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…

Question

asked Feb 13, 2022 167k views

1 Answer

Hedda · Answer 1 · 2022-02-18T03:09:14+0000

Answer:

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

$(\pi )/(2) [e^2 - 1 ]$ or 10.036

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 =
$(\pi )/(e^2) \int\limits^2_1 e^(2x)dx$

Volume =
$(\pi )/(e^2) [(e^(2x))/(2)]^2_1$

Volume =
$(\pi )/(2e^2) [e^4 - e^2 ]$

Volume =
$(\pi )/(2) [e^2 - 1 ]$ 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;

$(\pi )/(2) [e^2 - 1 ]$ or 10.036