Question

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

• A. $-\frac{1}{6}$
• B. $-\frac{1}{2}$
• C. $-\frac{5}{6}$
• D. $\frac{5}{6}$

Hint:

Whenever the limits of a double integral are constant and the integrand is of the form $f(x) \cdot g(y)$, always separate them into the product of two single integrals to save calculation steps.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To evaluate a double integral over a rectangular region where the limits of integration are constants, we can perform iterated integration.
Key Formula or Approach:
Since the limits of $x$ and $y$ are independent, and the integrand $f(x, y) = y^2 x$ is separable into $g(x)h(y)$, we can write: \[ \iint_{\text{R}} y^2 x \, dy \, dx = \left( \int_{x=a}^{b} x \, dx \right) \left( \int_{y=c}^{d} y^2 \, dy \right) \]

Step 2: Detailed Explanation:
The given limits are:
- $-3 \le x \le 2$
- $0 \le y \le 1$
Let us compute each integral individually.
First, the integral with respect to $x$: \[ \int_{-3}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{-3}^{2} \] \[ = \frac{2^2}{2} - \frac{(-3)^2}{2} = \frac{4}{2} - \frac{9}{2} = -\frac{5}{2} \] Second, the integral with respect to $y$: \[ \int_{0}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1} \] \[ = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \] Now, multiply the two independent results: \[ \iint_{\text{R}} y^2 x \, dy \, dx = \left( -\frac{5}{2} \right) \times \left( \frac{1}{3} \right) = -\frac{5}{6} \] Thus, the value of the double integral is $-\frac{5}{6}$.

Step 3: Final Answer:
The value is $-\frac{5}{6}$, which corresponds to Option (C).