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How do you find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis #y=e^(-x)#, y=0, x=0, x=1?

Answer 1

Christopher Agresta

I found:
#V=pi[1/2-e^-2/2]=1.36#

I used the Cylinder Method:

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Answer 2

Aurora Alvarado

To find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis y=e^(-x), y=0, x=0, x=1, you would use the method of cylindrical shells. The formula for the volume generated by revolving a region bounded by the curves y=f(x), y=0, x=a, and x=b about the x-axis is given by:

[ V = 2\pi \int_{a}^{b} x \cdot f(x) , dx ]

In this case, the region is bounded by y=e^(-x), y=0, x=0, and x=1. So, (a = 0) and (b = 1), and (f(x) = e^{-x}). Thus, the integral becomes:

[ V = 2\pi \int_{0}^{1} x \cdot e^{-x} , dx ]

You can evaluate this integral to find the volume of the solid generated by revolving the given region about the x-axis.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.