How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #y=e^x# on the interval #0

Question

How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #y=e^x# on the interval #0<=x<=1# ?

Samantha Adkison · Accepted Answer

The answer is #pi/2[e^2-1]#. Since you are only given a single function and we are rotating about the axis of the parameter, this requires the disk method.  The disk method is: #V=int_a^b Adx#
#=int_a^b pi r^2dx#
#=int_a^b pi [f(x)]^2dx# We have the known values: #f(x)=e^x#
#a=0#
#b=1# And now we can substitute: #V=int_0^1 pi (e^x)^2dx#
#=pi int_0^1 e^(2x)dx#
#=pi (e^(2x))/2|_0^1#
#=pi/2[e^2-e^0]#
#=pi/2[e^2-1]#

Scarlett Allison · Answer

undefined To find the surface area of the solid obtained by rotating the region bounded by $ y = e^x $ on the interval $ 0 \leq x \leq 1 $ about the x-axis, you would use the formula for surface area of revolution, which is $ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} \, dx $, where $ f(x) $ represents the function describing the curve and $ [a, b] $ is the interval of integration.

In this case, $ f(x) = e^x $, so $ f'(x) = e^x $. Thus, the integral becomes $ S = 2\pi \int_{0}^{1} e^x \sqrt{1 + e^{2x}} \, dx $.

This integral represents the surface area of the solid of revolution. By evaluating this integral using appropriate techniques, such as substitution or integration by parts, you can find the surface area of the solid obtained by rotating the given region about the x-axis.

How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #y=e^x# on the interval #0&lt;=x&lt;=1# ?

How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #y=e^x# on the interval #0<=x<=1# ?