Question

In Exercises \(43-50\), sketch the region \(R\) whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area. $$ \int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} d x d y $$

Short answer

Working with these integrals, it is clear that the area represented by both is \(\pi/2\), confirming that switching the order of integration doesn't change the computed area.

Step-by-step solution

Sketch the Region \(R\)

The integral ranges are \(0\leq y\leq 1\) and \(-\sqrt{1-y^{2}}\leq x\leq \sqrt{1-y^{2}}\). The inner integral is set by a half circle's equation \( x^2 + y^2 =1 \), but only for positive \(y\) because \(0\leq y\leq 1\). Thus, the region \(R\) described is the upper half of the unit circle.

Evaluate the Given Integral

The given iterated integral is: \(\int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} d x d y\). When evaluated, the result equals \(\pi/2\), which is the area of the upper half of the unit circle.

Switching the Order of Integration

To switch the order of integration, first observe the region \(R\): for \( -1 \leq x \leq 1\) the variable \( y \) varies between \( y=\sqrt{1-x^{2}} \) and \( y=1 \). Thus, the equivalent integral becomes \(\int_{-1}^{1} \int_{\sqrt{1-x^{2}}}^{1} dy dx\).

Evaluate the Switched Integral

Evaluate the integral from the previous step. It also results in \(\pi/2\), thus confirming both integrals give the same area for region \(R\).

Key concepts

Order of Integration

The order of integration refers to the sequence in which multiple integrals are performed in an iterated integral. In a double integral, for instance, we integrate with respect to one variable first, then with respect to the second variable. Changing the order requires understanding the geometry of the region being integrated over.

Take for example the integral provided in the exercise, \[\begin{equation}\begin{aligned}\int_{0}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} d x d y\end{aligned}\end{equation}\]. Here, the integration with respect to 'x' is done first, followed by 'y'. But switching the order of integration might simplify the integral or be necessary for particular applications. To correctly switch the order, one must consider the bounds of integration which are determined by the region of interest.

The step-by-step solution demonstrates switching the order of integration by considering the bounds for 'x' and 'y', showing that no matter which order you choose to integrate, the area under the curve remains unchanged. This property is crucial in calculus as it often allows for simplifications in complex computations.

Sketching Regions in Calculus

Sketching regions in calculus serves as a visual aid to understand the integration limits and the area or volume under consideration. When dealing with iterated integrals, a clear sketch can inform the correct set up of the integral limits.

In the given exercise, the region is part of a circle. Recognizing this shape is vital because it influences the limits for 'x' and 'y'. The region, upper half of the unit circle, can be depicted graphically by drawing a semicircle above the x-axis. This visual representation makes it evident why the integration limits for 'x' are within the range \[\begin{equation} -\sqrt{1-y^{2}} \text{and} \sqrt{1-y^{2}}\end{equation}\]
The visualization also helps in the process of switching the integration order. By sketching the region, you can see how 'y' depends on 'x' for the switched order, \((-1 \leq x \leq 1)\), and change the limits accordingly. Having a sketch as a reference can make it easier to avoid errors and can be especially helpful in exams or while learning new concepts.

Evaluating Double Integrals

Evaluating double integrals involves integrating a function over a two-dimensional region. The process is broken down into iterating single integrals, first with respect to one variable, then with respect to the other.

To evaluate the given double integral, one would integrate with respect to 'x' from \(-\sqrt{1-y^{2}}\) to \(\sqrt{1-y^{2}}\), and then with respect to 'y' from 0 to 1. The careful execution of these steps will yield the area of the region, which in our case is \(\pi/2\), representing the area of the upper half of the unit circle.

Evaluating double integrals becomes more straightforward with practice, particularly as one becomes familiar with certain standard forms and geometrical shapes. Both steps in the exercise, whether evaluating the original order or the switched order of integration, must be processed correctly to ensure the same result, confirming the consistency of double integrals across different integration orders.