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Chapter 16: Problem 43

Evaluating integrals Evaluate the following integrals. A sketch is helpful. \(\iint_{R} x y d A ; R\) is bounded by \(x=0, y=2 x+1,\) and \(y=-2 x+5\)

Short Answer

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Based on the solution above, write a short answer question about the topic or problem: Question: Evaluate the double integral of \(xy\) over the region \(R\), bounded by \(x=0, y=2x+1,\) and \(y=-2x+5\). Answer: 2

Step by step solution

01

Find the points of intersection of the boundary lines

The three boundary lines of the given region R are: 1. \(x = 0\) 2. \(y = 2x + 1\) 3. \(y = -2x + 5\) First, let's find the point of intersection between lines 2 and 3. Solve these lines' equations simultaneously: \begin{align*} 2x + 1 &= -2x + 5 \\ 4x &= 4 \\ x &= 1 \end{align*} Substitute \(x=1\) into any of the lines (let's take line 2): $$ y = 2(1) + 1 = 3 $$ So, the point of intersection between lines 2 and 3 is \((1, 3)\). Now, let's find the points of intersection of line 1 with lines 2 and 3. Since line 1 represents the y-axis, by plugging \(x=0\) into lines 2 and 3, we will get the y-coordinate of intersection points. For line 2: $$ y = 2(0) + 1 = 1 $$ Therefore, the intersection point between lines 1 and 2 is \((0,1)\). For line 3: $$ y = -2(0) + 5 = 5 $$ Thus, the intersection point between lines 1 and 3 is \((0, 5)\).
02

Set up the double integral

Now that we have identified the intersection points, we can determine the limits of integration for the given double integral. The region is bound by \(x\) from 0 to 1 and \(y\) from \(2x + 1\) to \(-2x + 5\). Thus, the double integral can be written as: $$ \iint_R xy\, dA = \int_0^1\int_{2x+1}^{-2x+5} xy\, dydx $$
03

Evaluate the double integral

Now we will evaluate the double integral by integrating first with respect to \(y\) and then with respect to \(x\). For the inner integral: $$ \int_{2x + 1}^{-2x + 5} xy\, dy = \frac{1}{2}xy^2\Big|_{2x + 1}^{-2x + 5} = \frac{1}{2}x((-2x + 5)^2 - (2x + 1)^2) $$ Now, we will evaluate the outer integral: \begin{align*} \int_0^1 \frac{1}{2}x((-2x + 5)^2 - (2x + 1)^2)\, dx &= \frac{1}{2}\int_0^1 x(((-2x + 5)^2 - (2x + 1)^2)\, dx \\ &= \frac{1}{2}\int_0^1 x((4x^2 - 20x + 25) - (4x^2 + 4x + 1))\, dx \\ &= \frac{1}{2}\int_0^1 x(-24x + 24)\, dx \\ &= \frac{1}{2}\int_0^1 (-24x^2 + 24x)\, dx \end{align*} Now, we will integrate the terms with respect to \(x\): $$ \frac{1}{2}\left[(-8x^3 + 12x^2)\Big|_0^1\right] = \frac{1}{2}((-8 + 12) - 0) = \frac{1}{2}(4) = 2 $$ Therefore, the value of the double integral is: $$ \iint_R xy\, dA = 2.

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Key Concepts

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

Double Integrals
When we talk about double integrals, we're delving into the realm of multivariable calculus. These integrals are used to sum up the effects of two varying quantities 鈥� offering a way to calculate volumes under surfaces or areas of regions in a plane, much like single integrals, but in two dimensions.

Imagine a flat surface with hills and valleys. If you wanted to fill those hills and valleys with water to make a perfectly flat lake, the amount of water needed would be the 'volume' under the surface. Mathematically, finding this volume is what a double integral allows us to compute.

