Question

Evaluate the given double integral by changing it to an iterated integral. \(\iint_{S} \frac{2}{1+x^{2}} d A ; S\) is the triangular region with vertices at \((0,0),(2,2)\), and \((0,2)\).

Short answer

The double integral evaluates to \(2 \arctan(2) - \ln(5)\).

Step-by-step solution

Identify the Region

The region \(S\) is a triangular area with vertices at \((0,0), (2,2), (0,2)\). This triangle is bounded by the lines \(x=0\), \(y=2\), and \(y=x\).

Set the Limits for Integration

Based on the triangle's boundaries, for every \(x\) between 0 and 2, \(y\) ranges from \(x\) to 2. This provides the limits for integration: \(x\) ranges from 0 to 2, and for each \(x\), \(y\) ranges from \(x\) to 2. Therefore, the double integral is transformed into an iterated integral with limits \(0 \leq x \leq 2\) and \(x \leq y \leq 2\).

Rewriting the Double Integral as Iterated Integral

Rewrite the double integral as an iterated integral: \[\int_{x=0}^{2} \int_{y=x}^{2} \frac{2}{1+x^2} \, dy \, dx.\]

Integrate with Respect to y

Integrate the inner integral with respect to \(y\): \[\int_{y=x}^{2} \frac{2}{1+x^2} \, dy = \left[ \frac{2}{1+x^2} y \right]_{y=x}^{y=2}\]Evaluating the integral, we get:\[\frac{2}{1+x^2}(2 - x).\]

Integrate with Respect to x

Integrate the result from Step 4 with respect to \(x\): \[\int_{x=0}^{2} \frac{2}{1+x^2}(2-x) \, dx.\]Split it into two separate integrals:\[2 \int_{x=0}^{2} \frac{1}{1+x^2} \, dx - \int_{x=0}^{2} \frac{2x}{1+x^2} \, dx.\]

Evaluate Both Integrals

Evaluate the first integral: \[2 \left( \arctan(x) \right) \bigg|_{0}^{2} = 2 \left( \arctan(2) - \arctan(0) \right) = 2 \arctan(2).\]Evaluate the second integral by substitution \(u = 1 + x^2\), \(du = 2x\, dx\):\[\int_{1}^{5} \frac{1}{u} \, du = \ln(u)\bigg|_{1}^{5} = \ln(5) - \ln(1) = \ln(5).\]

Combine Results

Combine the results of both integrals: \[2 \arctan(2) - \ln(5).\]

Finalize the Solution

The value of the double integral is \[2 \arctan(2) - \ln(5).\]

Key concepts

Iterated Integral

An iterated integral involves integrating a function multiple times over a given region, using more than one variable. In the context of a double integral, we first integrate with respect to one variable while holding the other constant. Afterward, we perform integration on the remaining variable.

This step-by-step process may seem daunting, but breaking it down makes it manageable. First, we identify the order of integration, which depends on the bounds of integration. We tackle the inner integral before handling the outer one. The double integral from our original exercise becomes two single integrals strung together:

\[ \int_{x=0}^{2} \int_{y=x}^{2} \frac{2}{1+x^2} \, dy \, dx. \]

In this expression, "dy" signifies the inner integration with respect to "y," while "dx" denotes the outer integration with respect to "x".

Integration Limits

The integration limits define the boundaries over which integration occurs. In a double integral situation, these limits are essential for precisely describing the region you'll operate in.

For our triangular region defined by the vertices \((0,0), (2,2), (0,2)\), the limits are derived from these points. With respect to "x," it ranges from 0 to 2, which is the horizontal span of the triangle. For each specific point 'x,' the variable 'y' ranges from the line 'y=x' up to 'y=2,' the upper boundary.

These defined bounds transform the double integral into the iterated form with limits for 'x' from:\[0 \leq x \leq 2\] and for 'y' ranging from: \[ x \leq y \leq 2.\]

Understanding these boundaries ensures that the integration covers the entire triangular area correctly.

Triangular Region

Triangular regions, such as the one described in the exercise, are crucial in calculus problems involving double integrals. These regions are determined by their vertices, and understanding their boundaries in terms of the coordinate axes is vital.

In our case, the triangle is formed by points \((0,0), (2,2), (0,2)\). Here is how it lies: the vertex at \((0,0)\) anchors the origin, while \((2,2)\) and \((0,2)\) define the other two vertices. The line \(y = x\) acts as a diagonal boundary inside the triangle.

This triangle is bound by:

• The vertical line \(x = 0\) (left edge)
• The horizontal line \(y = 2\) (top edge)
• And the line \(y = x\) (diagonal within triangle)

These boundaries establish the region where integration will happen, often referred to as the limits or bounds of integration.

Arctan Function

The arctan function, also known as the inverse tangent function, appears often in calculus, particularly in integrals involving rational expressions.When integrated, the expression \(\frac{1}{1+x^2}\) results in \(\arctan(x)\). This relationship is fundamental in deriving solutions involving trigonometric identities and can be especially helpful in simplifying integrals.In the example at hand, evaluating the integral \[2 \int_{x=0}^{2} \frac{1}{1+x^2} \, dx\] leads directly to \[2( \arctan(2) - \arctan(0)).\] This simplifies further to \(2 \arctan(2)\), since \(\arctan(0) = 0\). Understanding how the arctan function works can make tackling integrals much more intuitive, particularly when dealing with expressions that resemble the form \(\frac{a}{1+(bx)^2}\). This component becomes a critical part of calculating the final value of a double integral.