Let f (x) = sin(x-1) , 0 ≤ x ≤ 2 π + 1 , Find the volume of the solid formed when the region bounded by y =ƒ( x) , and the lines x = 0 , y = 0 and y = 1 is rotated by 2π about the y-axis.

Draw a rough sketch of the graph of f(x) = sin(x-1) to get an idea of what the region looks like. Realise that the normal volume of revolution for the area between x axis and the f(x) is given by V = pi* integral( f(x)f(x) * x dx) Look at the rough sketch of the graph and realise that in this case the area enclosed is the area under the graph but in the other coordinate i.e. area between y-axis and f(x). So the coordinates in the volume formula are reversed. Therefore, use the formula V=pi integral( x * x dy) x=arcsin(y)+1 V=pi* integral( (arcsin(y)+1) * (arcsin(y)+1) * dy) where the limits of integral are from y=0 to y=1 (look at the y coordinate limits of the enclosed graph) So this is now reduced to a simple integral problem which can be solved on the graphical calculator (recommended) or done the hard way using integeration by expanding the square and doing integeration by parts multiple times. V = 8.20