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: Alternative: use y as the variable
The region at height $y$ runs from $x = y^2/8$ to $x = \sqrt{9 - y^2}$ for $|y| \le 2\sqrt{2}$.

Step 2: Integrate
\[ \text{Area} = 2\int_0^{2\sqrt2}\left(\sqrt{9 - y^2} - \frac{y^2}{8}\right)dy \]

Step 3: Evaluate
$\int_0^{2\sqrt2}\sqrt{9 - y^2}\,dy = \frac{1}{2}\cdot2\sqrt{2}\cdot1 + \frac{9}{2}\sin^{-1}\frac{2\sqrt{2}}{3}$. Also $\int_0^{2\sqrt2}\frac{y^2}{8}dy = \frac{(2\sqrt2)^3}{24} = \frac{2\sqrt{2}}{3}$. Using $\sin^{-1}\frac{2\sqrt2}{3} = \frac{\pi}{2} - \sin^{-1}\frac{1}{3}$:

Step 4: Combine
Area $= 2\sqrt{2} + 9\left(\frac{\pi}{2} - \sin^{-1}\frac{1}{3}\right) - \frac{4\sqrt{2}}{3}$. This equals $\frac{2\sqrt{2}}{3} + \frac{9\pi}{2} - 9\sin^{-1}\frac{1}{3}$, which agrees with option (A) since $\frac{8\sqrt{2}}{3} - 2\sqrt{2} = \frac{2\sqrt{2}}{3}$.

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