How do I integrate the volume of revolution between 0 and pi of y=sin(x)?

To integrate the volume of revolution first imagine a thin disk around the x-axis which we want to know the volume of: Volume=area x height. The area of a circle is given by pi r^2 and in our case let us use the height of dx. Hence the volume= pi  r²  dx. Now we will use the radius at each point as the y-value at that point, hence volume = pi y^2 dx = pi sin^2(x) dx. We will integrate this between the limits using the identity sin^2(x)=1/2 (1 - cos(2x)). Hence the volume of integration is given by V=pi/2 integral{0->pi} (1-cos(2x))dx = pi/2*[x -1/2 sin (2x)]{x=0 -> pi} = pi/2*(pi - 0 - (0-0)) = pi^2/2