Question

The area (in square units) of the region bounded by the circle \(x^2+y^2 = 9\) and the parabola \(y^2\leq 8x\) is...

• A. \(\frac{8\sqrt{2}}{3}+\frac{9π}{2}-2\sqrt{2}-9sin^{-1}\frac{1}{3}\)
• B. \(\frac{8\sqrt{2}}{3}+\frac{9π}{2}+2\sqrt{2}+9sin^{-1}\frac{1}{3}\)
• C. \(\frac{4\sqrt{2}}{3}+\frac{9π}{4}-\sqrt{2}-\frac{9}{2}sin^{-1}\frac{1}{3}\)
• D. \(\frac{4\sqrt{2}}{3}+\frac{9π}{4}+\sqrt{2}+\frac{9}{2}sin^{-1}\frac{1}{3}\)

Hint:

Find where the circle and parabola cross, then add the parabola part and the circle part, each doubled for symmetry.

Answer 1

The Correct Option is A

Step 1: Understand the concept
The region is inside the circle and inside the parabola \(y^2 \le 8x\). Both curves are symmetric about the X-axis, so we compute the upper half and double it.

Step 2: Find the intersection
Substitute \(y^2 = 8x\) in \(x^2 + y^2 = 9\): \(x^2 + 8x - 9 = 0\), so \(x = 1\) (since \(x > 0\)). Then \(y = \pm2\sqrt{2}\).

Step 3: Area from x = 0 to 1 (parabola)
\[ 2\int_0^1\sqrt{8x}\,dx = 2\cdot2\sqrt{2}\cdot\frac{2}{3} = \frac{8\sqrt{2}}{3} \]

Step 4: Area from x = 1 to 3 (circle)
\[ 2\int_1^3\sqrt{9 - x^2}\,dx = 2\left[\frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\frac{x}{3}\right]_1^3 = \frac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}\frac{1}{3} \]
The total is \(\dfrac{8\sqrt{2}}{3} + \dfrac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}\dfrac{1}{3}\), option (A).

Final Answer:
The area is 8 sqrt2/3 + 9 pi/2 - 2 sqrt2 - 9 arcsin(1/3). This is option (A). \[ \boxed{\text{(A) }\frac{8\sqrt{2}}{3}+\frac{9\pi}{2}-2\sqrt{2}-9\sin^{-1}\frac{1}{3}} \]