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

Answer 1

#V=(pi(e^2-1))/(2e^2)#

At any distance x, they coordinate happens to be the radius of the revolved element #y=e^-x# radius is #r=e^-x# Circular area generated by revolving around x axis is #A=pir^2# volume of the elementary solid of thickness dx is #dV=Adx# The lower limit is given to be x=0 The upper limit is given to be x=1
#dV=Adx#
#dV=pir^2dx# #r^2=(e^-x)^2# #r^2=e^-(2x)#
#dV=pie^-(2x)dx# Integrating between x=0 and x=1
#int_0^1dV=int_0^1pie^-(2x)dx#
#V=-pi/2|e^(-2x)|_0^1#
#V=-pi/2(e^(-2xx1)-e^(-2xx0))# #V=-pi/2(e^(-2)-1)#
#V=-pi/2(1/e^2-1)#
#V=pi/2(1-1/e^2)#
#V=pi/2(e^2-1)/e^2#
#V=(pi(e^2-1))/(2e^2)#
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Answer 2

To find the volume of the solid generated by revolving the region bounded by the graphs ( y = e^{-x} ), ( y = 0 ), ( x = 0 ), and ( x = 1 ) about the x-axis, you use the disk method. The volume ( V ) is given by the integral ( V = \pi \int_{a}^{b} (f(x))^2 , dx ), where ( f(x) ) represents the height of the region. In this case, ( f(x) = e^{-x} ). Integrating from ( x = 0 ) to ( x = 1 ) gives ( V = \pi \int_{0}^{1} (e^{-x})^2 , dx ). Solving this integral yields ( V = \pi \int_{0}^{1} e^{-2x} , dx ). Applying the antiderivative and evaluating the limits of integration gives the volume of the solid.

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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.

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.

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.

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.

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