The volume of the solid of revolution is approximately 37439.394 cubic units.
How to find the solid of revolution enclosed by two functions
Let be [tex]f(x) = e^{\frac{x}{6} }[/tex] and [tex]g(x) = e^{\frac{35}{6} }[/tex], whose points of intersection are [tex](x_{1},y_{1}) =(0,1)[/tex], [tex](x_{2}, y_{2}) = (35, e^{35/6})[/tex], respectively. The formula for the solid of revolution generated about the y-axis is:
[tex]V = \pi \int\limits^{e^{35/6}}_{1} {f(y)} \, dy[/tex] (1)
Now we proceed to solve the integral: [tex]f(y) = 6\cdot \ln y[/tex]
[tex]V = \pi \int\limits^{e^{35/6}}_{1} {6\cdot \ln y} \, dy[/tex] (2)
[tex]V = 6\pi \int\limits^{e^{35/6}}_{1} {\ln y} \, dy[/tex]
[tex]V = 6\pi \left[(y-1)\cdot \ln y\right]\right|_{1}^{e^{35/6}}[/tex]
[tex]V = 6\pi \cdot \left[(e^{35/6}-1)\cdot \left(\frac{35}{6} \right)-(1-1)\cdot 0\right][/tex]
[tex]V = 35\pi\cdot (e^{35/6}-1)[/tex]
[tex]V \approx 37439.392[/tex]
The volume of the solid of revolution is approximately 37439.394 cubic units. [tex]\blacksquare[/tex]
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