let $r$ be the region enclosed by the line $y = \\frac{x}{3}$, the line $y = 4 - x$, and the $x$-axis.\na…

Explanation:

Step1: Use the method of cylindrical - shells

The volume (V) of the solid of revolution about the vertical line (x = a) using the cylindrical - shells method is given by (V=2\pi\int_{c}^{d}(a - x)h(x)dx), where ((a - x)) is the radius of the shell and (h(x)) is the height of the shell. First, we need to find the intersection of (y=\frac{x}{3}) and (y = 4 - x). Set (\frac{x}{3}=4 - x), then (\frac{x}{3}+x=4), (\frac{x + 3x}{3}=4), (\frac{4x}{3}=4), (x = 3). The region (R) is bounded by (x = 0) to (x = 3) for (y=\frac{x}{3}) and (x = 3) to (x = 4) for (y = 4 - x).

Step2: Set up the integral for the volume

For (0\leq x\leq3), the height of the shell (h(x)=\frac{x}{3}) and the radius of the shell (r(x)=5 - x). For (3\leq x\leq4), the height of the shell (h(x)=4 - x) and the radius of the shell (r(x)=5 - x). The volume (V=2\pi\int_{0}^{3}(5 - x)\frac{x}{3}dx+2\pi\int_{3}^{4}(5 - x)(4 - x)dx).

Step3: Expand and integrate the first integral

Expand ((5 - x)\frac{x}{3}=\frac{5x}{3}-\frac{x^{2}}{3}). Then (\int_{0}^{3}(\frac{5x}{3}-\frac{x^{2}}{3})dx=\left[\frac{5x^{2}}{6}-\frac{x^{3}}{9}\right]_{0}^{3}=\frac{5\times3^{2}}{6}-\frac{3^{3}}{9}=\frac{45}{6}- 3=\frac{45 - 18}{6}=\frac{27}{6}=\frac{9}{2}).

Step4: Expand and integrate the second integral

Expand ((5 - x)(4 - x)=20-9x + x^{2}). Then (\int_{3}^{4}(20-9x + x^{2})dx=\left[20x-\frac{9x^{2}}{2}+\frac{x^{3}}{3}\right]_{3}^{4}=(20\times4-\frac{9\times4^{2}}{2}+\frac{4^{3}}{3})-(20\times3-\frac{9\times3^{2}}{2}+\frac{3^{3}}{3})) [ \begin{align*} &=(80 - 72+\frac{64}{3})-(60-\frac{81}{2}+9)\ &=(8+\frac{64}{3})-(69-\frac{81}{2})\ &=\frac{24 + 64}{3}-\frac{138 - 81}{2}\ &=\frac{88}{3}-\frac{57}{2}\ &=\frac{176 - 171}{6}=\frac{5}{6} \end{align*} ]

Step5: Calculate the volume

(V = 2\pi\left(\frac{9}{2}\right)+2\pi\left(\frac{5}{6}\right)=9\pi+\frac{5\pi}{3}=\frac{27\pi+ 5\pi}{3}=\frac{32\pi}{3})

Answer:

(\frac{32\pi}{3})