Question

Set up an integral for both orders of integration, and use the more convenient order to evaluate the integral over the region \(R\). \(\int_{R} \int \frac{y}{1+x^{2}} d A\) \(R:\) region bounded by \(y=0, y=\sqrt{x}, x=4\)

Short answer

The value of the given double integral over region \(R\) is 0.

Step-by-step solution

Setting up Integral - Order \(dx~dy\)

Observe the region of integration \(R\), which is defined by \(y=0, y=\sqrt{x}, x=4\). Notice that \(y\) ranges from 0 to \(\sqrt{x}\) and \(x\) from 0 to 4. Therefore, the integral can be written as: \[ \int_{0}^{4} \int_{0}^{\sqrt{x}} \frac{y}{1+x^{2}} ~dy~dx \]

Setting up Integral - Order \(dy~dx\)

In this case, \(x\) ranges from \(y^2\) to 4 and \(y\) from 0 to 2. Therefore the integral can be written as: \[ \int_{0}^{2} \int_{y^2}^{4} \frac{y}{1+x^{2}} ~dx~dy \]

Choose convenient integration and Evaluate

It is evident that the integral in step 2 is more approachable because \(1/(1+x^2)\) is a known integration and it is easy to deal with. Now proceed with evaluating this integral: \[ \int_{0}^{2} \int_{y^2}^{4} \frac{y}{1+x^{2}} ~dx~dy = \int_{0}^{2} ~y~[arctan(x)]_{y^{2}}^{4} ~dy = 2[arctan(4) - arctan(y^{2})] \Big|_{0}^{2} = 2[arctan(4) - arctan(4)] = 0 \]

Key concepts

Iterative Integrals Setup

The setup of iterative integrals is a key component in evaluating double integrals. It involves deciding the sequence in which integration is performed over two variables. For the given integral,
\( \int_{R} \int \frac{y}{1+x^{2}} d A \),
the aim is to integrate the function \( \frac{y}{1+x^{2}} \) over a two-dimensional region 'R'. To do this, one needs to iterate the integral in terms of one variable first, while keeping the other variable constant, then integrate the result with respect to the second variable.

Importance of Correct Setup

Choosing the right order can greatly simplify the calculation. The setup is based on the given boundaries of the region within which integration is to take place. For instance, if the boundaries were expressed in terms of 'y', as is with the curve \( y = \sqrt{x} \), the integral would most naturally be written with 'dy' as the inner integral. However, if the boundaries were more naturally expressed in terms of 'x', the reverse might be more intuitive. The setup inherently defines the limits for each variable.

Integration Region

The integration region is the specific area in the two-dimensional plane over which the function is to be integrated. It's imperative for students to understand how to interpret and sketch this region, as it determines the limits of integration for each variable.
For the example at hand, region 'R' is defined by the curves \( y=0 \), \( y=\sqrt{x} \), and the vertical line \( x=4 \). The integration region lies between the x-axis (since \( y=0 \)) and the curve \( y=\sqrt{x} \), for 'x' values from 0 to 4. Visualizing this region is essential for setting up both orders of integration correctly.

Visual Representation

A sketch can greatly help in understanding the extent of integration. For our case, the region resembles a right triangle in the xy-plane with a curved hypotenuse. Correctly depicting and identifying the integration limits for 'x' and 'y' leads to establishing the proper bounds for the iterative integrals.

Integral Calculation

When calculating double integrals, the process involves two major steps: identifying the antiderivative of the inner integral and then integrating the result with respect to the outer variable.
For the given integral, evaluating the double integral requires finding an antiderivative of \( \frac{y}{1+x^2} \) with respect to 'x' first, then 'y'. Functions involving \( \frac{1}{1+x^2} \) often suggest the use of the arctangent function (\( \arctan(x) \)) in the antiderivative, which simplifies the evaluation process.

Execution of Calculation

Once the region and order of integration are set, you compute the inner integral, followed by the outer integral. If the integrand simplifies considerably, which is the case when integrating with respect to 'x' in our example, the process becomes much more straightforward. This simplification can result in easier final integration and, in this specific case, reveals that the value of the integral is zero.

Integration Order

The order of integration—whether to integrate with respect to 'x' before 'y' or vice versa—can greatly affect the difficulty level of the integral calculation.
In the example provided, the two possible orders were \( dx~dy \) and \( dy~dx \). The key factor to choosing the order is looking for the one that simplifies calculation.

Factors Affecting the Choice

Certain integrands or regions may lead to simpler antiderivatives with one order over the other. The chosen order \( dy~dx \) turns out to be more convenient because the corresponding antiderivative involves a simple constant 'y' and a well-known function of 'x' (\( \arctan(x) \)). When integrated, this leads to an immediate evaluation of the definite integral. On the contrary, if the order \( dx~dy \) is chosen, the antiderivative is not as straightforward, and it will lead to a more complex integration process.