Question

The area (in sq. units) of the part of the circle $x^2 + y^2 = 36$, which is outside the parabola $y^2 = 9x$, is :

• A. $24\pi + 3\sqrt{3}$
• B. $24\pi - 3\sqrt{3}$
• C. $12\pi + 3\sqrt{3}$
• D. $12\pi - 3\sqrt{3}$

Hint:

To find the area "outside," always calculate the smaller area bounded by the curves first, then subtract it from the total area of the circle ($\pi r^2$).

Answer 1

The Correct Option is A

Step 1: Find the intersection points of $x^2 + y^2 = 36$ and $y^2 = 9x$. $x^2 + 9x - 36 = 0 \Rightarrow (x+12)(x-3) = 0$. Since $y^2 \geq 0$, we take $x=3$. At $x=3$, $y^2 = 27 \Rightarrow y = \pm 3\sqrt{3}$.
Step 2: Area inside the parabola and circle is $A_{in} = 2 \left[ \int_0^3 \sqrt{9x} \, dx + \int_3^6 \sqrt{36-x^2} \, dx \right]$.
Step 3: Solving the integrals: $A_{in} = 12\pi - 3\sqrt{3}$.
Step 4: Area outside the parabola is Total Area - Area inside: $Area = 36\pi - (12\pi - 3\sqrt{3}) = 24\pi + 3\sqrt{3}$.