Question

Which of the following expressions will give the area of region bounded by the curve \( y = x^2 \) and line \( y = 16 \)?

• A. \( \int_0^4 x^2 dx \)
• B. \( 2 \int_0^4 x^2 dx \)
• C. \( \int_0^{16} \sqrt{y} dy \)
• D. \( 2 \int_0^{16} \sqrt{y} dy \)

Hint:

Always sketch the region first to determine if integration along \( x \) or \( y \) is more straightforward.
Integration along the \( x \)-axis would be \( \int_{-4}^{4} (16 - x^2) dx \), which is not listed as a simplified option.

Answer 1

The Correct Option is D

Concept:

• Area between curves: Area can be calculated by integrating along the \( x \)-axis or \( y \)-axis.
• Symmetry: The parabola \( y = x^2 \) is symmetric about the \( y \)-axis.
• Integration along \( y \)-axis: \( \text{Area} = \int_{c}^{d} (x_{\text{right}} - x_{\text{left}}) dy \).

Step 1: Identify the geometry and intersection points
The curve is \( y = x^2 \) and the line is \( y = 16 \).
The curves intersect when \( x^2 = 16 \implies x = \pm 4 \).
The region is bounded between the parabola and the horizontal line \( y = 16 \).

Step 2: Set up the area integral along the \( y \)-axis
Integrating along the \( y \)-axis is often easier for this shape.
Limits for \( y \) are from 0 to 16.
For a given \( y \), the \( x \)-values are \( x = \pm \sqrt{y} \).
The total width at height \( y \) is \( \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y} \).

Step 3: Formulate the final integral
Total Area \( = \int_0^{16} (x_{\text{right}} - x_{\text{left}}) dy \)
\[ \text{Area} = \int_0^{16} (\sqrt{y} - (-\sqrt{y})) dy \]
\[ \text{Area} = \int_0^{16} 2\sqrt{y} dy = 2 \int_0^{16} \sqrt{y} dy \]
This matches option (D).