Question

Domain \( A \) is bounded by the curve \( x^2 = 4y \), the ordinate \( x = 2 \), and the \( x \) axis.
The value of \( \iint_A y \, dx \, dy \) is

• A. \( \dfrac{1}{5} \)
• B. \( \dfrac{1}{3} \)
• C. \( \dfrac{5}{12} \)
• D. \( \dfrac{1}{2} \)

Hint:

Set the inner limit for \( y \) from \( 0 \) to \( x^2/4 \) and integrate over \( x \) from \( 0 \) to \( 2 \).

Answer 1

The Correct Option is A

Step 1: Set up the region of integration.
The curve \( x^2 = 4y \) gives \( y = \dfrac{x^2}{4} \), a parabola opening upward.
Region \( A \) sits between this parabola, the line \( x = 2 \), and the \( x \) axis, so \( x \) runs from \( 0 \) to \( 2 \).
For each fixed \( x \) in that range, \( y \) runs from \( 0 \) (the \( x \) axis) up to \( \dfrac{x^2}{4} \) (the parabola).

Step 2: Integrate with respect to \( y \) first.
\[ \int_0^{x^2/4} y \, dy = \left[ \frac{y^2}{2} \right]_0^{x^2/4} = \frac{x^4}{32} \]
This turns the double integral into a single integral in \( x \).

Step 3: Integrate with respect to \( x \).
\[ \int_0^2 \frac{x^4}{32} \, dx = \frac{1}{32}\left[\frac{x^5}{5}\right]_0^2 = \frac{1}{32}\cdot\frac{32}{5} = \frac{1}{5} \]

Final Answer:
The double integral works out to a clean fraction with denominator 5. \[ \boxed{\dfrac{1}{5}} \]