Question

Area lying in first quadrant and bounded by the circle \(x^2+ y^2 = 9\) and the lines x = 1 and x = 3 is:

• A. \(|\frac{9π}{2}-\sqrt 2-\frac{9}{4}sin^{-1}\frac{1}{3}|\) sq.units
• B. \(|9π-\sqrt 2-\frac{9}{2}sin^{-1}\frac{1}{3}|\) sq.units
• C. \(|\frac{9π}{4}+\sqrt 2-\frac{9}{2}sin^{-1}\frac{1}{3}|\) sq.units
• D. \(|\frac{9π}{4}-\sqrt 2-\frac{9}{2}sin^{-1}\frac{1}{3}|\) sq.units

Answer 1

The Correct Option is D

We need to find the area in the first quadrant bounded by the circle \(x^2 + y^2 = 9\) and the lines \(x = 1\) and \(x = 3\).

First, we determine the function for the circle in terms of \(y\). The circle's equation is rewritten as:

\[ y = \sqrt{9 - x^2} \]

Next, we'll evaluate the area using integration. Since we're dealing only with the first quadrant and bound by \(x = 1\) and \(x = 3\), we integrate as follows:

\[ \text{Area} = \int_1^3 \sqrt{9-x^2} \, dx \]

To solve this integral, we use trigonometric substitution. Set \(x = 3\sin\theta\), leading to:

\[ dx = 3\cos\theta \, d\theta \]

For the limits of integration:

• When \(x = 1\): \(1 = 3\sin\theta \Rightarrow \sin\theta = \frac{1}{3}\), providing \(\theta = \sin^{-1}\left(\frac{1}{3}\right)\)
• When \(x = 3\): \(3 = 3\sin\theta \Rightarrow \sin\theta = 1\), so \(\theta = \frac{\pi}{2}\)

Hence, the integral becomes:

\[ \text{Area} = \int_{\sin^{-1}(1/3)}^{\pi/2} \sqrt{9 - 9\sin^2\theta} \cdot 3\cos\theta \, d\theta \]

Simplify:

\[ = \int_{\sin^{-1}(1/3)}^{\pi/2} 9\cos^2\theta \, d\theta \]

Using the identity \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\), we have:

\[ = \int_{\sin^{-1}(1/3)}^{\pi/2} \frac{9}{2}(1 + \cos 2\theta) \, d\theta \]

\[ = \frac{9}{2} \left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\sin^{-1}(1/3)}^{\pi/2} \]

Evaluate this expression:

\[ = \frac{9}{2} \left[ \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right)\right) + 0 \right] \]

The calculation of \(\sin 2\theta\) from \(\theta = \sin^{-1}(1/3)\):

\[ \sin 2\theta = 2\sin\theta\cos\theta = 2 \times \frac{1}{3} \times \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{2\sqrt{8}}{9} = \frac{2\sqrt{2}}{3} \]

Thus the definite value substitution gives us:

\[ \frac{9}{2}\left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right)\right) - \frac{9}{4} \times \frac{2\sqrt{2}}{3} \]

Simplifying further:

\[ \frac{9\pi}{4} - \frac{9}{2}\sin^{-1}\left(\frac{1}{3}\right) - \sqrt{2} \]

The area in question is:

\(|\frac{9π}{4}-\sqrt 2-\frac{9}{2}sin^{-1}\frac{1}{3}|\) square units.