The area (in sq. units) of the smaller part of the circle \(x^{2}+y^{2}=36\) cut off by the line \(x = 3\sqrt{3}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(6\pi-9\sqrt{3}\)
The line \(x = 3\sqrt{3}\) cuts the circle at \(y = \pm\sqrt{36-27} = \pm3\). The smaller part lies between \(x = 3\sqrt{3}\) and x = 6, and the circle is symmetric about the x-axis, so the area is twice the part above the x-axis:
\( \displaystyle A = 2\int_{3\sqrt{3}}^{6}\sqrt{36-x^{2}}\,dx \)
\( \displaystyle = 2\left[\frac{x}{2}\sqrt{36-x^{2}}+18\sin^{-1}\frac{x}{6}\right]_{3\sqrt{3}}^{6} \)
At x = 6 the bracket is \(0+18\times\frac{\pi}{2} = 9\pi\). At \(x = 3\sqrt{3}\) it is \(\frac{3\sqrt{3}}{2}\times3+18\times\frac{\pi}{3} = \frac{9\sqrt{3}}{2}+6\pi\). So
\( \displaystyle A = 2\left[9\pi-\left(\frac{9\sqrt{3}}{2}+6\pi\right)\right] \)
\( \displaystyle = 2\left(3\pi-\frac{9\sqrt{3}}{2}\right) = 6\pi-9\sqrt{3} \)
Why the other options are incorrect
Option (2) \(3\pi-\displaystyle\frac{9\sqrt{3}}{2}\) — Incorrect. This is only the half above the x-axis; the factor 2 for the half below is left out.
Option (3) \(6\pi+9\sqrt{3}\) — Incorrect. The value \(\frac{9\sqrt{3}}{2}+6\pi\) at \(x = 3\sqrt{3}\) is subtracted as a whole, so the \(9\sqrt{3}\) term comes with a minus sign, not a plus sign.
Option (4) \(30\pi+9\sqrt{3}\) — Incorrect. This is the larger part: \(36\pi-(6\pi-9\sqrt{3}) = 30\pi+9\sqrt{3}\).