Chapter 13: Problem 50
Evaluate the following integrals. A sketch is helpful. \(\iint_{R}(x+y) d A ; R\) is bounded by \(y=|x|\) and \(y=4\).
Short Answer
Expert verified
The result of the double integral over the given region R is approximately 42.668.
Step by step solution
01
Sketch the region R
Before solving the integral, we need to sketch the region R. The region is bounded by \(y=|x|\) and \(y=4\). When \(x\) is positive, \(y=|x|\) is equal to \(y=x\). When \(x\) is negative, \(y=|x|\) is equal to \(y=-x\). The line \(y=4\) is a horizontal line.
02
Determine the limits of integration
To find the limits of integration, we need to find the points of intersection between the functions that bound the region R. When \(y=|x|\) intersects with \(y=4\):
For the positive side, \(x = y\), so the intersection point is \((4, 4)\). For the negative side, \(y = -x\), so the intersection point is \((-4, 4)\).
So, our limits of integration are as follows:
- For x, it ranges from \(-4\) to \(4\),
- For y, it ranges from \(|x|\) to \(4\).
03
Set up the integral
Now that the limits of integration have been determined, the double integral for the given region is:
$$\int_{-4}^{4}\int_{|x|}^{4}(x+y)dydx$$
04
Evaluate the inner integral
Let's evaluate the inner integral, keeping x constant:
$$\int_{|x|}^{4}(x+y)dy = \left[ xy + \frac{1}{2}y^2 \right]_{|x|}^4$$
Now, we substitute the limits of integration:
$$\left[ 4x + 8 - x|x| - \frac{1}{2}x^2 \right]$$
05
Evaluate the outer integral
Now, we will focus on the outer integral by plugging in the result of the inner integral:
$$\int_{-4}^{4} \left[ 4x + 8 - x|x| - \frac{1}{2}x^2 \right] dx$$
This integral can be separated into the sum of regular integrals:
$$\int_{-4}^{4} 4x dx + \int_{-4}^{4} 8 dx - \int_{-4}^{4} x|x| dx - \int_{-4}^{4} \frac{1}{2}x^2 dx$$
06
Evaluate the separated integrals
Now, let's evaluate each of these integrals one by one:
$$\int_{-4}^{4} 4x dx = \left[ 2x^2 \right]_{-4}^{4} = 32 - (-32) = 64$$
$$\int_{-4}^{4} 8 dx = \left[ 8x \right]_{-4}^{4} = 32 - (-32) = 64$$
For the third integral, we need to partition it due to the absolute value of x into two different intervals:
$$-\int_{-4}^{0} x(-x) dx + \int_{0}^{4} x(x) dx = -\int_{-4}^{0} x^2 dx + \int_{0}^{4} x^2 dx$$
$$= \left[ -\frac{1}{3}x^3 \right]_{-4}^{0} + \left[ \frac{1}{3}x^3 \right]_{0}^{4} = 21.333 + 21.333 = 42.666$$
$$\int_{-4}^{4} \frac{1}{2}x^2 dx = \left[ \frac{1}{6}x^3 \right]_{-4}^{4} = 21.333 -(-21.333) = 42.666$$
07
Combine the results and find the final answer
Now, we will combine these individual results to get the final answer:
$$64 + 64 - 42.666 - 42.666 = 128 - 85.332 = 42.668$$
So, the result of the double integral is approximately \(42.668\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrals Evaluation
The process of finding the integral of a function is known as integral evaluation. In the realm of double integrals, this involves calculating the volume under a surface over a given area. To perform the evaluation, you often need to compute an inner and an outer integral in succession. Each integral has a specific variable that you integrate with respect to, while treating other variables as constants.
The inner integral solves for one variable, often representing the function sliced at a certain point, while the outer integral sums up these slices across the region of interest. Steps such as substitution of integration limits and combining results are essential. In this context, the calculated number can represent various physical quantities, for instance, the mass of a material with a given density over an area.
The inner integral solves for one variable, often representing the function sliced at a certain point, while the outer integral sums up these slices across the region of interest. Steps such as substitution of integration limits and combining results are essential. In this context, the calculated number can represent various physical quantities, for instance, the mass of a material with a given density over an area.
Integration Limits
Choosing the correct integration limits is a critical step in double integral calculus. These limits define the boundaries of the region over which you are integrating. For definite integrals, the limits are numerical values that indicate where the integration begins and ends.
When setting up a double integral, the limits for the outer integral represent the starting and ending values along one axis, and the limits for the inner integral depend on the other axis, which can sometimes be a function of the outer variable. It's important to identify the right bounds to ensure the area being integrated over accurately represents the problem's physical or geometric context. Finding points of intersection between boundary curves, as in the case of the exercise, helps in establishing these boundaries.
When setting up a double integral, the limits for the outer integral represent the starting and ending values along one axis, and the limits for the inner integral depend on the other axis, which can sometimes be a function of the outer variable. It's important to identify the right bounds to ensure the area being integrated over accurately represents the problem's physical or geometric context. Finding points of intersection between boundary curves, as in the case of the exercise, helps in establishing these boundaries.
Absolute Value
Absolute value in calculus often indicates a distance from the origin or a non-negative value of a number, regardless of its original sign. In integration, the presence of an absolute value in the integrand or limits can signify a symmetry or a piecewise-defined function, which may need to be broken down into separate intervals for evaluation.
For example, when integrating a function that includes an absolute value, such as in the given exercise where the limit of integration was the absolute value of x, it's typical to split the integral at the point where the enclosed expression changes sign (usually at zero). From there, you handle each interval accordingly to its sign, which often simplifies the integration process.
For example, when integrating a function that includes an absolute value, such as in the given exercise where the limit of integration was the absolute value of x, it's typical to split the integral at the point where the enclosed expression changes sign (usually at zero). From there, you handle each interval accordingly to its sign, which often simplifies the integration process.
Area Under a Curve
In single-variable calculus, the area under a curve is found using definite integrals. The extension of this concept to functions of two variables uses double integrals. The 'curves' in question become surfaces, and the 'area under' a curve becomes the volume under a surface above a specified region in the plane.
The computation combines the individual areas of infinitesimally thin slices parallel to the y-axis (or x-axis, depending on the order of integration), over the interval defined by the integration limits. In the context of our exercise, the area under the curve was effectively calculated by the double integral over the region bounded by the lines defined by the absolute value of x and the constant value y = 4.
The computation combines the individual areas of infinitesimally thin slices parallel to the y-axis (or x-axis, depending on the order of integration), over the interval defined by the integration limits. In the context of our exercise, the area under the curve was effectively calculated by the double integral over the region bounded by the lines defined by the absolute value of x and the constant value y = 4.