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.

Short answer

The value of the integral \(\int_{0}^{a} \int_{0}^{b} e^{\max \left\{b^{2} x^{2}, a^{2} y^{2}\right\}} d y d x\) is \(((e^{a^2b^2}-1)/(ab))\).

Step-by-step solution

Identification of Ranges

Identify the ranges of the double integral and set them up. Here the range of \(x\) is from 0 to \(a\) and the range of \(y\) is from 0 to \(b\). So the integration takes place over this rectangular domain.

Break Integral into Cases

Break the integral into two cases based on the 'max' function. These two cases split the rectangular region into two halves. When \(b^2x^2 > a^2y^2\), we integrate \(e^{b^2x^2}\), and when \(a^2y^2 > b^2x^2\), we integrate \(e^{a^2y^2}\). These two cases correspond to triangular regions of the rectangle [0, a] x [0, b]. The first case corresponds to the triangle where \(x < y·(a/b)\), whereas the second case corresponds to the triangle where \(x > y·(a/b)\).

Solving the first half of the integral

Integrate \(e^{b^2x^2}\) first with respect to \(y\) and then \(x\) in the triangle where \(x < y·(a/b)\). This means that \(y\) ranges from 0 to \(x·(b/a)\), and \(x\) ranges from 0 to \(a\). The final result of this integral part should be \(((e^{a^2b^2}-1)/(2ab))\).

Solving the second half of the integral

Integrate \(e^{a^2y^2}\) first with respect to \(x\) and then \(y\) in the triangle where \(x > y·(a/b)\). This means that \(x\) ranges from 0 to \(y·(a/b)\), and \(y\) ranges from 0 to \(b\). The final result of this integral part should also be \(((e^{a^2b^2}-1)/(2ab))\)

Summation of Both Halves

The final answer is both parts summed, since they are identical. So, we multiply one part by 2 to get the result of the whole rectangular integration