Problem 12 - Calculus Early Transcendentals

Question

Sketch each region and write an iterated integral of a continuous function \(f\) over the region. Use the order \(d y d x\). $$R=\left\\{(x, y): 0 \leq x \leq 4, x^{\prime} \leq y \leq 8 \sqrt{x}\right\\}$$

Step-by-Step Solution

Answer

Based on the step-by-step solution provided, create a short answer: To find the iterated integral of a continuous function \(f(x, y)\) over the given region, we first need to analyze the boundaries for \(x\) and \(y\). We find that the region is represented by \((x, y): 0 \leq x \leq 4, x^{\prime} \leq y \leq 8\sqrt{x}\), with \(x'\) being \((y/8)^2\). Next, sketch the region by drawing the graphs of \(x'= (y/8)^2\) and \(y=8\sqrt{x}\) and identifying the region bounded by these curves and the given boundaries. Finally, we can write the iterated integral using the order \(d y d x\) as follows: $$\int_{0}^{4}\int_{(\frac{y}{8})^2}^{8\sqrt{x}} f(x, y) dy\, dx$$

Step 1: Finding \(x^\prime\)

First, we need to find the function represented by \(x'\). Since we are given that \(x'\leq y\leq8\sqrt{x}\), examining the upper bound for \(y\) might help us find \(x'\). To find \(x'\), let's consider the equality case: \(y=8\sqrt{x}\). Let's solve this equation for \(x\): \(y=8\sqrt{x}\\Rightarrow y/8=\sqrt{x}\\Rightarrow x=(y/8)^2\) Now, we have \(x'= (y/8)^2\).

Step 2: Sketch the Region R

To sketch the region, we need to draw the graphs of \(x'= (y/8)^2\) and \(y=8\sqrt{x}\), and identify the region bounded by these curves and the given boundaries for \(x\) and \(y\). The graph of \(y=8\sqrt{x}\) is a horizontal dilation of \(y=\sqrt{x}\) by a factor of 8, while the graph of \(x'=(y/8)^2\) is a vertical dilation of \(x=y^2\) by a factor of 1/8. Using the given boundaries: \(0\leq x\leq 4\) and \(x'\leq y\leq 8\sqrt{x}\), we can now sketch the region R.

Step 3: Writing the Iterated Integral

Now that we have the boundaries, we can write the iterated integral using the order \(d y d x\). The integral of a continuous function \(f(x, y)\) over the region R is: $$\int_{0}^{4}\int_{(\frac{y}{8})^2}^{8\sqrt{x}} f(x, y) dy\, dx$$

Key Concepts

Continuous Function

Understanding the behavior of continuous functions is crucial when dealing with calculus problems. A continuous function is one that does not have any abrupt changes in value—there are no holes, jumps, or asymptotes in its graph within its domain. Imagine drawing the graph of such a function without lifting your pen off the paper. This property is especially important when evaluating integrals, as the integral calculus is grounded in the concept of accumulating continuous quantities.

In the context of iterated integrals, a continuous function over a region implies that the function can be integrated smoothly across the entire area without encountering any discontinuities. When given an iterated integral, it's assumed the function f(x, y) mentioned in the exercise is continuous over the region of integration. This allows us to confidently proceed with integration without worrying about undefined values or need for partitioning the region.

Region of Integration

The region of integration is an essential concept when setting up iterated integrals. It is the area over which we integrate a function. To comprehend it, we visualize the 'space' or 'patch' on the xy-plane that we're interested in. In simple terms, think of it as your 'field of operation' for the integration process. You are being instructed where to 'paint' with your integration brush.

In this example, the region of integration is defined by inequalities involving x and y, specifically, 0 ≤ x ≤ 4 and (y/8)^2 ≤ y ≤ 8√x. The integral symbol along with its upper and lower limits, encapsulates the region of integration and tells us to 'sum up' the function's values within this specified area. The beauty of setting up an iterated integral as provided here, with the dy coming before dx, is that it nicely slices the region horizontally first, and then accumulates these slices vertically across the x-range.

Sketching Regions in Calculus

Effective sketching of regions in calculus is a skill that often helps visualize complex problems and consequently simplifies the integration process. When approaching a problem involving an iterated integral, the first step is often to sketch the region in the xy-plane so that the bounds of integration become clear. Remember, a good sketch doesn't have to be a piece of art, but it should correctly represent the relevant features of the region.

To sketch the given region for our exercise, you start by drawing the curves that form its boundaries. These curves are y = 8√x, which delineates the outer limit, and x = (y/8)^2, which provides the inner edge of the region. Once the curves are drawn, include the straight-line boundaries: the vertical lines at x = 0 and x = 4. The area that lies within all these bounding lines constitutes the region over which we integrate our function. Such sketches are invaluable, acting as a visual guide, ensuring that we set up the limits for the iterated integral correctly.