Question

Iterated integrals are given that compute the area of a region \(R\) in the \(x\) -y plane. Sketch the region \(R\), and give the iterated integral(s) that give the area of \(R\) with the opposite order of integration. $$ \int_{-2}^{2} \int_{0}^{4-x^{2}} d y d x $$

Short answer

Region \( R \) is under the parabola \( y = 4-x^2 \), and the integral with reversed order is \( \int_{0}^{4} \int_{-\sqrt{4-y}}^{\sqrt{4-y}} d x \, d y \).

Step-by-step solution

Interpret the Given Integral

The given iterated integral \( \int_{-2}^{2} \int_{0}^{4-x^{2}} d y d x \) represents the area of a region \( R \) in the \( x-y \) plane. The outer integral limits \( -2 \) to \( 2 \) apply to \( x \), and the inner integral limits \( 0 \) to \( 4 - x^2 \) apply to \( y \). This describes an area bound vertically from \( y = 0 \) to \( y = 4 - x^2 \).

Sketch the Region R

To sketch the region \( R \), note that \( y = 4 - x^2 \) is a downward-opening parabola factored for \( x \) values between \( -2 \) and \( 2 \). The line \( y = 0 \) is the x-axis. Thus, \( R \) is the area under the parabola \( y = 4 - x^2 \) and above the x-axis, for \( x \) from \( -2 \) to \( 2 \).

Determine Bounds for Opposite Order of Integration

The region \( R \) is defined by \( y \) ranging from \( 0 \) to \( 4 \). At any fixed \( y \) between these bounds, solve for \( x \) in \( y = 4 - x^2 \), giving \( x = \pm \sqrt{4-y} \). Since \( x \) ranges from \( -\sqrt{4-y} \) to \( \sqrt{4-y} \) for each \( y \), these are the new bounds for \( x \).

Write Iterated Integral for Opposite Order

Switching the order of integration involves integrating with respect to \( x \) first. So, the iterated integral becomes \( \int_{0}^{4} \int_{-\sqrt{4-y}}^{\sqrt{4-y}} d x \, d y \). This integral expresses the same region \( R \) but with \( y \) as the outside limit.

Key concepts

Iterated Integrals

Iterated integrals are a fundamental concept in multivariable calculus, specifically when dealing with double integrals. They are used to calculate quantities over a multi-dimensional region, such as area or volume. An iterated integral involves breaking down a multi-dimensional integral into a series of single integrals, evaluated one after the other. In the given problem, you've encountered an example of a double integral where the iterated integral format helps to find the area of the region. You start by integrating with respect to one variable, in this case, first with respect to \(y\), while holding the other variable \(x\) constant, then integrate the resulting function with respect to \(x\). This step-by-step integration is akin to peeling an onion in layers; you handle one piece at a time.The given integral \[ \int_{-2}^{2} \int_{0}^{4-x^{2}} dy \, dx \] shows the process of first integrating between the limits \(0\) and \(4 - x^2\) for \(y\), followed by integrating \(x\) from \(-2\) to \(2\). By breaking down this integration into smaller, more manageable parts, it makes it easier to calculate the area of complex regions.

Area of Regions

Calculating the area of regions is a common application of double integrals and is pivotal in understanding multivariable calculus. The area is found by setting up a double integral across the desired region. In the context of this exercise, the region \(R\) described is under a parabola given by the equation \(y = 4 - x^2\) and above the \(x\)-axis, within the limits of \(x = -2\) to \(x = 2\). The iterated integral helps encapsulate this computation by providing clearly defined boundaries for where and how the area should be calculated.To visualize it, sketch the curve \(y = 4 - x^2\), a downward-opening parabola intersecting the \(x\)-axis at \((x = -2, y = 0)\) and \((x = 2, y = 0)\). The area under the curve and above the \(x\)-axis forms the region whose area is calculated by the iterated integral given.This type of problem embodies the principle of "area under the curve" in two-dimensional space, which becomes essential in various applications, such as physics, engineering, and probability.

Order of Integration

The "order of integration" refers to the sequence in which you integrate the variables in an iterated integral. In double integrals, you can choose to integrate with respect to \(x\) first or \(y\) first, depending on how the region is described or sometimes for ease of calculation. In the initial integral provided, the integration is conducted in the order of \(dy \, dx\), which means \(y\) integration is performed first, followed by \(x\). Yet, certain problems or regions might be more straightforwardly approached by changing this order.For the opposite order of integration, \(dx \, dy\), switch the order to integrate with respect to \(x\) first, followed by \(y\). When changing the order, ensure the limits reflect the new orientation correctly. The solution adjusted the integral to \[ \int_{0}^{4} \int_{-\sqrt{4-y}}^{\sqrt{4-y}} dx \, dy \] to account for the boundaries that \(x\) moves from \(-\sqrt{4-y}\) to \(\sqrt{4-y}\) for every fixed \(y\) within the bounds of \(0\) to \(4\).Reordering integration can simplify complex expressions or make solving more feasible under certain constraints or conditions.