Setting Up Integrals to Find Volume of Solid Rotated Around Region R

[johnq2k7]

Indicate the method you use to set up the integrals (do not integrate) that give the volume of the solid generated by rotating the region R around:

The region R is bounded by the curves y=x, x= 2-y^2 and y=0

i.) the x-axis
ii.) the y-axis
iii.) the line x= -2
iv.) the line y= 1

work shown:

y=x ,

x= 2- y^2
y= sqrt(2-x)

i.) integral of Pi*[(x)- (sqrt(2-x))] dx as x goes from 0 to 1

ii.) integral of 2*Pi*x*[(x)-sqrt(2-x)]dx as y goes from 0 to 1

iii.) integral of 2*Pi*(-2-x)*[(x)-(sqrt(2-x))]dx as x goes from 0 to 1

iv.) integral of Pi*[(x-1)-(sqrt(2-x-1)]dx as y goes from 0 to 1

I think my definite integral limits are wrong... and in general my integral setup is wrong.. please help

 

[mighty2000]

!

I would use the method of cylindrical shells to set up the integrals for finding the volume of the solid generated by rotating the region R around different axes. This method involves integrating the product of the circumference of a cylindrical shell and its height, which represents the volume of the shell. The limits of integration would depend on the axis of rotation and the boundaries of the region R.

i.) For rotating around the x-axis, the limits of integration would be from 0 to 1, since 1 is the highest point of the curve x=2-y^2. The radius of the shell would be x, and the height would be the distance between the curves y=x and x=2-y^2, which is (2-x)-x = 2-2x. Therefore, the integral would be:

V = 2π∫(x)(2-2x)dx from 0 to 1

ii.) For rotating around the y-axis, the limits of integration would be from 0 to 2, since 2 is the highest point of the curve x=2-y^2. The radius of the shell would be y, and the height would be the distance between the curves y=x and x=0, which is x = 2-y^2. Therefore, the integral would be:

V = 2π∫(y)(2-y^2)dy from 0 to 2

iii.) For rotating around the line x=-2, the limits of integration would be from -2 to 0, since 0 is the highest point of the curve y=x. The radius of the shell would be (x+2), and the height would be the distance between the curves y=x and x=2-y^2, which is (2-x)-x = 2-2x. Therefore, the integral would be:

V = 2π∫(x+2)(2-2x)dx from -2 to 0

iv.) For rotating around the line y=1, the limits of integration would be from 0 to 1, since 1 is the highest point of the curve x=2-y^2. The radius of the shell would be (1-y), and the height would be the distance between the curves y=x and x=0, which is x = 2-y^2. Therefore, the integral would be:

V