Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify your results using the integration capabilities of a graphing utility. y = e^(x-1), y=0, x=1, x=2

Answer 1

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$  or 10.036

Step-by-step explanation:

Given the data in the question;

y = $y = e^{(x - 1 )$, y = 0, x = 1, x = 2.

Now, using the integration capabilities of a graphing utility

y = $y = e_2}^{(x - 1 )_$, y = 0

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx$

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2 dx$

Volume = $\pi \int\limits^2_1 e^{2x-2}dx$

Volume = $\frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx$

Volume = $\frac{\pi }{e^2} [\frac{e^{2x}}{2}]^2_1$

Volume = $\frac{\pi }{2e^2} [e^4 - e^2 ]$  

Volume = $\frac{\pi }{2} [e^2 - 1 ]$  or 10.036

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$  or 10.036