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

Explanation:

Step1: Find intersection points

Set $3x^{\frac{1}{3}}=\frac{3}{4}x$. Let $t = x^{\frac{1}{3}}$, then $3t=\frac{3}{4}t^{3}$. Rearranging gives $t^{3}-4t = 0$, factoring out $t$ we get $t(t^{2}-4)=0$, so $t = 0, t = 2,t=-2$. Since we are dealing with real - valued functions in the context of the area between curves in the first - quadrant (from the graph), we consider $x = 0$ and $x = 8$ (because when $t = 2,x=t^{3}=8$).

Step2: Determine the base of the cross - section

The base of the cross - section perpendicular to the $x$ - axis is $b=3x^{\frac{1}{3}}-\frac{3}{4}x$.

Step3: Determine the height of the cross - section

The height $h$ of the cross - section is $h = 2b=2\left(3x^{\frac{1}{3}}-\frac{3}{4}x\right)$.

Step4: Find the area of the cross - section

The area of the rectangle cross - section $A(x)=b\times h=(3x^{\frac{1}{3}}-\frac{3}{4}x)\times2(3x^{\frac{1}{3}}-\frac{3}{4}x)=2\left(3x^{\frac{1}{3}}-\frac{3}{4}x\right)^{2}$. Expand it: $A(x)=2\left(9x^{\frac{2}{3}}-\frac{9}{2}x^{\frac{4}{3}}+\frac{9}{16}x^{2}\right)=18x^{\frac{2}{3}} - 9x^{\frac{4}{3}}+\frac{9}{8}x^{2}$.

Step5: Calculate the volume using the integral

The volume $V=\int_{a}^{b}A(x)dx$, where $a = 0$ and $b = 8$. So $V=\int_{0}^{8}\left(18x^{\frac{2}{3}}-9x^{\frac{4}{3}}+\frac{9}{8}x^{2}\right)dx$. Integrating term - by - term: $\int 18x^{\frac{2}{3}}dx=18\times\frac{3}{5}x^{\frac{5}{3}}+C_1=\frac{54}{5}x^{\frac{5}{3}}+C_1$, $\int - 9x^{\frac{4}{3}}dx=-9\times\frac{3}{7}x^{\frac{7}{3}}+C_2=-\frac{27}{7}x^{\frac{7}{3}}+C_2$, $\int\frac{9}{8}x^{2}dx=\frac{9}{8}\times\frac{1}{3}x^{3}+C_3=\frac{3}{8}x^{3}+C_3$. Then $V=\left[\frac{54}{5}x^{\frac{5}{3}}-\frac{27}{7}x^{\frac{7}{3}}+\frac{3}{8}x^{3}\right]_{0}^{8}$.

Step6: Evaluate the definite integral

$V=\frac{54}{5}(8)^{\frac{5}{3}}-\frac{27}{7}(8)^{\frac{7}{3}}+\frac{3}{8}(8)^{3}$. Since $(8)^{\frac{5}{3}}=(2^{3})^{\frac{5}{3}} = 32$, $(8)^{\frac{7}{3}}=(2^{3})^{\frac{7}{3}} = 128$, and $(8)^{3}=512$. $V=\frac{54}{5}\times32-\frac{27}{7}\times128+\frac{3}{8}\times512$. $V=\frac{1728}{5}-\frac{3456}{7}+192$. Find a common denominator of 35: $V=\frac{1728\times7 - 3456\times5+192\times35}{35}=\frac{12096-17280 + 6720}{35}=\frac{12096+6720 - 17280}{35}=\frac{18816 - 17280}{35}=\frac{1536}{35}\approx43.886$.

Answer:

$43.886$