Essentially, a double integral, denoted by the symbol \( \iint \) or \( \iint_R \) when referring to a specific region R, works by accumulating quantities over a two-dimensional region. This is achieved by taking an integral inside an integral: the inner integral sums across one dimension, and the outer integral adds up those sums across the second dimension.
Limits of Integration
Determining the limits of integration for a double integral is akin to drawing the boundaries on a map within which a treasure hunt is confined. These limits indicate the precise region where our calculations will play out. In double integrals, we have two sets of limits: one for each variable.

With our given problem, we start with the outer limits for the variable \( x \) that run from 0 to 1. These values were derived from the intersection of the curves or lines that bound our region R in the \( x \) direction. Inside this, we find the inner limits for \( y \) that depend on \( x \) and are given by the equations \( 2x+1 \) and \( -2x+5 \) 鈥� these are drawn from the specific constraints of the problem, showing where the region is cut off in the \( y \) direction.

A crucial point here is that these limits can sometimes be swapped, a technique known as changing the order of integration, which can simplify the process or make it possible to solve some more complex integrals. However, this also involves reassessing the limits according to the new order, which must always reflect the actual physical or geometric bounds of the problem at hand.
Area Bounded by Curves
When we evaluate double integrals, particularly in problems where we are finding the area of a region, we need to clearly understand the area bounded by curves. Think of this as outlining the edges of a piece of oddly shaped land on a map. These borders, given by mathematical functions or lines, tell us exactly where the region starts and ends.

In the problem given, we have three lines that come together to form a triangular region. This is our defined 'piece of land'. By solving for where these lines intersect, we essentially mark the corners of our triangular region. The space inside this triangle is R, the region of integration.

The art of sketching this area cannot be understated. It lets students visualize the problem, converting abstract ideas into concrete shapes. Once drawn, it becomes easier to determine how to set up our double integral, because we can see just how the region lies in the plane, ensuring that we correctly apply the limits of integration.

In conclusion, identifying the area bounded by curves necessitates a grasp of geometry and algebra to find intersection points which are integral to establishing the framework for our double integral calculations.

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Most popular questions from this chapter

Solids bounded by hyperboloids Find the volume of the solid below the hyperboloid \(z=5-\sqrt{1+x^{2}+y^{2}}\) and above the following polar rectangles. $$R=\\{(r, \theta): \sqrt{3} \leq r \leq \sqrt{15},-\pi / 2 \leq \theta \leq \pi\\}$$

Sketch the following regions \(R\). Then express \(\iint_{R} g(r, \theta) d A\) as an iterated integral over \(R\) in polar coordinates. The region outside the circle \(r=1 / 2\) and inside the cardioid \(r=1+\cos \theta\)

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 wedge cut from the cardioid cylinder \(r=1+\cos \theta\) by the planes \(z=2-x\) and \(z=x-2\).

Mass from density data The following table gives the density (in units of \(\mathrm{g} / \mathrm{cm}^{2}\) ) at selected points (in polar coordinates) of a thin semicircular plate of radius \(3 .\) Estimate the mass of the plate and explain your method. $$\begin{array}{|c|c|c|c|c|c|}\hline & \boldsymbol{\theta}=\mathbf{0} & \boldsymbol{\theta}=\pi / 4 & \boldsymbol{\theta}=\pi / 2 &\boldsymbol{\theta}=3 \pi / 4 & \boldsymbol{\theta}=\pi \\\\\hline r=1 & 2.0 & 2.1 & 2.2 & 2.3 & 2.4 \\\\\hline r=2 & 2.5 & 2.7 & 2.9 & 3.1 & 3.3 \\\\\hline r=3 & 3.2 & 3.4 & 3.5 & 3.6 & 3.7 \\\\\hline\end{array}$$

Improper integrals Improper integrals arise in polar coordinates when the radial coordinate r becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} g(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} g(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\iint_{R} \frac{d A}{\left(x^{2}+y^{2}\right)^{5 / 2}} ; R=\\{(r, \theta): 1 \leq r<\infty, 0 \leq \theta \leq 2 \pi\\}$$
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