Calculus/Calculus

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Expert: - 4/15/2011


Question
Let r be the region in quadrant one enclosed by the graph of y=e^x and the line x=1. set up an integral expression to find the volume of the solid generated when r is revolved about the x-axis.

Answer
Revolving a point on the graph around the x axis makes a circle at that s value.
The left side is at x=0 and the right sides is at x=1.

To find the volume we need to integrate over all the circle areas from 0 to 1.
The area of a circle is pi*r^2 where r is y.

This means that r = e^x.  The area of a circle is pi*r^2, and here, r = e^x.

That comes down to integrating pi(e^x)^2 = pi(e^(2x)).
That integrates to F(x) = pi(e^(2x))/2, and this is used to find how much there is from x=0 to x=1.

The answer is gives by F(1) - F(0).

F(1) = pi(e^2/2) and F(0) = pi/2.

Factoring out a pi/2 gives (pi/2)((e^2/2) - 1).

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