Problem 21 - Advanced Engineering Mathematics

Question

Evaluate the double integral over the region \(R\) that is bounded by the graphs of the given equations. Choose the most convenient order of integration. $$ \iint_{R} \sqrt{x^{2}+1} d A ; x=y, x=-y, x=\sqrt{3} $$

Step-by-Step Solution

Answer

The double integral evaluates to \( \frac{14}{3} \).

Step 1: Understand the Region of Integration

First, determine the region \( R \) over which we will integrate. The region is bounded by the lines \( x = y \), \( x = -y \), and \( x = \sqrt{3} \). This forms a triangular region in the first quadrant, where \( y \) ranges from \( -x \) to \( x \) and \( x \) ranges from \( 0 \) to \( \sqrt{3} \).

Step 2: Choose the Order of Integration

Choose an order of integration that makes the computation easier. We'll integrate first with respect to \( y \) and then \( x \). This choice is convenient because for a fixed \( x \), \( y \) varies from \( -x \) to \( x \).

Step 3: Set up the Integral

Write down the double integral with the order of integration determined:\[ \int_{0}^{\sqrt{3}} \int_{-x}^{x} \sqrt{x^{2}+1} \, dy \, dx \]

Step 4: Evaluate the Inner Integral

Evaluate the integral with respect to \( y \). Since \( \sqrt{x^2 + 1} \) is constant with respect to \( y \), the inner integral becomes:\[ \int_{-x}^{x} \sqrt{x^2 + 1} \, dy = \sqrt{x^2 + 1} [y]_{-x}^{x} = \sqrt{x^2 + 1} (x - (-x)) = 2x \sqrt{x^2 + 1}. \]

Step 5: Evaluate the Outer Integral

Now integrate with respect to \( x \):\[ \int_{0}^{\sqrt{3}} 2x \sqrt{x^2 + 1} \, dx. \]Use substitution: let \( u = x^2 + 1 \), then \( du = 2x \, dx \).When \( x = 0, \, u = 1 \), and when \( x = \sqrt{3}, \, u = 4 \).The integral becomes:\[ \int_{1}^{4} \sqrt{u} \, du. \]

Step 6: Finish the Integration

Evaluate \( \int \sqrt{u} \, du \):\[ \int u^{1/2} \, du = \frac{2}{3} u^{3/2}. \]Calculate from \( 1 \) to \( 4 \):\[ \frac{2}{3} [u^{3/2}]_{1}^{4} = \frac{2}{3} (4^{3/2} - 1^{3/2}) = \frac{2}{3} (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3}. \]

Step 7: Conclusion

The value of the double integral \( \iint_R \sqrt{x^2 + 1} \, dA \) over the region \( R \) is \( \frac{14}{3} \).

Key Concepts

Integration Order

When solving double integrals, choosing the correct order of integration is crucial to simplifying the computation. This involves deciding whether to integrate with respect to one variable first and then the other. In the given exercise, we have a region bounded by the lines \( x = y \), \( x = -y \), and \( x = \sqrt{3} \). For convenience and ease of calculation, the order of integration we chose is to integrate first with respect to \( y \), followed by \( x \).

• By setting \( y \) as the inner integral, \( y \) ranges from \( -x \) to \( x \) for a fixed \( x \).
• This order allows us to handle the symmetry and limits more efficiently, resulting in simpler calculations.

After integrating with respect to \( y \), we must evaluate the result in terms of \( x \), following up with integrating it over the specified region \( x = 0 \) to \( x = \sqrt{3} \).
Understanding integration order not only aids in reducing computational complexity but also clarifies the problem setting, helping to see which variable influences more at different steps.

Region of Integration

Before performing any integration, understanding the region of integration is key. In this problem, the area \( R \) is bounded by the equations \( x = y \), \( x = -y \), and \( x = \sqrt{3} \). This forms a triangular region in the plane.

• The boundaries given (\( x = y \) and \( x = -y \)) suggest symmetry with respect to the origin.
• The line \( x = \sqrt{3} \) confines the region to the first quadrant.

The region of integration is crucial because it determines the limits of the integral. For our specific case:

• \( y \) limits are from \( -x \) to \( x \), reflecting the diagonal bounds.
• \( x \) limits are from \( 0 \) to \( \sqrt{3} \), dictated by the vertical boundary along \( x = \sqrt{3} \).

Grasping this bounded area precisely ensures that we apply the correct limits and understand the nature of the area we are evaluating.

Integration by Substitution

Integration by substitution is a powerful technique to simplify integrals, especially when dealing with complex expressions. In our problem, after setting up the double integral and evaluating the inner integral, the outer integral requires simplification.

• The expression \( 2x \sqrt{x^2 + 1} \) can be complicated to manage directly.
• Using substitution, we let \( u = x^2 + 1 \) which gives \( du = 2x \, dx \) simplifying the whole expression.

Substitution transforms the integral bounds as well:

• When \( x = 0 \), \( u = 1 \) and when \( x = \sqrt{3} \), \( u = 4 \).

This converts our integral into a simpler form \( \int_{1}^{4} \sqrt{u} \, du \) which is much easier to evaluate.
The substitution method not only aids in simplifying but also provides a more straightforward pathway to the solution, making such problems less daunting.