question let the region r be the area enclosed by the function f(x)=2√x and g(x)=2x. if the region r is the…

Explanation:

Step1: Find intersection points

Set $2\sqrt{x}=2x$. Square both sides: $4x = 4x^{2}$, which simplifies to $x - x^{2}=0$, factoring gives $x(1 - x)=0$. So $x = 0$ and $x = 1$ are the intersection - points.

Step2: Determine the diameter of the semi - circle

The diameter $d$ of the semi - circle at a given $x$ is $d=2\sqrt{x}-2x$.

Step3: Find the radius of the semi - circle

The radius $r=\frac{d}{2}=\sqrt{x}-x$.

Step4: Find the area of the semi - circle

The area of a semi - circle is $A=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(\sqrt{x}-x)^{2}=\frac{1}{2}\pi(x - 2x^{\frac{3}{2}}+x^{2})$.

Step5: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is given by $V=\int_{a}^{b}A(x)dx$. Here, $a = 0$, $b = 1$, and $A(x)=\frac{1}{2}\pi(x - 2x^{\frac{3}{2}}+x^{2})$. So $V=\frac{\pi}{2}\int_{0}^{1}(x - 2x^{\frac{3}{2}}+x^{2})dx$. Integrating term - by - term: $\int(x - 2x^{\frac{3}{2}}+x^{2})dx=\frac{1}{2}x^{2}-\frac{4}{5}x^{\frac{5}{2}}+\frac{1}{3}x^{3}+C$. Evaluating the definite integral: $\frac{\pi}{2}\left[\frac{1}{2}x^{2}-\frac{4}{5}x^{\frac{5}{2}}+\frac{1}{3}x^{3}\right]_{0}^{1}=\frac{\pi}{2}\left(\frac{1}{2}-\frac{4}{5}+\frac{1}{3}\right)$. Simplify the expression inside the parentheses: $\frac{1}{2}-\frac{4}{5}+\frac{1}{3}=\frac{15 - 24 + 10}{30}=\frac{1}{30}$. So $V=\frac{\pi}{60}\approx0.052$.

Answer:

$0.052$