Question

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

Short answer

The most convenient order to solve the integral is Order 2 (dydx). The final integral after choosing this order becomes \( \int_0^2 y(tan^{-1}(4) - tan^{-1}(y^2))dy \), which can be solved using ordinary integration methods to obtain the result.

Step-by-step solution

Understand the region R

The region \(R\) is given by the bounds \(y=0\), \(y=\sqrt{x}\), and \(x=4\). So, it is the area under the curve \(y=\sqrt{x}\), above the x-axis (since \(y=0\)), and to the left of the line \(x=4\). This is a right-ward opening parabola.

Write the integral in two orders

To integrate, we can either integrate first with respect to \(x\) inside the integral and then \(y\), or vice versa. These are the two different orders as asked in the question.Order 1(dxdy): \[\int_0^4 \int_0^{\sqrt{x}} \frac{y}{1+x^2} dy\, dx\] Order 2 (dydx): \[\int_0^2 \int_{y^2}^4 \frac{y}{1+x^2} dx\, dy\]

Choose the more convenient order

In this case, Order 2 (dydx) seems to be simpler as the integral with respect to \(x\) can be directly computed in terms of \(y\) and \(x\) without involving \(y\) as a function of \(x\). So, we proceed with Order 2 to evaluate the integral.

Evaluate the integral

\[\int_0^2 \int_{y^2}^4 \frac{y}{1+x^2} dx\, dy\] = \int_0^2 y \left[ tan^{-1}(x) \right]_{y^2}^4 dy = \int_0^2 y(tan^{-1}(4) - tan^{-1}(y^2)) dy\]This can be evaluated as an ordinary integral with a standard integration method (u-substitution, by-parts, etc.) to get the final answer.

Key concepts

Double Integration

Double integration refers to the process of integrating a function of two variables over a two-dimensional region. Think of it as finding the accumulated quantity, like area or volume, within a given space. When we perform double integration, we break this process into two separate integrations, hence the term 'double'.

In the context of the given problem, we want to determine the integral of the function \(\frac{y}{1+x^2}\) over the region \(R\). The problem's given conditions limit this region, and it is our goal to calculate the total integral value within these boundaries. This involves assessing the area under the curve \(y=\sqrt{x}\), above the x-axis, and to the left of the line \(x=4\). It's like summing up infinitesimally small 'slices' of area from the entire region.

Integral Bounds

The integral bounds are limits between which we evaluate the integral. They define the region of integration. For our two-dimensional integral, we have bounds for both \(x\) and \(y\), effectively framing our region of interest.

In the case of our exercise, the bounds are determined by the intersection of the curves and lines that enclose region \(R\). Specifically, for the first order, the y-bounds are from \(0\) to \(\sqrt{x}\), and the x-bounds are from \(0\) to \(4\). For the second order, the x-bounds are from \(y^2\) to \(4\), and the y-bounds are from \(0\) to \(2\). These bounds ensure that we are only considering the region specified in the problem.

Iterated Integrals

Iterated integrals are a sequence of integrations performed one after the other. Each integration corresponds to one variable while keeping the other variables as constants.

The idea is to integrate with respect to one variable first, which will produce an intermediate function, and then integrate this function with respect to the second variable. Our example involves iterating in either \(dy\) or \(dx\) first, but the principle is the same regardless of the order. The operation reduces a complex, two-dimensional problem into simpler, one-dimensional problems.

Order of Integration

The order of integration is the sequence in which we integrate with respect to the variables. In a two-dimensional setting, this sequence can significantly affect the complexity of the calculations.

For example, the given problem can be integrated in two ways: first with respect to \(x\) followed by \(y\) (\(dxdy\)), or first with respect to \(y\) followed by \(x\) (\(dydx\)). While mathematically equivalent, one order might be more algebraically convenient than the other. As highlighted in the solution, Order 2 was chosen for its simplicity. By considering the form of the integrand and the bounds of integration, we can often identify the more convenient order to simplify our work.