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Magazine Chapter 15: Problem 56
Find the average value of \(f\) over region \(D .\) $$\begin{array}{l}{f(x, y)=x \sin y, \quad D \text { is enclosed by the curves } y=0} \\ {y=x^{2}, \text { and } x=1}\end{array}$$
Short Answer
Expert verified
\(\frac{3}{2}(1 - \sin(1))\)
Step by step solution
01
Understand the Problem and Define Region D
The region \( D \) is enclosed by the curves: \( y = 0 \), \( y = x^2 \), and \( x = 1 \). Therefore, \( D \) is the area between the parabola \( y = x^2 \) and the x-axis (\( y = 0 \)) from \( x = 0 \) to \( x = 1 \).
02
Set Up the Double Integral for Average Value
The formula for the average value of \( f \) over the region \( D \) is \( \frac{1}{A(D)} \int \int_D f(x, y) \, dA \), where \( A(D) \) is the area of region \( D \). First, find \( A(D) = \int_{0}^{1} (x^2 - 0) \, dx \).
03
Calculate the Area of Region D
Calculate \( A(D) \):\[A(D) = \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3}.\]
04
Set Up the Double Integral to Find \( \int \int_D f(x, y) \, dA \)
Integrate \( f(x, y) = x \sin y \) over \( D \):\[\int_0^1 \int_0^{x^2} x \sin y \, dy \, dx.\]
05
Evaluate the Inner Integral
Evaluate the inner integral with respect to \( y \):\[\int_0^{x^2} x \sin y \, dy = x \left[-\cos y \right]_0^{x^2} = x[-\cos(x^2) + 1].\]
06
Evaluate the Outer Integral
Evaluate the outer integral:\[\int_0^1 x(1 - \cos(x^2)) \, dx.\]Which can be split into two parts:\[\int_0^1 x \, dx - \int_0^1 x \cos(x^2) \, dx.\]
07
Solve the Integral Parts
Calculate:1. \( \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \).2. For \( \int_0^1 x \cos(x^2) \, dx \), perform substitution: let \( u = x^2 \), \( du = 2x \, dx \). The integral changes to:\[\frac{1}{2} \int_0^1 \cos(u) \, du = \frac{1}{2} \left[ \sin(u) \right]_0^1 = \frac{1}{2}\sin(1).\]So, the value is \( \frac{1}{2} \sin(1).\)
08
Calculate the Double Integral
Combine the results:\[\left(\frac{1}{2} - \frac{1}{2} \sin(1) \right) = \frac{1}{2}(1 - \sin(1)).\]
09
Calculate the Average Value of f
Divide the double integral by the area of the region:\[\text{Average value} = \frac{1}{A(D)} \int \int_D f(x, y) \, dA = 3(\frac{1}{2}(1 - \sin(1))) = \frac{3}{2}(1 - \sin(1)).\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integrals
Double integrals are a method used to compute volumes under surfaces or, more generally, to integrate over two-dimensional regions.
They work similarly to single-variable integrals but extend the concept to include an extra variable. With single-variable integrals, we find the area under a curve; with double integrals, we find volumes under surfaces or even more complex applications.
The process involves integrating a function of two variables, like \( f(x, y) \), over a specified region \( D \) in the xy-plane.
They work similarly to single-variable integrals but extend the concept to include an extra variable. With single-variable integrals, we find the area under a curve; with double integrals, we find volumes under surfaces or even more complex applications.
The process involves integrating a function of two variables, like \( f(x, y) \), over a specified region \( D \) in the xy-plane.
- The order of integration 鈥� whether first with respect to \( x \) or \( y \) 鈥� can be chosen based on the region's shape or simplifying the math.
- Double integrals can be represented with the notation \( \int \int_D f(x, y) \, dA \), where \( dA \) signifies a small area element within the region \( D \).
- They are particularly useful in calculating areas, volumes, and finding average values over regions.
Region of Integration
The region of integration refers to the specific area under which we want to evaluate a double integral.
Understanding the region is crucial since it dictates the limits of integration and the order.
In this problem, the region \( D \) is defined by the curves \( y = 0 \) (the x-axis), \( y = x^2 \) (a parabola), and the vertical line \( x = 1 \).
Understanding the region is crucial since it dictates the limits of integration and the order.
In this problem, the region \( D \) is defined by the curves \( y = 0 \) (the x-axis), \( y = x^2 \) (a parabola), and the vertical line \( x = 1 \).
- This creates a parabolic region where the integration takes place from \( x = 0 \) to \( x = 1 \), while \( y \) varies from \( 0 \) to \( x^2 \).
- The area \( A(D) \) of this region is given by integrating the difference between the top and bottom curves along the x-axis, \( \int_0^1 (x^2 - 0) \, dx \).
- This understanding helps us set up the boundaries for the double integration and is key to solving such problems correctly.
Parabolic Region
A parabolic region, such as the one described in this problem, is one bounded by a parabola and other lines/curves.
The parabola in this example is given by \( y = x^2 \), with other boundaries defined by the x-axis and a vertical line at \( x = 1 \).
Parabolic regions can uniquely characterize the limits of integration due to their symmetry and shape.
The parabola in this example is given by \( y = x^2 \), with other boundaries defined by the x-axis and a vertical line at \( x = 1 \).
Parabolic regions can uniquely characterize the limits of integration due to their symmetry and shape.
- These symmetries can sometimes simplify the integral calculations due to predictable patterns.
- Understanding how a parabola defines an area helps in integrating along its bounds, especially when finding the area under or bounded by it.
- By integrating from the x-axis up to the parabola, we measure the area encased within these bounding functions over the interval \( 0 \leq x \leq 1 \).
Trigonometric Integrals
Trigonometric integrals involve integrating trigonometric functions such as sine or cosine.
They often appear when the function to be integrated involves trigonometric terms, which is common in physics and engineering problems.
In our double integral, the function \( f(x, y) = x \sin y \) creates a need for integrating a trigonometric function with respect to \( y \), leading to solutions involving sine and cosine functions.
They often appear when the function to be integrated involves trigonometric terms, which is common in physics and engineering problems.
In our double integral, the function \( f(x, y) = x \sin y \) creates a need for integrating a trigonometric function with respect to \( y \), leading to solutions involving sine and cosine functions.
- For example, \( \int \sin y \, dy = -\cos y + C \), representing a standard result from integrating trigonometric functions.
- Solutions often require using specific techniques like substitution or transformation to simplify complex trigonometric integrals.
- These integrals often come together with chain or product rule derivatives when differentiating.
