let the region r be the area enclosed by the function f(x)=x^(1/3) and g(x)=(1/3)x. find the volume of the…

Explanation:

Step1: Find intersection points

Set $x^{\frac{1}{3}}=\frac{1}{3}x$. Let $t = x^{\frac{1}{3}}$, then $t=\frac{1}{3}t^{3}$, $t^{3}-3t = 0$, $t(t^{2}-3)=0$. So $t = 0,\pm\sqrt{3}$, and $x = 0, \pm3\sqrt{3}$. Since the region in the graph is in the first - quadrant, we consider $x = 0$ and $x = 3\sqrt{3}$.

Step2: Use the washer - method formula

The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the 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, $R(x)=x^{\frac{1}{3}} + 1$ and $r(x)=\frac{1}{3}x + 1$, $a = 0$, $b = 3\sqrt{3}$.

Step3: Set up the integral

$V=\pi\int_{0}^{3\sqrt{3}}((x^{\frac{1}{3}} + 1)^{2}-(\frac{1}{3}x + 1)^{2})dx$. Expand the integrand: $(x^{\frac{1}{3}}+1)^{2}=x^{\frac{2}{3}} + 2x^{\frac{1}{3}}+1$ and $(\frac{1}{3}x + 1)^{2}=\frac{1}{9}x^{2}+\frac{2}{3}x + 1$. Then the integrand is $x^{\frac{2}{3}}+2x^{\frac{1}{3}}+1-(\frac{1}{9}x^{2}+\frac{2}{3}x + 1)=x^{\frac{2}{3}}+2x^{\frac{1}{3}}-\frac{1}{9}x^{2}-\frac{2}{3}x$.

Step4: Integrate term - by - term

$\int(x^{\frac{2}{3}}+2x^{\frac{1}{3}}-\frac{1}{9}x^{2}-\frac{2}{3}x)dx=\frac{3}{5}x^{\frac{5}{3}}+\frac{3}{2}x^{\frac{4}{3}}-\frac{1}{27}x^{3}-\frac{1}{3}x^{2}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{3}{5}x^{\frac{5}{3}}+\frac{3}{2}x^{\frac{4}{3}}-\frac{1}{27}x^{3}-\frac{1}{3}x^{2}\right]_{0}^{3\sqrt{3}}$. $x = 3\sqrt{3}=3^{\frac{3}{2}}$. $\frac{3}{5}(3^{\frac{3}{2}})^{\frac{5}{3}}+\frac{3}{2}(3^{\frac{3}{2}})^{\frac{4}{3}}-\frac{1}{27}(3^{\frac{3}{2}})^{3}-\frac{1}{3}(3^{\frac{3}{2}})^{2}$ $=\frac{3}{5}\times3^{\frac{5}{2}}+\frac{3}{2}\times3^{2}-\frac{1}{27}\times3^{\frac{9}{2}}-\frac{1}{3}\times3^{3}$ $=\frac{3}{5}\times3^{2}\times3^{\frac{1}{2}}+\frac{27}{2}-\frac{1}{27}\times3^{4}\times3^{\frac{1}{2}} - 9$ $=\frac{27}{5}\sqrt{3}+\frac{27}{2}-3\sqrt{3}-9$ $=(\frac{27}{5}-3)\sqrt{3}+\frac{27 - 18}{2}$ $=\frac{12}{5}\sqrt{3}+\frac{9}{2}$. $V=\pi(\frac{12}{5}\sqrt{3}+\frac{9}{2})\approx\pi(4.157+4.5)\approx27.185$.

Answer:

$27.185$