For the region bounded by y=2x, the x-axis, and x=1, determine which of the following is greater: the volume of the solid generated when the region is revolved about the x-axis or about the y-axis.When the region is revolved about the x-axis, the volume is(Type an exact answer, using piπ as needed.)When the region is revolved about the y-axis, the volume is(Type an exact answer, using piπ as needed.)Which volume is greater?A.The volume about the y dash axis is greater.The volume about the y-axis is greater.B.The volumes are equal.The volumes are equal.C.The volume about the x dash axis is greater.

The area of a disk of revolution at any x about the x- axis is πy² where y=2x. If we integrate this area on the given range of values of x from x=0 to x=1 , we will get the volume of revolution about the x-axis, which here equals,

[tex]\int\limits^1_0 {(pi)y^{2} } \, dx[/tex]

which when evaluated gives 4pi/3.

Now we have to calculate the volume of revolution about the y-axis. For that we have to first see by drawing the diagram that the area of the CD like disk centered about the y-axis for any y, as we rotate the triangular area given in the question would be pi - pi*x². if we integrate this area over the range of value of y that is from y=0 to y=2 , we will obtain the volume of revolution about the y-axis, which is given by,

[tex]\int\limits^2_0 {pi - pix^{2} } \, dy = \int\limits^2_0 {1 - y^{2}/4 } \, dy[/tex]

If we just evaluate the integral as usual we will get 4pi/3 again(In the second step i have just replaced x with y/2 as given by the equation of the line), which is the same answer we got for the volume of revolution about the x-axis. Which means that the answer B) is correct.