The region under the curves #y=e^(1-2x), 0<=x<=2# is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?

1 Answer
Jan 8, 2018

See below.

Explanation:

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It can be seen from the diagram, that if we form rectangles with a width of #deltax# and a height of #f(x)# and revolve these around the x axis through an angle #pi# radians, we will form a series of discs. These discs will have a radius f(x) and thicness #deltax#. The volume of a disc will therfore be:

#pi(f(x))^2*deltax#

When these discs are summed in the given interval, they will give the volume of revolution.

We need to use the interval #[0 , 2]#

so our integral will be:

#pi*int_(0)^(2)(e^(1-2x))^2 dx#

#(e^(1-2x))^2=e^(2-4x)#

#pi*int_(0)^(2)(e^(2-4x)) dx=pi*[-1/4e^(2-4x)]_(0)^(2)#

#pi*{[-1/4e^(2-4x)]^(2)-[-1/4e^(2-4x)]_(0)}#

Plugging in upper and lower bounds:

#pi*{[-1/4e^(2-4(2))]^(2)-[-1/4e^(2-4(0))]_(0)}#

#pi*{[-1/(4e^6) ]^(2)-[-1/4e^(2)]_(0)}#

#pi*{(-1+e^8)/(4e^6)}=((-1+e^8)/(4e^6))*picolor(white)(88)# cubic units

Volume #= ((-1+e^8)/(4e^6))*pi~~5.8014#

Volume of Revolution:

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