Area Under the Curve
Area between ellipse and line
Grade 12
Question:
<p>The area of the region bounded by the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1\) in the first quadrant is: [MAU010]</p>
<li>\(3\pi\)</li>
<li>\(\dfrac{3\pi}{2}\)</li>
<li>\(6\pi\)</li>
<li>\(\dfrac{\pi}{2}\)</li>
Step-by-Step Solution
Key Concept: Area of full ellipse = \piab = \pi \cdot 2 \cdot 3 = 6\pi. First quadrant = 6\pi/4 = 3\pi/2.
<div class='solution'>
<p>Full ellipse area $=\pi ab=\pi\cdot2\cdot3=6\pi$.</p>
<p>First-quadrant area $=\frac{6\pi}{4}=\boxed{\frac{3\pi}{2}}$.</p>
<p><em>Direct:</em> $\int_0^2\frac{3}{2}\sqrt{4-x^2}\,dx=\frac{3}{2}\cdot\frac{\pi\cdot4}{4}=\frac{3\pi}{2}$.</p>
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Correct Answer: B