Question

use a geometric formula to find the area between the graphs of ( y = f(x) ) and ( y = g(x) ) over the indicated interval.
( f(x)=\frac{sqrt{3}}{3} x, g(x)=sqrt{25 - x^{2}} ;left0, \frac{5 sqrt{3}}{2}
ight )
the area is (type an exact answer, using ( pi ) as needed.)

Explanation:

Step1: Analyze the functions

The function \(y = g(x)=\sqrt{25 - x^{2}}\) represents the upper - half of the circle \(x^{2}+y^{2}=25\) (center \((0,0)\) and radius \(r = 5\)). The function \(y = f(x)=\frac{\sqrt{3}}{3}x\) is a straight line passing through the origin with slope \(m=\frac{\sqrt{3}}{3}\).

Step2: Calculate the area of the sector

The area of a sector of a circle with radius \(r\) and central angle \(\theta\) (in radians) is \(A_{sector}=\frac{1}{2}r^{2}\theta\).
For the circle \(x^{2}+y^{2}=25\), \(r = 5\).
The slope of the line \(y=\frac{\sqrt{3}}{3}x\) is \(m=\tan\theta=\frac{\sqrt{3}}{3}\), so \(\theta=\frac{\pi}{6}\) (since \(\tan\theta=\frac{\sqrt{3}}{3}\) and \(\theta\in[0,\frac{\pi}{2}]\)).
The area of the sector of the circle \(x^{2}+y^{2}=25\) with radius \(r = 5\) and \(\theta=\frac{\pi}{6}\) is \(A_{sector}=\frac{1}{2}\times5^{2}\times\frac{\pi}{6}=\frac{25\pi}{12}\).

Step3: Calculate the area of the triangle

The area of a triangle with base \(b\) and height \(h\) is \(A_{triangle}=\frac{1}{2}bh\).
For the line \(y = \frac{\sqrt{3}}{3}x\), when \(x=\frac{5\sqrt{3}}{2}\), \(y=\frac{5}{2}\).
The area of the triangle formed by the line \(y=\frac{\sqrt{3}}{3}x\), the \(x\) - axis and \(x = \frac{5\sqrt{3}}{2}\) is \(A_{triangle}=\frac{1}{2}\times\frac{5\sqrt{3}}{2}\times\frac{5}{2}=\frac{25\sqrt{3}}{8}\).

Step4: Calculate the area between the two curves

The area \(A\) between the two curves \(y = g(x)\) and \(y = f(x)\) is the area of the sector of the circle minus the area of the triangle.

$$
LATEXBLOCK0
$$

Answer:

\(\frac{25\pi}{12}-\frac{25\sqrt{3}}{8}\)