes.

Visualizing this: Imagine layers of expanding triangular regions along the \( z \)-direction, contained by a 'roof' bounded by \( z \), while the 'walls' taper through parabolas.

Order of Integration

The order of integration is a crucial choice in constructing an iterated integral for a triple integral. Depending on the region \( S \), some orders can simplify the problem, while others might complicate it. In our exercise, the chosen order was \( dx \, dy \, dz \, \). Let's look at how this affects the integration process:

• By selecting \( dx \) as the innermost integration, we first solve for variable \( x \) within its constraints \( 0 \leq x \leq y^2 \). This choice often handles functions with simpler x-dependencies.


• The next layer, \( dy \), fixes \( y \) within \( 0 \leq y \leq \sqrt{z} \). Here, \( dy \) evaluates over the triangular region in the \( yz \)-plane.
• Finally, \( dz \) captures the remaining variation in \( z \) from 0 to 1, encompassing the block of space built by all prior integrations.

Choosing the right order of integration helps in simplifying the problem and reduces computational complexity. It’s often based on the ease of the resulting integral computations or the symmetry and format of region \( S \). When solving similar problems, consider altering the order if the initial integration seems complex.