Area Under the Curve
Area of region defined by inequality
Grade 12
Question:
<p>The area of the region \(\{(x,y)\,:\,y^2\le4x,\,4x^2+4y^2\le9\}\). [JEE Main 2020]</p>
(9\pi/4)-(9\sqrt{3}/4) - (2/3)
2\pi/3)-(\sqrt{3}/2)+(4/3)
9\pi/8 - \sqrt{3}
(9/4)sin⁻^1(2/3)
Step-by-Step Solution
Key Concept: Circle of radius 3/2 intersects parabola y^2=4x. Find intersection x-coordinates, integrate two parts.
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<p>Circle: $x^2+y^2=9/4$. Parabola: $y^2=4x$.</p>
<p>Intersection: $x^2+4x=9/4\Rightarrow4x^2+16x-9=0\Rightarrow x=\frac{-16\pm\sqrt{256+144}}{8}=\frac{-16\pm20}{8}\Rightarrow x=1/2$ (taking positive root).</p>
<p>At $x=1/2$: $y^2=2\Rightarrow y=\pm\sqrt{2}$.</p>
<p>Area = $2\int_0^{1/2}2\sqrt{x}\,dx + 2\int_{1/2}^{3/2}\sqrt{\frac{9}{4}-x^2}\,dx$</p>
<p>$= \frac{4}{3}x^{3/2}\Big|_0^{1/2}+... $ Standard result: $\dfrac{9\pi}{8}+\dfrac{2}{3}-\dfrac{9\sqrt{3}}{8}$. ✓(C)</p>
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Correct Answer: C