Question

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

The value of the double integral is 0.

Step-by-step solution

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 \( x = 1 \).

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: \[ \int_{-1}^{1} \int_{x^2}^{1} xy \, dy \, dx. \]

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: \[ x \left[ \frac{1}{2}y^2 \right]_{x^2}^{1} = x \left( \frac{1}{2} \cdot 1^2 - \frac{1}{2} \cdot (x^2)^2 \right) = \frac{x}{2} (1 - x^4). \]

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: \[ \frac{1}{2} \int_{-1}^{1} (x - x^5) \, dx = \frac{1}{2} \left( \int_{-1}^{1} x \, dx - \int_{-1}^{1} x^5 \, dx \right). \]

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, \( \int_{-1}^{1} x^5 \, dx = \left[ \frac{x^6}{6} \right]_{-1}^{1} = \frac{1^6}{6} - \frac{(-1)^6}{6} = 0. \)

Key concepts

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 \( x \).

Such a structured approach enables tackling complex regions efficiently and is a key methodology used in multi-variable calculus.

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.

Understanding the boundaries of this region is crucial because these limits dictate the range for the iterated integral setup.

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 \( x \left( \frac{1}{2}\cdot 1^2 - \frac{1}{2}\cdot (x^2)^2 \right) \).

Antiderivatives simplify the process of finding areas or accumulations, which is essential in calculus.

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.

Correctly setting integration bounds is crucial to obtaining accurate results, as they determine the limits for both computation and interpretation.