Let R be the region in the first quadrant enclosed by the graphs of
#y=e^(-x^2)#, #y=1-cosx#, and the y axis, how do you find the volume of the solid generated when the region R is revolved about the x axis?

Question

Let R be the region in the first quadrant enclosed by the graphs of 
#y=e^(-x^2)#, #y=1-cosx#, and the y axis, how do you find the volume of the solid generated when the region R is revolved about the x axis?

Emma Aldridge · Accepted Answer

See the answer below:
Credits: Thanks to web site https://tutor.hix.ai to help us with the special integral Gauss error function.

William Abdalla · Answer

undefined To find the volume of the solid generated when the region $ R $ is revolved about the $ x $-axis, you can use the method of cylindrical shells. First, find the points of intersection between the curves $ y = e^{-x^2} $ and $ y = 1 - \cos(x) $. These points represent the limits of integration for $ x $. Then, integrate the circumference of cylindrical shells from $ x = 0 $ to the point of intersection, multiplied by the height of the shell, which is the difference between the $ y $-coordinates of the curves at that $ x $-value. This integral will give you the volume of the solid of revolution. The integral expression for the volume is:

$$ V = \int_{a}^{b} 2\pi x(f(x) - g(x)) \, dx $$

where $ a $ and $ b $ are the points of intersection between the curves, $ f(x) $ is the upper curve (in this case $ 1 - \cos(x) $), and $ g(x) $ is the lower curve (in this case $ e^{-x^2} $).

Let R be the region in the first quadrant enclosed by the graphs of #y=e^(-x^2)#, #y=1-cosx#, and the y axis, how do you find the volume of the solid generated when the region R is revolved about the x axis?