x = x/2 makes no sense, so I'll go with x = π/2
Also, you do not specify the axis, but This looks like a good candidate for rotating around the x-axis. So, using discs of thickness dx, you get
v = ∫[0,π/2] πr^2 dx
where r = y = cos(x)√sin(x)
v = ∫[0,π/2] π(cos(x)√sin(x))^2 dx
= ∫[0,π/2] π*cos^2(x)*sin(x) dx = π/3
What is the volume of the solid obtained by rotating y=cos(x)√sin(x) from x=o to x=x/2 around the axis
2 answers
your function is undefined for all x's which make sinx negative, but for
your domain of rotation we don't have to worry about that
vol = π∫ y^2 dx
= π∫ sinx(cos^2 x) dx from 0 to π/2
= π[ -(1/3)(cosx)^3] from 0 to π/2
= π( (-1/3)(0) - (-1/3)(1) )
= (1/3)π
your domain of rotation we don't have to worry about that
vol = π∫ y^2 dx
= π∫ sinx(cos^2 x) dx from 0 to π/2
= π[ -(1/3)(cosx)^3] from 0 to π/2
= π( (-1/3)(0) - (-1/3)(1) )
= (1/3)π