MHT CET · Maths · Area Under Curves
The area bounded by the circle \(x^{2}+y^{2}=16\) and lines x=0 and x=2 is
- A \(\left[4 \sqrt{3}+\frac{8 \pi}{3}\right]\) sq. units
- B \(\frac{1}{2}\left[4 \sqrt{3}+\frac{8 \pi}{3}\right]\) sq. units
- C \(\left[4 \sqrt{3}-\frac{8 \pi}{3}\right]\) sq. units
- D \(\frac{1}{2}\left[4 \sqrt{3}-\frac{8 \pi}{3}\right]\) sq. units
Answer & Solution
Correct Answer
(A) \(\left[4 \sqrt{3}+\frac{8 \pi}{3}\right]\) sq. units
Step-by-step Solution
Detailed explanation
Given equation of circle is \(x^{2}+y^{2}=16\)
\(\therefore y^{2}=16-x^{2} \therefore y=\sqrt{16-x^{2}}\)
Required area is shaded.
\(A=2 \mathrm{~A}(\mathrm{OABCO})\)
\(\text {Area } =2 \int_{0}^{2} \sqrt{16-x^{2}} \mathrm{dx} \)
\( =2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{0}^{2} \)
\( =2\left\{\frac{2}{2} \sqrt{12}+8 \sin ^{-1} \frac{1}{2}-0\right\}=2[2 \sqrt{3}+\) \(8\left(\frac{\pi}{6}\right)]=4 \sqrt{3}+\frac{8 \pi}{3}\)
\(\therefore y^{2}=16-x^{2} \therefore y=\sqrt{16-x^{2}}\)
Required area is shaded.
\(A=2 \mathrm{~A}(\mathrm{OABCO})\)
\(\text {Area } =2 \int_{0}^{2} \sqrt{16-x^{2}} \mathrm{dx} \)
\( =2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{0}^{2} \)
\( =2\left\{\frac{2}{2} \sqrt{12}+8 \sin ^{-1} \frac{1}{2}-0\right\}=2[2 \sqrt{3}+\) \(8\left(\frac{\pi}{6}\right)]=4 \sqrt{3}+\frac{8 \pi}{3}\)
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