question\nlet the region r be the area enclosed by the function f(x)=2x^{\frac{1}{2}} and g(x)=\frac{1}{2}x…

Explanation:

Step1: Find intersection points

Set $2x^{\frac{1}{2}}=\frac{1}{2}x$. Rearranging gives $\frac{1}{2}x - 2x^{\frac{1}{2}}=0$. Let $u = x^{\frac{1}{2}}$, then $\frac{1}{2}u^{2}-2u = 0$, $u(\frac{1}{2}u - 2)=0$. So $u = 0$ or $u = 4$. Back - substituting $u = x^{\frac{1}{2}}$, we get $x = 0$ and $x = 16$.

Step2: Determine base of cross - section

The base of the rectangular cross - section perpendicular to the $x$ - axis is $b=2x^{\frac{1}{2}}-\frac{1}{2}x$.

Step3: Determine height of cross - section

The height of the rectangular cross - section is $h=\frac{1}{2}(2x^{\frac{1}{2}}-\frac{1}{2}x)$.

Step4: Find area of cross - section

The area of the rectangular cross - section $A(x)=b\times h=(2x^{\frac{1}{2}}-\frac{1}{2}x)\times\frac{1}{2}(2x^{\frac{1}{2}}-\frac{1}{2}x)=\frac{1}{2}(2x^{\frac{1}{2}}-\frac{1}{2}x)^{2}$. Expand it: $A(x)=\frac{1}{2}(4x - 2x^{\frac{3}{2}}+\frac{1}{4}x^{2}) = 2x - x^{\frac{3}{2}}+\frac{1}{8}x^{2}$.

Step5: Calculate the volume

The volume $V=\int_{a}^{b}A(x)dx=\int_{0}^{16}(2x - x^{\frac{3}{2}}+\frac{1}{8}x^{2})dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $V=\left[x^{2}-\frac{2}{5}x^{\frac{5}{2}}+\frac{1}{24}x^{3}\right]_{0}^{16}$. $V = 16^{2}-\frac{2}{5}\times16^{\frac{5}{2}}+\frac{1}{24}\times16^{3}$ $V = 256-\frac{2}{5}\times1024+\frac{1}{24}\times4096$ $V = 256-\frac{2048}{5}+\frac{512}{3}$ $V=\frac{3840 - 6144+2560}{15}=\frac{256}{15}\approx17.067$

Answer:

$17.067$