Question

(a) Sketch the region \(R\) given by the problem. (b) Set up the iterated integrals, in both orders, that evaluate the given double integral for the described region \(R\) (c) Evaluate one of the iterated integrals to find the signed volume under the surface \(z=f(x, y)\) over the region \(R .\) \(\iint_{R} x^{2}-y^{2} d A,\) where \(R\) is the rectangle with corners (-1,-1),(1,-1),(1,1) and (-1,1)

Short answer

The signed volume is 0.

Step-by-step solution

Sketch the Region

The region \( R \) is defined as a rectangle with corners at \((-1, -1)\), \((1, -1)\), \((1, 1)\), and \((-1, 1)\). Plot these points in the xy-plane to form the rectangular region. The sides are parallel to the coordinate axes, with the rectangle having a width of 2 units along the x-axis and a height of 2 units along the y-axis.

Set Up the Iterated Integral (x as Outer Integral)

For an iterated integral where the x-variable is the outer integral, the limits for x are [-1, 1], and the limits for y, being dependent within the x bounds, are [-1, 1] as well. Thus, the integral can be written as:\[ \int_{-1}^{1} \int_{-1}^{1} (x^2 - y^2) \, dy \, dx \]

Set Up the Iterated Integral (y as Outer Integral)

Conversely, with the y-variable as the outer integral, the limits for y are [-1, 1], and the limits for each x, dependent within those limits, are also [-1, 1]. Thus, the integral can be expressed as:\[ \int_{-1}^{1} \int_{-1}^{1} (x^2 - y^2) \, dx \, dy \]

Evaluate the Iterated Integral

Choose one of the iterated integral setups to evaluate. Let's evaluate \( \int_{-1}^{1} \int_{-1}^{1} (x^2 - y^2) \, dy \, dx \).**Inner Integral** (w.r.t. \( y \)):\[ \int_{-1}^{1} (x^2 - y^2) \, dy = \left[ x^2y - \frac{y^3}{3} \right]_{-1}^{1} \]Evaluating yields:\[ x^2(1) - \frac{1^3}{3} - (x^2(-1) - \frac{(-1)^3}{3}) = x^2 - \frac{1}{3} + x^2 - \left(-\frac{1}{3}\right) = 2x^2 - \frac{2}{3} \]**Outer Integral** (w.r.t. \( x \)):\[ \int_{-1}^{1} \left(2x^2 - \frac{2}{3}\right) \, dx = \left[ \frac{2x^3}{3} - \frac{2x}{3} \right]_{-1}^{1} \]Evaluating yields:\[ \left( \frac{2(1)^3}{3} - \frac{2(1)}{3} \right) - \left( \frac{2(-1)^3}{3} - \frac{2(-1)}{3} \right) = 0 \]

Key concepts

Iterated Integrals

In calculus, a double integral allows us to compute the volume under a surface in a three-dimensional space. To evaluate a double integral, we use iterated integrals. This method involves computing one integral at a time, following a specific order. For example, when evaluating a double integral over a region, you first integrate with respect to one variable while treating the other as constant. Then, you integrate the result with respect to the second variable.

The order of integration depends on the problem and the region over which the integration occurs. You can choose to integrate with respect to either variable first, depending on which setup simplifies the evaluation. In our exercise, we had two possibilities:

• First integrating with respect to \( y \) (inner integral) and then with respect to \( x \) (outer integral)
• Or first integrating with respect to \( x \) (inner integral) and then with respect to \( y \) (outer integral)

Whichever order you choose, the limits of integration must match the bounds of your defined region.

Rectangular Region

A rectangular region in the xy-plane is defined by two pairs of parallel lines. The corners of the rectangle correspond to the limits of our integration. For instance, in this exercise, the region \( R \) is outlined by the points \((-1, -1)\), \((1, -1)\), \((1, 1)\), and \((-1, 1)\). These points form a 2x2 square centered at the origin.

This rectangular setup simplifies the evaluation of the double integral because its sides align perfectly with the coordinate axes. When working with rectangular regions, the limits of integration for each variable correspond directly to the interval between these points. This alignment makes it easier to visualize the area over which you are integrating.

Signed Volume

The double integral over a region can be interpreted as calculating the signed volume under a surface \( z = f(x, y) \) over that region. In this context, 'signed volume' means the result can represent both positive and negative sums. Above the xy-plane, the volume is positive, while below, it is negative. This duality accounts for both the height and the depth beneath the surface regarding the xy-plane.

In the exercise, the surface \( z = x^2 - y^2 \) represents a saddle shape. By integrating this function over the rectangular region \( R \), the result of zero indicates a balance between positive and negative areas. The positive and negative volumes cancel each other out, leading to a total signed volume of zero.

Limits of Integration

When setting up an iterated integral, the limits of integration are crucial as they define the boundaries of the region over which you are integrating. In this context, the limits specify which part of the plane you are considering as your area of interest.

The limits are derived from the specific lines or curves that define the boundaries. Since our region \( R \) is rectangular, the limits are constant and straightforward. For both orders of integration considered in the exercise, the limits were \([-1, 1]\) for the variables \( x \) and \( y \). This symmetry around the origin simplifies the problem considerably.

Moreover, understanding and correctly setting these limits is fundamental because wrong limits will lead to incorrect results. Therefore, visualize the region clearly and extract the correct bounds to ensure precise evaluations.