Question

Find the area of the region bounded by the curve \( y^2 = x \) and the line \( x = 4 \):

• A. \( \frac{32}{3} \)
• B. \( \frac{16}{3} \)
• C. \( \frac{8}{3} \)
• D. \( 16 \)

Hint:

Parabolas like \( y^2 = 4ax \) always contain symmetric upper and lower regions. Always sketch a rough graph so you don't forget to multiply by the factor of 2!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The area is bounded by a parabola $y^2 = x$ and a vertical line $x = 4$. The parabola is symmetric about the $x$-axis.
Step 2: Detailed Explanation:
Area = $2 \int_0^4 y dx = 2 \int_0^4 \sqrt{x} dx$.
$= 2 [\frac{x^{3/2}}{3/2}]_0^4 = 2 \cdot \frac{2}{3} [x^{3/2}]_0^4$.
$= \frac{4}{3} (4^{3/2}) = \frac{4}{3} (8) = \frac{32}{3}$.
Step 3: Final Answer:
The area is $\frac{32}{3}$ sq units.