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 base of a…

Explanation:

Step1: Find intersection points

Set $2\sqrt{x}=2x$. Divide both sides by 2: $\sqrt{x}=x$. Square both sides: $x = x^{2}$, which gives $x^{2}-x=0$, factoring out $x$ we get $x(x - 1)=0$. So the intersection points are $x = 0$ and $x=1$.

Step2: Determine the diameter of semi - circle

The diameter $d$ of each semi - circle perpendicular to the $x$ - axis is $d=2\sqrt{x}-2x$.

Step3: Find the radius of semi - circle

The radius $r$ of the semi - circle is $r=\frac{2\sqrt{x}-2x}{2}=\sqrt{x}-x$.

Step4: Find the area of semi - circle

The area formula for a semi - circle is $A=\frac{1}{2}\pi r^{2}$. Substituting $r=\sqrt{x}-x$ into the formula, we have $A(x)=\frac{1}{2}\pi(\sqrt{x}-x)^{2}=\frac{1}{2}\pi(x - 2x^{\frac{3}{2}}+x^{2})$.

Step5: Calculate the volume using 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, $a = 0$, $b = 1$, so $V=\int_{0}^{1}\frac{1}{2}\pi(x - 2x^{\frac{3}{2}}+x^{2})dx$. We know that $\int_{0}^{1}\frac{1}{2}\pi(x - 2x^{\frac{3}{2}}+x^{2})dx=\frac{\pi}{2}\left[\frac{x^{2}}{2}-2\times\frac{2}{5}x^{\frac{5}{2}}+\frac{x^{3}}{3}\right]_{0}^{1}$. Evaluating the definite integral: [ \begin{align*} \frac{\pi}{2}\left(\frac{1}{2}-\frac{4}{5}+\frac{1}{3}\right)&=\frac{\pi}{2}\left(\frac{15 - 24+10}{30}\right)\ &=\frac{\pi}{2}\times\frac{1}{30}\ &=\frac{\pi}{60}\approx 0.052 \end{align*} ]

Answer:

$0.052$