Let $D$ be the region in $R^2$ bounded by the parabola $y^2 = 2x$ and the line $y = x$. Then $\iint_D 3xy \, dx \, dy$ equals _______

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Question

Let $D$ be the region in $R^2$ bounded by the parabola $y^2 = 2x$ and the line $y = x$. Then $\iint_D 3xy \, dx \, dy$ equals ________.

Region D Boundaries

The problem asks for the evaluation of the double integral $\iint_D 3xy \, dx \, dy$ over a region $D$ defined by the boundaries $y^2 = 2x$ (a parabola) and $y = x$ (a line).

First, find the intersection points of the boundaries:

Set $y^2 = 2x$ and $y = x$ .
Substituting $y = x$ into the parabola equation gives $x^2 = 2x$ .
Solving $x^2 - 2x = 0$ yields $x(x - 2) = 0$ , so $x = 0$ or $x = 2$ .
The corresponding points are $(0, 0)$ and $(2, 2)$ .

To set up the iterated integral, we express $x$ in terms of $y$ :

Parabola: $x = \frac{y^2}{2}$
Line: $x = y$

In the region $D$ , for $y$ between 0 and 2, the line $x = y$ is to the right of the parabola $x = \frac{y^2}{2}$ . Thus, the bounds are $\frac{y^2}{2} \le x \le y$ and $0 \le y \le 2$ .

Integral Calculation

The double integral is set up as an iterated integral:

$\iint_D 3xy \, dx \, dy = \int_0^2 \int_{y^2/2}^y 3xy \, dx \, dy$

Step 1: Evaluate the inner integral $\int_{y^2/2}^y 3xy \, dx$ :

$3y \left[ \frac{x^2}{2} \right]_{x=y^2/2}^{x=y} = 3y \left( \frac{y^2}{2} - \frac{(y^2/2)^2}{2} \right) = 3y \left( \frac{y^2}{2} - \frac{y^4}{8} \right) = \frac{3y^3}{2} - \frac{3y^5}{8}$

Step 2: Evaluate the outer integral $\int_0^2 \left( \frac{3y^3}{2} - \frac{3y^5}{8} \right) dy$ :

$\left[ \frac{3}{2} \frac{y^4}{4} - \frac{3}{8} \frac{y^6}{6} \right]_0^2 = \left[ \frac{3y^4}{8} - \frac{y^6}{16} \right]_0^2$

Substitute the limits:

$= \left( \frac{3(2^4)}{8} - \frac{2^6}{16} \right) - (0) = \left( \frac{3(16)}{8} - \frac{64}{16} \right) = (6 - 4) = 2$

Integral Result

The value of the integral $\iint_D 3xy \, dx \, dy$ is 2.

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