let the region r be the area enclosed by the function f(x) = √x - 1 and g(x) = 1/2x - 1. find the volume of…

Explanation:

Step1: Find intersection points

Set $\sqrt{x}-1=\frac{1}{2}x - 1$. Then $\sqrt{x}=\frac{1}{2}x$. Let $t = \sqrt{x}(t\geq0)$, so $t=\frac{1}{2}t^{2}$. Rearranging gives $\frac{1}{2}t^{2}-t = 0$, factoring out $t$ we get $t(\frac{1}{2}t - 1)=0$. So $t = 0$ or $t = 2$. When $t = 0,x = 0$; when $t = 2,x = 4$. The intersection - points are $x = 0$ and $x = 4$.

Step2: Use the disk - washer method

The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the disk - washer method is $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer radius and $r(x)$ is the inner radius. Here, $R(x)=\sqrt{x}-1 + 2=\sqrt{x}+1$ and $r(x)=\frac{1}{2}x - 1+2=\frac{1}{2}x + 1$, and $a = 0$, $b = 4$.

Step3: Set up the integral

$V=\pi\int_{0}^{4}[(\sqrt{x}+1)^{2}-(\frac{1}{2}x + 1)^{2}]dx$. Expand the integrand: $(\sqrt{x}+1)^{2}=x + 2\sqrt{x}+1$ and $(\frac{1}{2}x + 1)^{2}=\frac{1}{4}x^{2}+x + 1$. Then the integrand is $x + 2\sqrt{x}+1-(\frac{1}{4}x^{2}+x + 1)=2\sqrt{x}-\frac{1}{4}x^{2}$.

Step4: Evaluate the integral

$V=\pi\int_{0}^{4}(2x^{\frac{1}{2}}-\frac{1}{4}x^{2})dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\pi\left[2\times\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{4}\times\frac{1}{3}x^{3}\right]{0}^{4}$. $=\pi\left[\frac{4}{3}x^{\frac{3}{2}}-\frac{1}{12}x^{3}\right]{0}^{4}=\pi\left(\frac{4}{3}\times4^{\frac{3}{2}}-\frac{1}{12}\times4^{3}\right)$. $=\pi\left(\frac{4}{3}\times8-\frac{1}{12}\times64\right)=\pi\left(\frac{32}{3}-\frac{16}{3}\right)=\frac{16\pi}{3}\approx16.755$.

Answer:

$16.755$