Question

The value of $\iint_S \cos(x) \sin(y) \, dx \, dy$ is, where $S$ is $\left[0, \frac{\pi}{2}\right] \times \left[0, \frac{\pi}{2}\right]$

• A. 1
• B. 0
• C. $\frac{\pi}{2}$
• D. $-1$

Hint:

Whenever double integral limits are constants and the integrand separates into $g(x)h(y)$, always split it into two single 1D integrals immediately to save computation time!

Answer 1

The Correct Option is A

Step 1: Concept
This problem involves evaluating a double integral over a rectangular domain $S = [a, b] \times [c, d]$. When the integrand is separable, i.e., $f(x, y) = g(x) h(y)$, Fubini's Theorem allows splitting the double integral into the product of two independent single integrals.

Step 2: Key Formulas and Approach
For $S = [a, b] \times [c, d]$ and $f(x, y) = g(x) h(y)$: \[ \iint_S g(x) h(y) \, dx \, dy = \left( \int_a^b g(x) \, dx \right) \cdot \left( \int_c^d h(y) \, dy \right) \]

Step 3: Step-by-step Explanation

• Here $g(x) = \cos x$ on $[0, \pi/2]$ and $h(y) = \sin y$ on $[0, \pi/2]$.
• Separate the double integral into two single integrals: \[ I = \left( \int_0^{\pi/2} \cos x \, dx \right) \cdot \left( \int_0^{\pi/2} \sin y \, dy \right) \]
• Evaluate the first integral with respect to $x$: \[ \int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]
• Evaluate the second integral with respect to $y$: \[ \int_0^{\pi/2} \sin y \, dy = [-\cos y]_0^{\pi/2} = -\cos\left(\frac{\pi}{2}\right) - (-\cos 0) = 0 + 1 = 1 \]
• Multiply the results: \[ I = 1 \times 1 = 1 \]

Step 4: Final Answer
The value of the double integral is 1. Thus, Option (A) is correct.