Respostas

2013-11-30T20:03:12-02:00
 \int\limits^3_0 {y^2} \, dx =  \int\limits^3_0 {(x+1)^2} \, dx= \int\limits^3_0 {(x^2+2x+1)} \, dx=F(3) - F(0)
F(x)= (x^3/3)+(2x^2/2)+(x)
F(0)=0
F(3)=  (3^3/3)+(3^2)+(3)=21
V=π(F(3) -F(0))
V= 21π