Problem 19

Question

Set up the iterated integral for evaluating $\iiint_{D} f(r, \theta, z) d z r d r d \theta$ over the given region $D .$ $D$ is the prism whose base is the triangle in the $x y$ -plane bounded by the $x$ -axis and the lines $y=x$ and $x=1$ and whose top lies in the plane $z=2-y$

Step-by-Step Solution

Verified

Answer

Use the iterated integral $ \int_{0}^{\pi/4} \int_{0}^{\sec \theta} \int_{0}^{2 - r \sin \theta} f(r, \theta, z) \; dz \, r \, dr \, d\theta $.

1Step 1: Understand the Problem

We need to set up a triple integral to evaluate $\iiint_{D} f(r, \theta, z) \; d z \, r \, d r \, d \theta$ over the region $D$. $D$ is the region defined as a prism with a triangular base on the $xy$-plane.

2Step 2: Describe the Triangular Base in Polar Coordinates

The base of the prism is bounded by the lines $y=x$ (or $r \sin \theta = r \cos \theta$), $x=1$ (or $r \cos \theta = 1$), and the $x$-axis ($y=0$ or $\theta = 0$). In polar coordinates, the base is expressed as:- $0 \leq \theta \leq \frac{\pi}{4}$, - $0 \leq r \leq \sec \theta$.

3Step 3: Determine the Bounds for $z$

The upper boundary of $z$ is given by the plane $z = 2 - y$. In terms of polar coordinates, this becomes $z = 2 - r \sin \theta$. Since $z$ ranges from the $xy$-plane to this plane, we have:- $0 \leq z \leq 2 - r \sin \theta$.

4Step 4: Write the Iterated Integral

Combine the bounds into one iterated integral. The integral in terms of $r, \theta, z$ is:\[ \int_{0}^{\pi/4} \int_{0}^{\sec \theta} \int_{0}^{2 - r \sin \theta} f(r, \theta, z) \; dz \, r \, dr \, d\theta \]

Key Concepts

Polar CoordinatesTriple IntegralBounds of Integration

Polar Coordinates

Polar coordinates offer an alternative way to describe points in a plane, using distance and angle rather than horizontal and vertical positions like in Cartesian coordinates. In this system, any point can be described simply by:

Radial distance, $r$, from the origin.
Angular coordinate, $\theta$, measured from the positive $x$-axis.

For example, consider a point on the line $ y = x $ in Cartesian coordinates. In polar coordinates, this line is equivalent to $ r \sin \theta = r \cos \theta $, simplifying to $ \tan \theta = 1 $, which implies $ \theta = \frac{\pi}{4} $.
This conversion is particularly useful for regions that are symmetrical around a point or involve circular shapes.
When evaluating integrals over such regions, transforming into polar coordinates often simplifies the problem by aligning the coordinates with the geometry of the region.

Triple Integral

A triple integral, symbolized by $ \iiint $, allows us to integrate a function over a three-dimensional region. It generalizes the concept of iterated integrals to three dimensions, stacking one integral on top of another.
The process involves evaluating three nested integrals step by step:

An inner integral that considers variations in one dimension (e.g., $z$).
A middle integral that integrates the result over another dimension (e.g., $r$).
An outer integral that covers the last dimension (e.g., $\theta$).

In the specific exercise, our triple integral is used to evaluate the function $f(r, \theta, z)$ within a three-dimensional region. This prism-shaped region is defined by both its base in the $xy$-plane and its variable height described by $z = 2 - y$. This setup can handle a broad range of functions, giving us a flexible tool to calculate volumes or other quantities across complex shapes.

Bounds of Integration

Bounds of integration are critical in defining the limits over which we evaluate the integrals. They provide the necessary constraints that encompass the region of integration in each of the three dimensions.
In the context of our problem:

The angle $\theta$ ranges from $0$ to $\frac{\pi}{4}$, capturing the angular limits of the triangle in polar terms.
The radial boundary $r$ varies between $0$ and $\sec \theta$, describing the extent of the triangle from the origin outward.
The vertical limit $z$ is between $0$ and $2 - r \sin \theta$, representing the height from the base to the slanted top of the prism.

This comprehensive bounding system ensures that the entire region of integration is covered, enabling the correct computation of the integral across all dimensions. Understanding and setting these bounds correctly is essential to ensuring the accuracy of results.

Problem 19

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