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

Solution:

Given, the curve is x² + y² = 32 --- (1)

Given, the line is y = x --- (2)

We have to find the area of the region in the first quadrant enclosed by the x-axis, the line and the circle.

The standard form of equation of the circle is given by

x² + y² = r² --- (3)

Comparing (1) and (3)

Radius of circle, r = √32 = 4√2

From (1) and (2),

x² + x² = 32

2x² = 32

x² = 32/2

x² = 16

x = ±4

When x = 4,

(4)² + y² =32

16 + y² = 32

y² = 32 - 16

y² = 16

y = ±4

The points of intersection are (4, 4) and (-4, -4).

From the equation of the curve, y² = 32 - x²

y = √(32 - x²)

Required area = \(\int_{0}^{4}x\, dx+\int_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}\, dx\)

\(\int_{0}^{4}x\, dx =[\frac{x^{2}}{2}]_{0}^{4}=[\frac{4^{2}}{2}-0]=\frac{16}{2}=8\)

\(\int_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}\,dx=[\frac{x\sqrt{32-x^{2}}}{2}+\frac{32}{2}sin^{-1}(\frac{x}{4\sqrt{2}})]_{4}^{4\sqrt{2}}\\=[\frac{4\sqrt{2}\sqrt{32-(4\sqrt{2})^{2}}}{2}+\frac{32}{2}sin^{-1}(\frac{4\sqrt{2}}{4\sqrt{2}})]-[\frac{4\sqrt{32-(4)^{2}}}{2}+\frac{32}{2}sin^{-1}(\frac{4}{4\sqrt{2}})]\\=[0+16sin^{-1}(1)]-[8+16sin^{-1}(\frac{1}{\sqrt{2}})]\\=[16\times \frac{\pi }{2}-8+16\times \frac{\pi }{4}]\\=[8\pi -8-4\pi ]\\=[4\pi -8]\)

Required area = \(\int_{0}^{4}x\, dx+\int_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}\, dx\)

= 8+4π -8=4π

Therefore, the area is 4π square units.

---

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

Summary:

The area of the region in the first quadrant enclosed by the x-axis, the line y = x and and circle x² + y² =32 is 4π square units.