Evaluate ∫∫ ඥ(4xଶ − yଶ) dxdy over the triangle formed bystraight lines y = 0, x = 1, y = x.

1. Break Down the Problem

We need to evaluate the double integral

$\iint_{D} \sqrt{4x^{2} - y^{2}} \, dx \, dy$

over the triangular region $D$ formed by the lines $y = 0$, $x = 1$, and $y = x$.

2. Relevant Concepts

The triangular region is bounded by:

• The x-axis (line $y = 0$)
• The vertical line $x = 1$
• The line $y = x$

The limits of integration need to be determined based on the region of integration.

3. Analysis and Detail

The triangle has vertices at the points $(0, 0)$, $(1, 0)$, and $(1, 1)$.

To find the limits for $x$ and $y$:

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

Thus, the double integral can be written as:

$\int_{0}^{1} \int_{0}^{x} \sqrt{4x^{2} - y^{2}} \, dy \, dx$

4. Verify and Summarize

Now, we evaluate the inner integral first:

$I = \int_{0}^{x} \sqrt{4x^{2} - y^{2}} \, dy$

This integral can be evaluated using the substitution $y = 2x \sin(\theta)$, which changes $dy = 2x \cos(\theta) d\theta$. The limits will change accordingly when $y = 0$ and $y = x$.

Final Answer

However, I will simplify the evaluation without substitution for clarity. The formula for the integral $\int \sqrt{a^2 - y^2} \, dy$ is given by:

$\int \sqrt{a^2 - y^2} \, dy = \frac{y}{2} \sqrt{a^2 - y^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{y}{a}\right) + C$

Here, $a = 2x$.

Thus, we have:

1. Evaluate the inner integral: $I = \left[ \frac{y}{2} \sqrt{4x^2 - y^2} + 2x^2 \sin^{-1}\left(\frac{y}{2x}\right) \right]_{0}^{x}$ Evaluating this from $0$ to $x$: $I = \left[ \frac{x}{2} \sqrt{4x^2 - x^2} + 2x^2 \sin^{-1}(0) \right] - \left[ \frac{0}{2} + 2x^2 \sin^{-1}(0) \right] = \frac{x}{2} \sqrt{3x^2} = \frac{\sqrt{3}}{2} x^2$

2. Next, we evaluate the outer integral: $\int_{0}^{1} \frac{\sqrt{3}}{2} x^2 \, dx = \frac{\sqrt{3}}{2} \cdot \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}$

Final Result

Thus, the value of the double integral is

$\frac{\sqrt{3}}{6}.$