let the region r be the area enclosed by the function f(x) = √x - 2, the x - axis, and the y - axis. find…

Explanation:

Step1: Find the intersection with x - axis

Set $f(x)=\sqrt{x}-2 = 0$, then $\sqrt{x}=2$, so $x = 4$. The region is bounded by $x = 0$ and $x = 4$.

Step2: Use the disk - washer method formula

The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the disk - washer method is $V=\pi\int_{a}^{b}[(R(x))^{2}-(r(x))^{2}]dx$. Here, $a = 0$, $b = 4$, $R(x)=5+\sqrt{x}-2=3 + \sqrt{x}$ (distance from the line $y=-5$ to the curve $y = \sqrt{x}-2$) and $r(x)=5$ (distance from the line $y=-5$ to the $x$-axis).

Step3: Set up the integral

$V=\pi\int_{0}^{4}[(3 + \sqrt{x})^{2}-5^{2}]dx=\pi\int_{0}^{4}(9 + 6\sqrt{x}+x - 25)dx=\pi\int_{0}^{4}(x + 6x^{\frac{1}{2}}-16)dx$.

Step4: Integrate term - by - term

$\int(x + 6x^{\frac{1}{2}}-16)dx=\frac{1}{2}x^{2}+6\times\frac{2}{3}x^{\frac{3}{2}}-16x+C=\frac{1}{2}x^{2}+4x^{\frac{3}{2}}-16x+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{1}{2}x^{2}+4x^{\frac{3}{2}}-16x\right]_{0}^{4}=\pi\left(\frac{1}{2}(4)^{2}+4(4)^{\frac{3}{2}}-16(4)\right)=\pi(8 + 32-64)=\pi(-24)\approx - 75.398$. But volume is non - negative, so $V\approx75.398$.

Answer:

$75.398$