JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in square units) of the region enclosed by the ellipse \(x^2+3 y^2=18\) in the first quadrant below the line \(y=x\) is :
- A \(\sqrt{3} \pi+\frac{3}{4}\)
- B \(\sqrt{3} \pi\)
- C \(\sqrt{3} \pi-\frac{3}{4}\)
- D \(\sqrt{3} \pi+1\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3} \pi\)
Step-by-step Solution
Detailed explanation
\(\frac{x^2}{18}+\frac{y^2}{6}=1\) \( \frac{x^2}{18}+\frac{3 x^2}{18}=1 \Rightarrow 4 x^2=18 \Rightarrow x^2=\frac{9}{2} \) \( \int_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} d x \)…
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