definite-integral

Question: The area of the smaller segment cut off from the circle ${{x}^{2}}+{{y}^{2}}=9$ by $x=1$ is

1) $\frac{1}{2}(9{{\sec }^{-1}}3-\sqrt{8})$

2) $9{{\sec }^{-1}}(3)-\sqrt{8}$

3) $\sqrt{8}-9{{\sec }^{-1}}(3)$

4) None of these

Solution:

Explanation: No Explanation

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