Question

use geometric formulas to find the unsigned area between the graph of y = f(x) and the x - axis over the indicated interval.
f(x)=\sqrt{64 - x^{2}}; -8,8
the area between f(x) and the x - axis is \square. (type an exact answer, using \pi as needed.)

Explanation:

Step1: Recognize the function's graph shape

The function \(y = \sqrt{64 - x^{2}}\) can be rewritten as \(y^{2}+x^{2}=64\) (\(y\geq0\)). This is the upper - half of a circle with equation \(x^{2}+y^{2}=r^{2}\), where \(r = 8\).

Step2: Use the formula for the area of a semicircle

The area of a full circle is \(A=\pi r^{2}\). Since we have the upper - half of the circle (the region between \(y = \sqrt{64 - x^{2}}\) and the \(x\) - axis over the interval \([-8,8]\)), the area formula for a semicircle is \(A=\frac{1}{2}\pi r^{2}\).
Substitute \(r = 8\) into the formula: \(A=\frac{1}{2}\pi\times(8)^{2}\).
Calculate \((8)^{2}=64\), then \(A=\frac{1}{2}\pi\times64\).

Answer:

\(32\pi\)