CBSE 2024 · Region 5 · Set 1 · Q33 · 5 marks
Find the area of the region bounded by the curve $\displaystyle 4 x^{2}+y^{2}=36$ using integration.
Marking-scheme solution
The given equation can be written as: $\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{36}=1$, which is an ellipse.
Correct Graph Area of the region bounded by the curve\begin{aligned}
& =4 \times \frac{6}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x
& =8\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{0}^{3}
& =18 \pi
\end{aligned}
$$
Correct Graph Area of the region bounded by the curve\begin{aligned}
& =4 \times \frac{6}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x
& =8\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{0}^{3}
& =18 \pi
\end{aligned}
$$Application of IntegralsArea under Simple CurvesApplylong_answermedium
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