Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid S. Then write an iterated integral for $$ \iiint_{S} f(x, y, z) d V $$ \(S\) is the region in the first octant bounded by the cylinder \(y^{2}+z^{2}=1\) and the planes \(x=1\) and \(x=4 .\)

Step-by-Step Solution

Answer

The solid S is a cylinder sliced between x=1 and x=4, with iterated integral: \[ \int_{1}^{4} \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} f(x, y, z) \, dy \, dz \, dx. \]

Step 1: Understanding the region S

The region \( S \) is bounded by a cylinder with the equation \( y^2 + z^2 = 1 \). This represents a circular cylinder centered along the x-axis with a radius of 1. The first octant condition implies that \( x, y, z \geq 0 \). The planes \( x = 1 \) and \( x = 4 \) confine the region in the x-direction. Therefore, \( x \) varies between 1 and 4, while \( y^2 + z^2 \leq 1 \) limits the values of \( y \) and \( z \).

Step 2: Determine the limits for y and z

From the equation \( y^2 + z^2 \leq 1 \), it follows that the region in the \( yz \)-plane is a circle with radius 1. For any fixed \( x \), \( y \) ranges from \( -\sqrt{1-z^2} \) to \( \sqrt{1-z^2} \) (since \( z \geq 0 \), this effectively means \( y \) ranges from 0 to \( \sqrt{1-z^2} \)).\( z \) ranges from 0 to 1 due to the circle's radius.

Step 3: Set up the iterated integral

The iterated integral for \( \iiint_S f(x, y, z) \, dV \) is set up with \( x \) as the outermost integral because our x-boundaries are constants. Thus, the integral becomes:\[ \int_{1}^{4} \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} f(x, y, z) \, dy \, dz \, dx. \]This order of integration provides the most straightforward approach in this context because the limits of \( y \) and \( z \) depend on each other and the cylinder's equation.

Key Concepts

Iterated Integral

An iterated integral is a way to evaluate a triple integral by performing one integral at a time. It breaks down a 3D integration into a sequence of 1-dimensional integrals. In our exercise, we are integrating over a complex region defined by a cylinder and two planes. To do this effectively, we first integrate with respect to one variable while treating the others as constants. We then repeat this process for the subsequent variables.

For the given region, we set up the iterated integral as:

• The outermost integral: With respect to x, ranging from 1 to 4.
• The middle integral: With respect to z, with limits 0 to 1 due to the circle's radius formed by the cylinder in the yz-plane.
• The innermost integral: With respect to y, affected by the value of z, ranging from 0 to \(\sqrt{1-z^2}\).

This approach simplifies the computation by maintaining constant limits where possible and addressing dependencies between variables.

Cylindrical Coordinates

When dealing with regions related to cylinders or circular shapes, cylindrical coordinates can be extremely useful. These coordinates are particularly advantageous in this problem because they naturally align with the geometry of a circular cylinder.

The standard transition to cylindrical coordinates involves:

• \( x = r\cos\theta \)
• \( y = r\sin\theta \)
• \( z = z \)

For a cylinder like \( y^2 + z^2 = 1 \), using \( y = r\cos\theta \) and \( z = r\sin\theta \) helps express the circular cross-section.This substitution can simplify integration over circular boundaries. However, in this exercise, since the x-ranges are constants, reconstructing the cylindrical form is not necessary for the triple integral setup. Nevertheless, it's essential to understand when and why these transformations could be beneficial.

First Octant

The first octant in 3D space is a "section" where all three coordinate axes are positive. In other words, any point (x, y, z) in this octant satisfies:

• \(x \geq 0\)
• \(y \geq 0\)
• \(z \geq 0\)

In the provided exercise, the restriction to the first octant ensures that both y and z, parts of the cylinder's boundary, cannot be negative. This simplifies sketching the region, ensuring we consider only the positive portions of the coordinate system.

This condition influences our integral boundaries for y and z. For example, y ranges from 0 to the top of the cylinder's cross-section in the yz-plane. Whenever dealing with complex shapes bounded in 3D space, identifying the octant simplifies the integration process.

Boundary Conditions

Understanding the boundary conditions is crucial for setting up integrals. In this exercise, the boundary limits allow us to form the integral boundaries and define the region in space over which to integrate. Let’s break these down step-by-step:

• The plane \(x=1\) serves as the lower limit of x.
• The plane \(x=4\) acts as the upper limit of x. Together, they define a finite section of the cylinder along the x-axis.
• The equation \(y^2 + z^2 = 1\) is the boundary of the circular base of the cylinder in the yz-plane. It outlines the top edge where y and z interact within a fixed radius.

Boundary conditions circumscribe the space for the iterated integral, restricting the calculation to realistic regions according to the given problem setup. Properly understanding these limits ensures the desired portion of space is accurately depicted and evaluated in the triple integral.