Question

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

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

Answer 1

y = x                                           ...(1)

x² + y² = 32                                ...(2)

The region enclosed by y = x and x² + y² = 32 is shown in the following figure:

On solving (1) and (2) we find that the given line and circle meet at B(4, 4) in the first quadrant. Let us draw BM perpendicular to the x-axis.

Now, required area = area of triangle BOM + area of region BMAB

Area of triangle BOM ${\displaystyle ={\int }_0^{4}ydx={\int }_0^{4}xdx=\frac{1}{2}{\left[\frac{x^2}{2}\right]}_0^4=8.........\left(3\right)}$

Area of region BMAB= ${\displaystyle {\int }_0^{\sqrt{32}}ydx={\int }_0^{\sqrt{32}}\sqrt{32-x^2}}$

${\displaystyle ={[\frac{1}{2}\times \sqrt{32-x^2}+\frac{1}{2}\times 32\times {\sin }^{-1}\left(\frac{x}{\sqrt{32}}\right)]}_4^{\sqrt{32}}}$

${\displaystyle =(\frac{1}{2}\times \sqrt{32}\times 0+\frac{1}{2}\times 32\times {\sin }^{-1}\left(1\right))-(\frac{1}{2}\times 4\times 4+\frac{1}{2}\times 32\times {\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right))}$

${\displaystyle =8\pi -8-4\pi }$

∴ Area of triangle BOM=4π−8   ... (4)

On adding (3) and (4), we have:

Required area =${\displaystyle 8+4\pi -8=4\pi }$

Answer 2

Put y = x in ${\displaystyle x^2+y^2=32}$

${\displaystyle \therefore x^2 +x^2=32}$

${\displaystyle 2x^2=32}$

${\displaystyle x^2=16}$

x = 4

${\displaystyle A={\int }_0^4{y}_{\text{line}}dx+{\int }_4^{\sqrt{32}}{y}_{\text{circle}}dx}$

${\displaystyle A={\int }_0^{4}xdx+{\int }_4^{\sqrt{32}}\left(\sqrt{32-x^2}\right)dx}$

${\displaystyle ={\left(\frac{x^2}{2}\right)}_0^4+{\int }_4^{\sqrt{32}}\sqrt{{\left(\sqrt{32}\right)}^2-x^2}dx}$

${\displaystyle =\left(8\right)+{(\frac{x}{2}\sqrt{32-x^2}+\frac{32}{2}{\sin }^{-1}\left(\frac{x}{\sqrt{32}}\right))}^{\sqrt{32}}}$

${\displaystyle =\left(8\right)+(0+16\times \frac{\pi }{2}-(2\sqrt{16}+16{\sin }^{-1}\left(\frac{4}{\sqrt{32}}\right)))}$

${\displaystyle =8+8\pi -8-16{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)}$

${\displaystyle =8\pi -16\times \frac{\pi }{4}= 8\pi -4\pi =4\pi sq unit}$