The volume of the solid of revolution is [tex]\frac{16\pi}{3}[/tex] cubic units.
First, we determine the limits between the two curves. ([tex]f(x) = 4\cdot x -2[/tex], [tex]g(x) = x^{2}+1[/tex])
[tex]f(x) = g(x)[/tex] (1)
[tex]4\cdot x - 2 = x^{2}+1[/tex]
[tex]x^{2}-4\cdot x +3 = 0[/tex]
[tex](x-1)\cdot (x-3) = 0[/tex]
The lower and upper bounds are 1 and 3, respectively. It is to notice that [tex]f(x) > g(x)[/tex] for [tex]x \in (1, 3)[/tex]. Thus, we determine the volume of the solid of revolution by shell method, that is to say:
[tex]V = 2\pi \int\limits^a_b {x\cdot |f(x) - g(x) |} \, dx[/tex] (2)
If we know that [tex]a = 3[/tex], [tex]b = 1[/tex], [tex]f(x) = 4\cdot x - 2[/tex] and [tex]g(x) = x^{2}+1[/tex], then the volume of the solid of revolution is:
[tex]V = 2\pi \int\limits^3_1 {|4\cdot x^{2}-2\cdot x -x^{3}-x|} \, dx[/tex]
[tex]V = 2\pi\int\limits^3_1 {(4\cdot x^{2}-x^{3}-3\cdot x)} \, dx[/tex]
[tex]V = 8\pi \int\limits^3_1 {x^{2}} \, dx - 2\pi \int\limits^3_1 {x^{3}} \, dx -6\pi \int\limits^3_1 {x} \, dx[/tex]
[tex]V = 8\pi\cdot \left(\frac{3^{3}}{3}-\frac{1^{3}}{3} \right)-2\pi \cdot \left(\frac{3^{4}}{4}-\frac{1^{4}}{4} \right) - 6\pi\cdot \left(\frac{3^{2}}{2}-\frac{1^{2}}{2} \right)[/tex]
[tex]V = \frac{208\pi}{3} - 40\pi -24\pi[/tex]
[tex]V = \frac{16\pi}{3}[/tex]
The volume of the solid of revolution is [tex]\frac{16\pi}{3}[/tex] cubic units.
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