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Magazine Chapter 13: Problem 15
In Problems \(15-20\), evaluate the given double integral by changing it to an iterated integral. $$ \iint_{S} x y d A ; S \text { is the region bounded by } y=x^{2} \text { and } y=1 \text { . } $$
Short Answer
Expert verified
The value of the double integral is 0.
Step by step solution
01
Understand the Region of Integration
We are given a region \( S \) bounded by \( y = x^2 \) and \( y = 1 \). This is a parabola and a horizontal line intersecting at points where \( x^2 = 1 \), i.e., \( x = -1 \) and \( x = 1 \). Thus, \( S \) is the region between these curves from \( x = -1 \) to \( x = 1 \).
02
Set Up the Iterated Integral
The double integral can be set up as an iterated integral. Since \( y = x^2 \) is below and \( y = 1 \) is above, for each fixed \( x \), \( y \) goes from \( x^2 \) to \( 1 \). Thus, the iterated integral is: \[ \int_{-1}^{1} \int_{x^2}^{1} xy \, dy \, dx. \]
03
Integrate with Respect to y
First, compute the inner integral with respect to \( y \): \[ \int_{x^2}^{1} xy \, dy. \]The antiderivative of \( y \) with respect to \( y \) is \( \frac{1}{2} y^2 \). Thus, the integral becomes: \[ x \left[ \frac{1}{2}y^2 \right]_{x^2}^{1} = x \left( \frac{1}{2} \cdot 1^2 - \frac{1}{2} \cdot (x^2)^2 \right) = \frac{x}{2} (1 - x^4). \]
04
Integrate with Respect to x
Now, integrate the result from the previous step with respect to \( x \): \[ \int_{-1}^{1} \frac{x}{2} (1 - x^4) \, dx. \]Distribute the \( \frac{x}{2} \) and evaluate: \[ \frac{1}{2} \int_{-1}^{1} (x - x^5) \, dx = \frac{1}{2} \left( \int_{-1}^{1} x \, dx - \int_{-1}^{1} x^5 \, dx \right). \]
05
Solve Each Integral Separately
Compute each integral separately.First, \( \int_{-1}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{1^2}{2} - \frac{(-1)^2}{2} = 0. \)Second, \( \int_{-1}^{1} x^5 \, dx = \left[ \frac{x^6}{6} \right]_{-1}^{1} = \frac{1^6}{6} - \frac{(-1)^6}{6} = 0. \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Iterated Integral
A crucial aspect of evaluating a double integral is transforming it into an iterated integral. This involves breaking down a complex double integral into simpler steps of single integrals, which are computed sequentially.
- The double integral in this case is given as \( \iint_{S} xy \, dA \) where \( S \) is a specific region in the plane.
- By converting the double integral into an iterated form, we can handle two separate single integral calculations. This process simplifies computation significantly.
- In our exercise, we set up the iterative integral as \( \int_{-1}^{1} \int_{x^2}^{1} xy \, dy \, dx \). Here, the integration with respect to \( y \) happens first, and then with respect to \( x \).
Region of Integration
For evaluating double integrals, identifying the correct region of integration is critical.
- A region of integration, \( S \), is the 2D area over which the integration process is applied.
- In our example, \( S \) is bounded by the parabola \( y = x^2 \) and the line \( y = 1 \), creating a segment between the parabola and the line.
- The intersection of these boundaries determines our limits for \( x \), which are from \( -1 \) to \( 1 \), as these are the \( x \)-values where the curve and line intersect.
Antiderivative
The concept of an antiderivative is foundational in integrating functions, particularly in finding the integral over a specific range or region.
- An antiderivative is essentially the reverse of the derivative of a function, often known as the indefinite integral.
- For instance, in our solution, we found the antiderivative of \( y \), which is \( \frac{1}{2}y^2 \).
- Once the antiderivative is known, the definite integral can be computed by applying it to the stated bounds. Here, it transforms \( \int_{x^2}^{1} xy \, dy \) into an evaluation of \( x \left( \frac{1}{2}\cdot 1^2 - \frac{1}{2}\cdot (x^2)^2 \right) \).
Integration Bounds
Integration bounds determine the limits across which a function is integrated and are vital for solving integrals accurately.
- These bounds specify where the integration starts, ends, and how the area of interest is confined within a specific region.
- Our example uses two bounds: the inner integral covers \( y \) from \( x^2 \) to \( 1 \), while the outer limits for \( x \) are from \( -1 \) to \( 1 \).
- When performing the integration, it's important to respect the bounds, ensuring that calculations encompass all intended regions but no more.
