question let the region r be the area enclosed by the function f(x)=3x^1/2 and g(x)=1/4x. if the region r is…

Explanation:

Step1: Find intersection points

Set $3\sqrt{x}=\frac{1}{4}x$. Let $t = \sqrt{x}(t\geq0)$, then $3t=\frac{1}{4}t^{2}$, which gives $t^{2}-12t = 0$, so $t(t - 12)=0$. Thus $t = 0$ or $t=12$, and $x = 0$ or $x = 144$.

Step2: Determine base - height of cross - section

The base of each rectangular cross - section perpendicular to the $x$ - axis is $b=\frac{1}{4}x-3\sqrt{x}$, and the height $h = 2(\frac{1}{4}x - 3\sqrt{x})$.

Step3: Set up volume integral

The volume $V$ of the solid using the cross - sectional area formula $A(x)=b\times h$ and integrating from $x = 0$ to $x = 144$ is $V=\int_{0}^{144}A(x)dx=\int_{0}^{144}2(\frac{1}{4}x - 3\sqrt{x})^{2}dx$. Expand $(\frac{1}{4}x - 3\sqrt{x})^{2}=\frac{1}{16}x^{2}-\frac{3}{2}x^{\frac{3}{2}} + 9x$. Then $A(x)=2(\frac{1}{16}x^{2}-\frac{3}{2}x^{\frac{3}{2}}+9x)=\frac{1}{8}x^{2}-3x^{\frac{3}{2}} + 18x$.

Step4: Evaluate the integral

$\int_{0}^{144}(\frac{1}{8}x^{2}-3x^{\frac{3}{2}}+18x)dx=\left[\frac{1}{8}\times\frac{1}{3}x^{3}-3\times\frac{2}{5}x^{\frac{5}{2}}+18\times\frac{1}{2}x^{2}\right]_{0}^{144}$ $=\frac{1}{24}(144)^{3}-\frac{6}{5}(144)^{\frac{5}{2}} + 9(144)^{2}$ $=\frac{1}{24}\times144\times144\times144-\frac{6}{5}\times144^{2}\times12+9\times144^{2}$ $=6\times144^{2}-\frac{6\times144^{2}\times12}{5}+9\times144^{2}$ $=144^{2}(6-\frac{72}{5}+9)$ $=144^{2}(\frac{30 - 72+45}{5})$ $=144^{2}\times\frac{3}{5}$ $=\frac{20736\times3}{5}=12441.6$

Answer:

$12441.6000$