KCET · Maths · Area Under Curves
The area of the region bounded by \(y=-\sqrt{16-x^{2}}\) and \(X\)-axis is
- A \(8 \pi \mathrm{sq}\) units
- B \(20 \pi\) sq units
- C \(16 \pi\) sq units
- D \(256 \pi\) sq units
Answer & Solution
Correct Answer
(A) \(8 \pi \mathrm{sq}\) units
Step-by-step Solution
Detailed explanation
The area bounded by \(y=\sqrt{16-x^{2}}\) and \(X\)-axis is shown in the figure.
\(\operatorname{Area}=\int_{-4}^{4} y d x\)
\(=\int_{-4}^{4} \sqrt{4^{2}-x^{2}} d x\)
\(=\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{-4}^{4}\)
\(=\left[0+8 \sin ^{-1}(1)\right]-\left[\left(8 \sin ^{-1}(-1)\right)\right]\)
\(=8\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right]\)
\(=8 \pi\)
\(\operatorname{Area}=\int_{-4}^{4} y d x\)
\(=\int_{-4}^{4} \sqrt{4^{2}-x^{2}} d x\)
\(=\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{-4}^{4}\)
\(=\left[0+8 \sin ^{-1}(1)\right]-\left[\left(8 \sin ^{-1}(-1)\right)\right]\)
\(=8\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right]\)
\(=8 \pi\)
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