Calculo

1) 2)
f(x)=x√(x^3+1) [0,1] Y=√X [4,9]
∫▒〖π r^2 〗 Y=√X
π∫_0^1▒〖(X√(X^3+1))〗^2Y^,=1/(2√X)
π∫_0^1▒〖X^2 (X^3+1)〗 S=2π∫_4^9▒√X √(1+〖(1/(2√X))〗^2 )=
π∫_0^1▒X^5 +X^2 2π∫_4^9▒√X√(1+1/4X=)
π x^6/6+x^3/3 ■(1@↑@0) 2π∫_4^9▒√X √((4X+1)/4X)=
π 1^6/6+1^3/3-π 0^6/0+0^3/3 2π∫_4^9▒√X √(4X+1)/√4X=1.57μ^3 2π∫_4^9▒√X √(4X+1)/(2√X)=
π∫_4^9▒√(4X+1)=μ=4X+1 μ^,=4□(24&dx) μ^,/4 □(24&dx)
1/4 π∫_4^9▒μ^(1/2) □(24&dx)1/4 π μ^(3/2)/(3/2) ■(9@↑@4)
1/4 π 2/3 μ^(3/2) ■(9@↑@4)= π 1/6 μ^(3/2)■(9@↑@4)
π 1/6 〖(4X+1)〗^(3/2) ■(9@↑@4)
π 1/6 〖(4(9)+1)〗^(3/2) - π 1/6 〖(4(4)+1)〗^(3/2)=

π 1/6 〖(37)〗^(3/2)- π 1/6 〖17〗^(3/2)=117.84-36.70 =81.14μ^2
3) Y^2=X^3 [0,4] 4) F(X)=√3X [0,3]
Y=√(X^3 )∫▒〖πr^2 〗=
∫▒〖πr^2 〗 π∫_0^3▒〖(√3X)〗^2 =
π∫_0^4▒〖(√(X^3 ))〗^2 = π∫_0^3▒〖3X=〗
π∫_0^4▒X^3π 〖3X〗^2/2 ■(3@↑@0)
π X^4/4 ■(4@↑@0) π (3〖(3)〗^2)/2- π 〖3(0)〗^2/2=
π 〖(4)〗^4/4 - π 〖(0)〗^4/4= 42.41π^3...