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Chapter 16: Problem 32

Regions of integration Sketch each region \(R\) and write an iterated integral of a continuous function \(f\) over \(R\). Use the order dy dr. \(R\) is the triangular region with vertices \((0,0),(0,2),\) and (1,1)

Short Answer

Expert verified
Question: Set up the iterated integral of the continuous function f over the triangular region R with vertices (0,0), (0,2), and (1,1) in the order dy dr. Answer: $$\int_{0}^{1}\int_{0}^{-x+2}f(r, y) \, dy \, dr$$

Step by step solution

01

Sketch the triangular region R

First, we will sketch the triangular region R with vertices (0,0), (0,2), and (1,1). To do this, plot the given points on a coordinate plane and connect them to form a closed triangular region.
02

Find the equations of the lines bounding R

Next, we need to find the equations of the lines that form the sides of the triangular region R. The lines are: 1. Line connecting (0, 0) and (0, 2): This line is vertical and passes through x = 0, so the equation of this line is \(x = 0\). 2. Line connecting (0, 2) and (1, 1): First, find the slope: \((1-2)/(1-0) = -1\). The equation for this line will be in the form \(y = mx + b\), so \(y = -1x + b\). To find the y-intercept (b), plug in the point (1,1): \(1 = -1(1) + b\) gives us \(b = 2\). So, the equation of this line is \(y = -x + 2\). 3. Line connecting (0, 0) and (1,1): This line has slope \((1-0)/(1-0) = 1\) and passes through the point (0, 0), so its equation is \(y = x\).
03

Set up the iterated integral in the order dy dr

Now we'll set up the iterated integral of the continuous function f over the triangular region R in the order dy dr. To do this, we need to determine the limits of integration based on the triangular region's boundaries: 1. The outer integral: \(R\) is bounded by the lines \(x = 0\) and \(y = x\). So the limits of integration for \(dr\), the outer integral, are from 0 to y. 2. The inner integral: To find the limits of integration for \(dy\), the inner integral, we look at the vertical extent of the region bounded by the lines \(y = -x + 2\) and \(y = 0\). So the limits of integration for \(dy\) are from 0 to \(-x+2\). Hence, the iterated integral of the continuous function f over the triangular region R in the order dy dr can be written as: $$\int_{0}^{1}\int_{0}^{-x+2}f(r, y) \, dy \, dr$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Region of Integration
When working with iterated integrals, identifying the region of integration is paramount. In our case, we are dealing with a triangular region, which means the region that defines where the integration happens is shaped like a triangle.
To fully understand this region:
  • Visualize the vertices: Here, we have the points \( (0,0), (0,2), (1,1) \), which define the corners of our triangle.
  • Sketching the region: Plot these points on a coordinate plane. Connecting them creates the triangular region where the integration will take place.
  • Closed region: The area within these connected points is the closed region that gives us our boundary for integration.
The region of integration confines the variables within set boundaries, essential for setting up the integral correctly.
Limits of Integration
Once the region of integration is established, the next step is determining the limits of integration. This will guide how integration is performed over the given region.
The limits are derived from the boundaries of the integration region. For a triangular region:
  • Vertical Extents: We look at how the region stretches vertically, from bottom to top. For each point along the x-axis, the value of y changes.
  • y-integral limits: In our example, we observe the range in the y-direction, going from the bottom line \( y = 0 \) to the top line \( y = -x + 2 \).
  • x-integral limits: The outer integral spans from \( x = 0 \) to \( x = 1 \), covering the horizontal extent of the triangular region.
These limits of integration enable us to set boundaries for iterating through the calculations over the specific region.
Triangular Region
The triangular region in integration problems is a common and significant concept. Defining such regions requires careful plotting and understanding of the geometric shape.
  • Vertices Coordinates: The triangular region is entirely defined by its vertices. For our example, the coordinates are \( (0,0), (0,2), (1,1) \).
  • Defining Lines: Each side of the triangle comes from a line formed between two vertices. One line may be vertical, while others might have positive or negative slopes, defining the outer structure of the triangle.
  • Geometric Representation: Visualising the triangle helps in understanding its structure, which is crucial for setting up integrals as it dictates the limits of the integration.
Understanding triangular regions is important, especially in integration tasks, as it impacts how the function interacts with the boundaries.

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Most popular questions from this chapter

Intersecting spheres One sphere is centered at the origin and has a radius of \(R\). Another sphere is centered at \((0.0, r)\) and has a radius of \(r,\) where \(r>R / 2 .\) What is the volume of the region common to the two spheres?

Normal distribution An important integral in statistics associated with the normal distribution is \(I=\int_{-\infty}^{\infty} e^{-x^{2}} d x .\) It is evaluated in the following steps. a. In Section \(8.9,\) it is shown that \(\int_{0}^{\infty} e^{-x^{2}} d x\) converges (in the narrative following Example 7 ). Use this result to explain why \(\int_{-\infty}^{\infty} e^{-x^{2}} d x\) converges. b. Assume $$I^{2}=\left(\int_{-\infty}^{\infty} e^{-x^{2}} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} d y\right)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y,$$ where we have chosen the variables of integration to be \(x\) and \(y\) and then written the product as an iterated integral. Evaluate this integral in polar coordinates and show that \(I=\sqrt{\pi}\) Why is the solution \(I=-\sqrt{\pi}\) rejected? c. Evaluate \(\int_{0}^{\infty} e^{-x^{2}} d x, \int_{0}^{\infty} x e^{-x^{2}} d x,\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\) (using part (a) if needed).

Choose the best coordinate system and find the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. The solid inside the cylinder \(r=2 \cos \theta,\) for \(0 \leq z \leq 4-x\).

Density distribution A right circular cylinder with height \(8 \mathrm{cm}\) and radius \(2 \mathrm{cm}\) is filled with water. A heated filament running along its axis produces a variable density in the water given by \(\rho(r)=1-0.05 e^{-0.01 r^{2}} \mathrm{g} / \mathrm{cm}^{3}(\rho\) stands for density here, not for the radial spherical coordinate). Find the mass of the water in the cylinder. Neglect the volume of the filament.

Consider the following two- and three-dimensional regions with variable dimensions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid is enclosed by a hemisphere of radius \(a\). How far from the base is the center of mass?
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