Question

The area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² = 32 is
(a) 16π sq. units
(b) 4π sq. units
(c) 32π sq. u nits
(d) 24 sq. units

Answer 1

To find: area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² = 32

y = x ..(1)
x² + y² = 32 ..(2)

Solving (1) and (2), we find the coordinates of the point of intersection A.
i.e., A(4, 4)

Draw perpendicular AC to x-axis.



The required area of the region AOBA = Area of the region bounded by AOCA + Area of the region bounded by ACBA

Thus,
$$\begin{gathered} \mathrm{Required}\mathrm{area}={\int }_0^{4}xdx+{\int }_4^{4\sqrt{2}}\sqrt{32-x^2}dx \\ ={\left(\frac{x^2}{2}\right)}_0^4+{\left(\frac{x}{2}\sqrt{32-x^2}+\frac{32}{2}{\sin }^{-1}\frac{x}{4\sqrt{2}}\right)}_4^{4\sqrt{2}} \\ =\left(\frac{4^2}{2}-\frac{0^2}{2}\right)+\left(\frac{4\sqrt{2}}{2}\sqrt{32-{\left(4\sqrt{2}\right)}^2}+\frac{32}{2}{\sin }^{-1}\frac{4\sqrt{2}}{4\sqrt{2}}-\frac{4}{2}\sqrt{32-4^2}-\frac{32}{2}{\sin }^{-1}\frac{4}{4\sqrt{2}}\right) \\ =\left(\frac{16}{2}\right)+\left(2\sqrt{2}\sqrt{32-32}+16{\sin }^{-1}1-2\sqrt{32-16}-16{\sin }^{-1}\frac{1}{\sqrt{2}}\right) \\ =\left(8\right)+\left(16\times \frac{\mathrm{\pi }}{2}-2\sqrt{16}-16\times \frac{\mathrm{\pi }}{4}\right) \\ =\left(8\right)+\left(8\mathrm{\pi }-2\times 4-4\mathrm{\pi }\right) \\ =\left(8\right)+\left(4\mathrm{\pi }-8\right) \\ =4\mathrm{\pi } \\ \mathrm{Thus},\mathrm{Area}=4\mathrm{\pi }\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, the correct option is (b).