Thread: volume of a solid rotated about the y-axis

The problem is: If the region under the following curve from x = 0 to x = 1 is rotated about the y-axis, what is its volume?

y= 6/(x^2+3x+2)


In order to evaluate this, I use the cylindrical shell method by multiplying 2 pi by the function, right? please help!!

Originally Posted by DarthPipsqueak

The problem is: If the region under the following curve from x = 0 to x = 1 is rotated about the y-axis, what is its volume?

y= 6/(x^2+3x+2)


In order to evaluate this, I use the cylindrical shell method by multiplying 2 pi by the function, right? please help!!

The curved surface area of a cylinder is

h is the value of the function for each value of x.
x is the radius of the cylindrical shells.

Integrating the cylindrical shell surface area from x=0 to x=1 will find the volume of revolution about the y axis.

Hence, you must also multiply by x
in order to get the curved surface area of all the shells.