To evaluate the given integral over the specified triangular region, we first need to set up the limits of integration and handle the integrand. Here's the problem explicitly:

$I = \iint_{R} \frac{2xy}{5\sqrt{1 + x^2y^2 - y^4}} \, dx\, dy$

Step 1: Define the Region

The triangular region is defined by the vertices $(0,0)$, $(1,1)$, and $(0,1)$. The lines forming the boundaries are:

1. $y = 0$ (horizontal line at the base),
2. $x = 0$ (vertical line),
3. $y = x$ (diagonal line).

From this, the limits of integration are:

• $y$ varies from $0$ to $1$,
• For a fixed $y$, $x$ varies from $0$ to $y$.

Step 2: Write the Integral with Limits

The integral becomes: $I = \int_{y=0}^{1} \int_{x=0}^{y} \frac{2xy}{5\sqrt{1 + x^2y^2 - y^4}} \, dx\, dy.$

Step 3: Simplify and Solve

We cannot directly integrate due to the square root in the denominator, so let's analyze the steps:

(i) Inner integral (with respect to $x$):

$\int_{x=0}^{y} \frac{2xy}{5\sqrt{1 + x^2y^2 - y^4}} \, dx.$ Here, $y$ is treated as a constant. Let’s substitute $u = 1 + x^2y^2 - y^4$:

• $du = 2x y^2 dx$,
• When $x = 0$, $u = 1 - y^4$,
• When $x = y$, $u = 1 + y^4 - y^4 = 1$.

Rewriting: $\int_{x=0}^{y} \frac{2xy}{5\sqrt{1 + x^2y^2 - y^4}} \, dx = \int_{u=1-y^4}^{1} \frac{1}{5\sqrt{u}} \cdot \frac{1}{y} \, du.$

This simplifies to: $\frac{1}{5y} \int_{u=1-y^4}^{1} u^{-1/2} \, du.$

(ii) Solving the $u$-integral:

The antiderivative of $u^{-1/2}$ is $2u^{1/2}$. Thus: $\frac{1}{5y} \int_{u=1-y^4}^{1} u^{-1/2} \, du = \frac{1}{5y} \left[ 2u^{1/2} \right]_{u=1-y^4}^{u=1}.$

Evaluate the limits: [ \frac{1}{5y} \left[ 2\sqrt{1} - 2\sqrt{1 - y^4} \right] = \frac{2}{5y} \left[ 1 - \sqrt{1 - y^4} \right]. ]

(iii) Outer integral (with respect to $y$):

The integral now becomes: $I = \int_{y=0}^{1} \frac{2}{5y} \left[ 1 - \sqrt{1 - y^4} \right] \, dy.$

(iv) Simplify the remaining expression:

Split the integral into two parts: $I = \frac{2}{5} \int_{y=0}^{1} \frac{1}{y} \, dy - \frac{2}{5} \int_{y=0}^{1} \frac{\sqrt{1 - y^4}}{y} \, dy.$

• The first term, $\int_{y=0}^{1} \frac{1}{y} \, dy$, diverges due to a singularity at $y=0$. Hence, this integral is improper.

Final Answer:

The integral diverges due to the term $\frac{1}{y}$ as $y \to 0$.