Question

Let $f(x, y) = x^2 + y^2$ and $R = [-1, 1] \times [0, 1]$, then $\iint_R f(x, y) \, dx \, dy$ is

• A. $\frac{1}{3}$
• B. $\frac{4}{3}$
• C. $\frac{2}{3}$
• D. 1

Hint:

Symmetry trick: Since $x^2$ is an even function of $x$ on $[-1, 1]$, $\int_{-1}^1 (x^2 + y^2) dx = 2 \int_0^1 (x^2 + y^2) dx = 2 [\frac{1}{3} + y^2]$. Then $\int_0^1 2 (\frac{1}{3} + y^2) dy = 2 (\frac{1}{3} + \frac{1}{3}) = \frac{4}{3}$.

Answer 1

The Correct Option is B

Step 1: Concept
To evaluate the double integral of a function $f(x, y)$ over a rectangle $R = [a, b] \times [c, d]$, we write it as an iterated integral: \[ \iint_R f(x, y) \, dx \, dy = \int_c^d \left( \int_a^b f(x, y) \, dx \right) dy \]

Step 2: Key Formulas and Approach
Set the limits of integration: $x \in [-1, 1]$ and $y \in [0, 1]$. Apply standard integration power rules: $\int x^n dx = \frac{x^{n+1}}{n+1}$.

Step 3: Step-by-step Explanation

• Formulate the iterated integral: \[ I = \int_{y=0}^1 \int_{x=-1}^1 (x^2 + y^2) \, dx \, dy \]
• Perform inner integration with respect to $x$ (treating $y$ as a constant): \[ \int_{-1}^1 (x^2 + y^2) \, dx = \left[ \frac{x^3}{3} + x y^2 \right]_{x=-1}^{x=1} \] \[ = \left( \frac{1^3}{3} + (1)y^2 \right) - \left( \frac{(-1)^3}{3} + (-1)y^2 \right) \] \[ = \left( \frac{1}{3} + y^2 \right) - \left( -\frac{1}{3} - y^2 \right) = \frac{2}{3} + 2y^2 \]
• Perform outer integration with respect to $y$: \[ I = \int_0^1 \left( \frac{2}{3} + 2y^2 \right) dy = \left[ \frac{2}{3}y + \frac{2y^3}{3} \right]_0^1 \] \[ = \left( \frac{2}{3}(1) + \frac{2(1)^3}{3} \right) - 0 = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

Step 4: Final Answer
The value of the double integral is $\frac{4}{3}$. Thus, Option (B) is correct.