Question

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?

Answer 1

See below.

Explanation:

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: