Question:
<p>The area (in sq. units) of the part of the circle x<sup>2</sup> + y<sup>2</sup> = 36, which is outside the parabola y<sup>2</sup> = 9x, is:</p>
<p style="display:inline"><span class="math-tex">\(12 \pi-3 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(24 \pi-3 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(12 \pi+3 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(24 \pi+3 \sqrt{3}\)</span></p>
Step-by-Step Solution
Key Concept: The area outside the parabola is calculated by subtracting the area shared by the circle and the parabola from the total area of the circle, splitting the integral at their intersection point.
<p><img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1757488893-u9eamy.jpg" style="height:196px; width:250px" /><br />
Required area<br />
<span class="math-tex">\( =\pi \times(6)^2-2 \int\limits_0^3 \sqrt{9} {xdx}-\int\limits_3^6 \sqrt{36-{x}^2} {dx} \)</span><br />
<span class="math-tex">\( =36 \pi-12 \sqrt{3}-2\left(\frac{{x}}{2} \sqrt{36-{x}^2}+18 \sin ^{-1} \frac{{x}}{6}\right)_3^6 \)</span><br />
<span class="math-tex">\( =36 \pi-12 \sqrt{3}-2\left(9 \pi-3 \pi-\frac{9 \sqrt{3}}{2}\right) \)</span><br />
<span class="math-tex">\( =24 \pi-3 \sqrt{3} \)</span></p>
Correct Answer: B