How do you use substitution to integrate $2 π y (8 - \frac{y^{2}}{3}) d y$ $2piy(8-y^2/3)dy$ ?

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Answer 1 · 2015-08-31T23:26:59

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

$V = 2 π \int x f (x) d x$ $V = 2piintxf(x)dx$

Switch out $x$ $x$ for $y$ $y$ . Looks like $f (y) = 8 - \frac{y^{2}}{3}$ $f(y) = 8 - y^2/3$ .

$= 2 π \int_{a}^{b} 8 y - \frac{y^{3}}{3} d y$ $= 2piint_a^b 8y - y^3/3dy$

$= 2 π [4 y^{2} - \frac{y^{4}}{12}] {∣ ∣ ∣}_{a}^{b}$ $= 2pi[4y^2 - y^4/12]|_(a)^(b)$

$= 2 π [(4 b^{2} - \frac{b^{4}}{12}) - (4 a^{2} - \frac{a^{4}}{12})]$ $= color(blue)(2pi[(4b^2 - b^4/12) - (4a^2 - a^4/12)])$

How do you use substitution to integrate 2piy(8-y^2/3)dy2πy(8−y23)dy2piy(8-y^2/3)dy?