Problem 19 - Thomas Calculus

Question

Evaluate $$ \int_{0}^{a} \int_{0}^{b} e^{\max \left(b^{2} x^{2}, a^{2} y^{2}\right)} d y d x $$ where \(a\) and \(b\) are positive numbers and $$\quad \max \left(b^{2} x^{2}, a^{2} y^{2}\right)=\left\\{\begin{array}{ll}{b^{2} x^{2}} & {\text { if } b^{2} x^{2} \geq a^{2} y^{2}} \\ {a^{2} y^{2}} & {\text { if } b^{2} x^{2}

Step-by-Step Solution

Answer

The integral evaluates to a combination of exponential integrals related to the regions formed by the max function.

Step 1: Understand the Max Function

The function \( \max(b^2x^2, a^2y^2) \) chooses the larger of \( b^2x^2 \) and \( a^2y^2 \). This effectively splits the integration into regions where either \( b^2x^2 \geq a^2y^2 \) or \( b^2x^2 < a^2y^2 \).

Step 2: Set up the Regions

For the condition \( b^2x^2 \geq a^2y^2 \), we solve for the inequality: \( y \leq \frac{b}{a}x \). The region where \( b^2x^2 < a^2y^2 \) is given by \( y > \frac{b}{a}x \).

Step 3: Divide the Integral

Divide the integral into two parts:1. \( \int_0^a \int_0^{(b/a)x} e^{b^2x^2} \, dy \, dx \)2. \( \int_0^a \int_{(b/a)x}^b e^{a^2y^2} \, dy \, dx \)

Step 4: Evaluate the First Integral

For the integral \( \int_0^a \int_0^{(b/a)x} e^{b^2x^2} \, dy \, dx \), the inner integral with respect to \( y \) simply multiplies the expression by its limit since the integrand is constant in \( y \). Thus, it becomes \( \int_0^a \left[ e^{b^2x^2} y \right]_0^{(b/a)x} \, dx = \int_0^a e^{b^2x^2} \frac{b}{a}x \, dx \).

Step 5: Integrate the Result

Now integrate \( \int_0^a e^{b^2x^2} \frac{b}{a}x \, dx \). Let \( u = b^2x^2 \), then \( du = 2b^2x \, dx \). Substitute to get:\( \int_0^{b^2a^2} e^u \frac{1}{2b^2} \, du \), which simplifies to \( \frac{1}{2b^2} \left[ e^u \right]_0^{b^2a^2} \).

Step 6: Solve the Second Integral

Consider \( \int_0^a \int_{(b/a)x}^b e^{a^2y^2} \, dy \, dx \). Perform the inner integral with respect to \( y \). Use substitution \( v = a^2y^2 \). Then:\( \int_{(b/a)x}^b e^{a^2y^2} \, dy = \frac{1}{2a^2} \left[ e^{a^2y^2} \right]_{(b/a)x}^b \).

Step 7: Combine Results

After solving both smaller integrals:\( \text{First part: } \frac{1}{2b^2} \left( e^{b^2a^2} - 1 \right) \)and\( \text{Second part: } \frac{1}{2a^2} \int_0^a \left( e^{a^2b^2} - e^{a^4x^2/b^2} \right) \, dx \).Compute this final integral for complete solution.

Key Concepts

Region of Integration

In the world of double integrals, the region of integration can often determine how you evaluate the integral. For the given problem, the integral is defined over a rectangle in the xy-plane, with sides extending from 0 to \( a \) on the x-axis and from 0 to \( b \) on the y-axis. This forms a rectangular region. However, because of the max function in the integrand, this region is further divided based on the comparison between \( b^2x^2 \) and \( a^2y^2 \).

The first region, where \( b^2x^2 \geq a^2y^2 \), corresponds to points below the line \( y = \frac{b}{a}x \). Conversely, the second region, where \( b^2x^2 < a^2y^2 \), consists of points above this line. This division allows the integral to be split into two separate evaluations, making it easier to handle the max function in the integrand.

Max Function

The max function \( \max(b^2x^2, a^2y^2) \) plays a critical role in this problem. It essentially selects the larger value between \( b^2x^2 \) and \( a^2y^2 \) at each point \( (x, y) \) within the region of integration. This condition results in the creation of two sub-regions with different integrands for each.

This function allows you to switch between different expressions based on geometric constraints. For example, when \( b^2x^2 \geq a^2y^2 \), you use the expression \( e^{b^2x^2} \) in the integrand, highlighting the greater influence of the x-terms. In the other region, where \( b^2x^2 < a^2y^2 \), \( e^{a^2y^2} \) is used, emphasizing the y-terms. This ensures that you compute the integral correctly over divided regions.

Change of Variables

Change of variables is a powerful technique essential in evaluating complex integrals. It involves substituting new variables for the variables in the integrand, which simplifies integration. In this exercise, during the evaluation of the integrals, the change of variables transforms the integrands to a more feasible form.

For instance, when evaluating the first integral component, let \( u = b^2x^2 \), and thus \( du = 2b^2x \, dx \). This substitution converts the integral into a simpler exponential form with respect to \( u \). Such substitutions are often strategic, helping to streamline the integral into a standard format that can be easily integrated.

Definite Integrals

Definite integrals are the calculated areas under a curve within specified bounds. In the context of double integrals, definite integrals help evaluate areas within a certain region of the xy-plane.

In the solution to the given problem, each sub-region from the integration is evaluated using definite integrals. For example, after determining the expressions and limits for each region (\( \int_0^a \int_0^{(b/a)x} e^{b^2x^2} \, dy \, dx \) and \( \int_0^a \int_{(b/a)x}^b e^{a^2y^2} \, dy \, dx \)), you find the definite integral for each setup. This helps in calculating the precise area considering the constraints imposed by the max function. These calculated areas are the summation of these evaluated definite integrals, providing the final result for the double integral.