The area of the region bounded by the curve x = \(\rm \sqrt{9-y^2}\) and y-axis is 

Concept:

\(\rm \int\sqrt{a^2-x^2}=\frac x 2\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac x a\)

Calculations:

Given, the curve is  \(\rm x = \sqrt {9 - y^2}\)

Now, equation of Y-axis x = 0, 

⇒ y² = 9 

⇒ y = -3, 3

Area bounded by  the curve y = \(\rm \sqrt{9-x^2}\) and y-axis 

\(=\rm 2\int_0^3ydx\)

\(=\rm 2\int_0^3\rm \sqrt{9-x^2}dx\)

\(\rm =2[\frac x2\sqrt{9-x^2}+\frac{9}{2}sin^{-1}\frac x 3]_0^3\)

\(\rm =2[\frac x2\sqrt{9-x^2}+\frac{9}{2}sin^{-1}\frac x 3]_0^3\)

= 2[\(\rm \frac{9π}{4}-0\)]

= 4.5 π sq. units 

Hence, option 4 is correct.