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How do you find the volume bounded by #y = e^x# , y=0 , x = 0, x = ln2 revolved about the x=axis?

1 Answer

Jun 19, 2018

#(3pi)/2~~4.71#

Explanation:

#V=piint_a^by^2dx#

#y^2=(e^x)^2=e^(2x)#

#int_0^ln2e^(2x)dx=[e^(2x)/2]_0^ln2#

#pi[e^(2x)/2]_0^ln2=pi(e^(2ln2)/2-e^(2(0))/2)=pi(4/2-1/2)=(3pi)/2~~4.71#

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