I don't understand how to set up integrals using the shell, disc, or washer method please help

Question

1. Using the shell method, set up but do not evaluate the definite integral for the volume of the rotated solid: the region in the first quadrant bounded by y = -x³+3x² is rotated about the y-axis
2. Using the shell method, set up but do not evaluate the definite integral for the volume of the solid obtained when rotating the region enclosed by y = √x and y = x/2 about the line y = 2
3. Set up but do not evaluate the definite integral that represents the volume of the solid: The region is bounded by y = e^x , y = e² and x = 0 is rotated about the line y = -1
4. Set up but do not evaluate the integral(s) for the volume generated when the region in the 1st quadrant bounded by y = x, x = 2-y² and the positive x-axis is rotated about the y-axis
5. Set up but do not evaluate the integral that represents the volume of the solid: The base of a solid is bounded by y = x³, y = 0 and x = 1 and the cross-sections perpendicular to the x-axis are equilateral triangles.
6. Set up but do not evaluate an integral for the volume of the solid obtained by rotating the region bounded by the given curves y = x²-4 and y = 4-x² about the line x = 4. So not simplify your integrand.

Answer 1

desmos.com/calculator/zau8znqcx2

Answer 2

All right Chelsea Q.,

A picture is worth a thousand words. I'll try to help you in fewer words than that. The idea behind shells, discs, washers (a type of disc) etc. is how to conveniently slice a volume up into pieces that can be easily visualized AND mathematically measured, because you know the equations for their edges.

So, sketch that function of yours: it runs from the origin, upwards for a while, crests, and then plunges across the x-axis at x=3. So when you rotate that around the y-axis, it makes a hill that's been punched flat in the middle (at (x,y)=(0,0)).

OK, how do you set it up for integration as shells? Each piece of shell is a cylinder, (at some value of x!), centered on the y-axis. The lateral thickness is dx ; the height is from the x-axis (y=0) to the value of the function at that value of x, i.e. -x^3+3x^2 . The circumference of the cylinder is 2.pi.r, i.e. 2.pi.x. The volume of a shell(slice) is the product of the circumference, the thickness, and the height of the slice, that is 2.pi.x(-x^3+3x^2)dx . so you need to multiply that out, then integrate it from x = 0 to x = 3 (because that's where it crosses the axis, and thus leaves Quadrant 1). Note: you won't need the arbitrary constant of the indefinite integral, because you would just add it for the x=3 limit, then subtract it out again for the x=0 limit!

-- Cheers, --Mr. d.