Question

set up a definite integral that represents the area bbounded by the graphs of the indicated equations over the given interval. compute answers to three decimal places. hint: a circle of radius r, with center at the origin, has equation ( x^{2}+y^{2}=r^{2} ) and area ( pi r^{2} ).
( y=sqrt{81 - x^{2}} ; y=-sqrt{81 - x^{2}} ; -9 leq x leq 9 )
set up a definite integral that represents the area bounded by the graphs of the indicated equations over the given interval.
( int_{-9}^{9} 2 sqrt{81 - x^{2}} d x )
what is the area bounded by these graphs, rounded to three decimal places?
( square ) square units

Explanation:

Step1: Recognize the geometric shape

The equations \(y = \sqrt{81 - x^{2}}\) and \(y=-\sqrt{81 - x^{2}}\) with \(-9\leq x\leq9\) represent the upper - half and lower - half of a circle. The general equation of a circle is \(x^{2}+y^{2}=r^{2}\). Here, \(y^{2}=81 - x^{2}\) or \(x^{2}+y^{2}=81\), so \(r = 9\).

Step2: Recall the area formula for a circle

The area formula for a circle is \(A=\pi r^{2}\). Substitute \(r = 9\) into the formula.

$$A=\pi\times9^{2}$$

$$A = 81\pi$$

$$A\approx81\times3.141592654$$

Answer:

\(254.469\) square units