Question

Let S denote the plane region bounded by the following curves:

Answer 1

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 $f(x) = e^{\frac{x}{6} }$ and $g(x) = e^{\frac{35}{6} }$, whose points of intersection are $(x_{1},y_{1}) =(0,1)$, $(x_{2}, y_{2}) = (35, e^{35/6})$, respectively. The formula for the solid of revolution generated about the y-axis is:

$V = \pi \int\limits^{e^{35/6}}_{1} {f(y)} \, dy$ (1)

Now we proceed to solve the integral: $f(y) = 6\cdot \ln y$

$V = \pi \int\limits^{e^{35/6}}_{1} {6\cdot \ln y} \, dy$ (2)

$V = 6\pi \int\limits^{e^{35/6}}_{1} {\ln y} \, dy$

$V = 6\pi \left[(y-1)\cdot \ln y\right]\right|_{1}^{e^{35/6}}$

$V = 6\pi \cdot \left[(e^{35/6}-1)\cdot \left(\frac{35}{6} \right)-(1-1)\cdot 0\right]$

$V = 35\pi\cdot (e^{35/6}-1)$

$V \approx 37439.392$

The volume of the solid of revolution is approximately 37439.394 cubic units. $\blacksquare$

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