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How do you use substitution to integrate #2piy(8-y^2/3)dy#?

1 Answer

Aug 31, 2015

You can simply multiply it out. No complicated substitution required. Looks like the Shell Method?

#V = 2piintxf(x)dx#

Switch out #x# for #y#. Looks like #f(y) = 8 - y^2/3#.

#= 2piint_a^b 8y - y^3/3dy#

#= 2pi[4y^2 - y^4/12]|_(a)^(b)#

#= color(blue)(2pi[(4b^2 - b^4/12) - (4a^2 - a^4/12)])#

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