Using the cylindrical shell method

Question

Use the cylindrical shell method to find the exact volume of the solid generated by revolving the region by the curve y=Square root x²+1 (the whole equation x² +1 is squared), x=square root 3, x=0, y=0 about the y-axis.

Answer 1

The upper curve is y = sqrt(x²+1), bottom is the x axis (y = 0) and the right edge is x = sqrt(3). The intersection of the right edge and the top curve happens at sqrt((sqrt(3)^2 +1) = sqrt(10).

The cylinder shell will have a height = y = sqrt(x²+1) - 0

The cylinder shell will have a radius of x

The width of the shell will be dx , so we integrate in x

Volume = int(2πrhdr) = int(2πx(sqrt(x²+1)dx) for 0 to sqrt(3) This is integrable with a u sub of u = x²

Please consider a tutor.