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Sketch the following regions and write an iterated integral of a continuous function \(f\) over the region. Use the order \(d y d x\) \(R\) is the region in the first quadrant bounded by the \(y\) -axis and the parabolas \(y=x^{2}\) and \(y=1-x^{2}\)

Short Answer

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Question: Set up the iterated integral for a continuous function \(f\) over the region \(R\) enclosed by the parabolas \(y = x^2\) and \(y = 1 - x^2\) in the order dy dx. Found the region R by sketching and determining the intersection points of the two parabolas. Solution: \(\int_{0}^{\frac{1}{\sqrt{2}}} \int_{x^2}^{1-x^2} f(x, y) \, dy \, dx\)

Step by step solution

01

To find the intersection points, we set the two equations equal to each other: \(x^2 = 1-x^2\) Add \(x^2\) to both sides: \(2x^2 = 1\) Divide by 2: \(x^2 = \frac{1}{2}\) Take the square root of both sides: \(x = \pm \frac{1}{\sqrt{2}}\) These are the x-coordinates of the intersection points. To find the y-coordinates, plug the x-coordinates back into either of the parabolas. We'll use the first one: \(y = x^2\). \(y = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}\) So, the intersection points are \((-\frac{1}{\sqrt{2}}, \frac{1}{2})\) and \((\frac{1}{\sqrt{2}}, \frac{1}{2})\). Now, sketch the region R: it is the part of the first quadrant enclosed by the two parabolas and the y-axis. #Step 2: Set up the iterated integral of a continuous function f over the region R using the order dy dx#

Since we are using the order dy dx, we must first find the limits for y, which are determined by the given parabolas, and then find the limits for x, which are determined by the intersection points. For x values between 0 and \(\frac{1}{\sqrt{2}}\), the lower curve is given by \(y = x^2\) and the upper curve is given by \(y = 1 - x^2\). Thus, the limits for y are \(x^2\) and \(1 - x^2\). The limits for x are the intersection points on the x-axis, which are 0 and \(\frac{1}{\sqrt{2}}\). Now we can set up the iterated integral: \(\int_{0}^{\frac{1}{\sqrt{2}}} \int_{x^2}^{1-x^2} f(x, y) \, dy \, dx\) This is the iterated integral of a continuous function \(f\) over the region R in the order dy dx.

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Key Concepts

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

Continuous Function
A continuous function is a type of function where small changes in the input result in small changes in the output. This means there are no sudden jumps or breaks in the function's graph.
In mathematics, a function is considered continuous if, for every point within its domain, its limit equals the function's value at that point. If we can draw the graph of a function without lifting our pencil, then likely, it's continuous!
Iterated integrals often involve continuous functions because they ensure smoothness over the range of integration. This smooth behavior allows for easy calculation of total areas or volumes using techniques like integration.
When dealing with iterated integrals, the assumption of continuity simplifies the process of setting up and evaluating the integral.
First Quadrant
The first quadrant is one of the four sections of a Cartesian coordinate plane. It is the top-right section where both x and y coordinates are positive. This makes it a very important region for solving mathematical problems, especially in calculus and geometry.
When dealing with iterated integrals, specifying the region of interest like the first quadrant helps identify the bounds for integration. Here, problems are often simplified because we only consider positive values.
In the exercise, the region of interest in the first quadrant is bounded by parabolas and the y-axis. Identifying and focusing on this specific quadrant allows students to concentrate on the relevant parts of the functions involved in the problem.
Intersection Points
Intersection points are where two or more curves meet or cross each other. Finding these points usually involves solving a system of equations.
In our exercise, the intersection points are determined by setting the parabolas equal to each other: \[x^2 = 1 - x^2.\]Solving this equation tells us where the two parabolas intersect each other.
These points were found to be \((\pm \frac{1}{\sqrt{2}}, \frac{1}{2})\). As these are within the first quadrant, only the positive x-coordinate is relevant for setting up the integral.
This process of finding intersection points is crucial in defining the boundaries for integrating functions over a region. It tells us where one function transitions into another over a specific range.
Parabolas
Parabolas are symmetrical, U-shaped curves that are defined by quadratic equations. In the exercise, the parabolas are described by the equations:- \(y = x^2\)- \(y = 1 - x^2\)
These are important in the exercise because they define the boundaries of the area, or region R, over which we are integrating.
The parabola \(y = x^2\) opens upwards and starts from the origin, while \(y = 1 - x^2\) opens downwards with its vertex at the point \((0, 1)\).
Understanding the nature of each parabola helps in visualizing and sketching the region of interest, making it easier to set up the iterated integral over this bounded region.

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Most popular questions from this chapter

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