Question

The value of $\iint_R y^2 x \, dy \, dx$ over the rectangle, $R = \{(x, y) : -3 \le x \le 2, 0 \le y \le 1\}$ is given by}

• A. $-\frac{1}{6}$
• B. $-\frac{5}{6}$
• C. $\frac{1}{6}$
• D. $\frac{5}{6}$

Hint:

This separation technique only works when the limits of both integration variables are constants. If the boundaries depend on another variable, we must integrate sequentially.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Over a rectangular region with constant boundaries, a double integral of a separable function $f(x, y) = g(x)h(y)$ can be computed as the product of two independent single integrals.
Key Formula or Approach:
For a rectangular region $R = [a, b] \times [c, d]$:
\[ \iint_R g(x)h(y) \, dy \, dx = \left( \int_a^b g(x) \, dx \right) \cdot \left( \int_c^d h(y) \, dy \right) \]

Step 2: Detailed Explanation:
We want to evaluate:
\[ I = \iint_R y^2 x \, dy \, dx \]
where the boundaries are $-3 \le x \le 2$ and $0 \le y \le 1$.
Since the limits of integration are constants and the integrand is separable, we split the double integral:
\[ I = \left( \int_{-3}^2 x \, dx \right) \cdot \left( \int_0^1 y^2 \, dy \right) \]
Let us evaluate each single integral:
1. Evaluate the first integral with respect to $x$:
\[ \int_{-3}^2 x \, dx = \left[ \frac{x^2}{2} \right]_{-3}^2 = \frac{2^2 - (-3)^2}{2} = \frac{4 - 9}{2} = -\frac{5}{2} \]
2. Evaluate the second integral with respect to $y$:
\[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1^3 - 0^3}{3} = \frac{1}{3} \]
Multiply the two results:
\[ I = \left(-\frac{5}{2}\right) \cdot \left(\frac{1}{3}\right) = -\frac{5}{6} \]
Therefore, the value of the double integral is $-\frac{5}{6}$.

Step 3: Final Answer
The correct option is (B).