3) We have given y=f(x)=xsin(x^3) and x=0 to (pi)^{1/3}
by using shell method
we know that the volume of the solid around y-axis is integration of (x=a to b)((2pi)*r*h)dx
we have radius r=x and height h=xsin(x^3) and thickness dx
So Volume V= integration of (x=0 to (pi)^{1/3})((2pi)*x*xsin(x^3) )dx
=integration of (x=0 to (pi)^{1/3})((2pi)x^{2}sin(x^3) )dx
substitute u=x^3,du=3x^2dx implies x^2dx=du/3
=(2pi)* integration of (x=0 to (pi)^{1/3})(sin(u))(du/3)
=(2pi)/3 *integration of (x=0 to (pi)^{1/3})(sin(u))du
=(2pi)/3*[-cos(u)] from x=0 to (pi)^{1/3}
resubstitute u=x^3
=(2pi)/3*[-cos(x^3)] from x=0 to (pi)^{1/3}
=(2pi)/3*[-cos(((pi)^{1/3})^3)-(-cos(0))]
=(2pi)/3*[-cos(pi)+1]
=(2pi)/3*[1+1]
=(4*pi)/3
Volume V=(4*pi)/3
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