Calculus Help; disk method?

This is basic. You have read the examples yes? There will be no trouble if you draw the diagrams.

Consider the solid of revolution of the curve f(x) about x axis, from x=a to x=b.

If you have a plane perpendicular to the x axis, then you can define V(x) as the volume of the solid that lies to the "left" of the plane, i.e. at those x which are less than the x of the plane.

Then if you move the plane a very small distance dx along the x axis, what will be the added volume? We must know our geometry formulas

dV = pi * (f(x))^2 dx

Clearly the total volume of the solid is V(b) - V(a).

The Newton-Leibniz formula (fundamental theorem of calculus) says that you can get this by integrating the derivative dV/dx from a to b.

V_tot = int[a to b] (dV/dx) dx

V_tot = int[a to b] (pi * (f(x))^2) dx

Intuitively we think of this as adding up the volumes of all those infinitesimally thin disks of which the solid consists.

1.

Since this solid's leftmost point is at x=0 and rightmost at x=4, then

V = pi * int[0 to 4] (sqrtx)^2 dx = 8pi

2.

Here we integrate along y. The solid goes from y=0 to y=4.

x = f(y) = sqrt(16 - y^2)

pi * int[0 to 4] (f(y))^2 dy

3.

Ambiguous bounds. Ignoring the bound x=6,

V = pi * int[0 to 4] ( 6^2 - (6-y)^2 ) dy

Using the method of shells you get

1.

A horizontal slice through the region at y, when rotated, will produce a cylindrical shell of radius y and height (length) 4-y^2

2pi * int[0 to 2] y(4-y^2) dy

2.

A vertical slice at x produces a shell of radius x and height sqrt(16-x^2)

2pi * int[0 to 4] x sqrt(16-x^2) dx

3.

A vertical slice at x will produce a shell of radius 6-x and height 4-x.

2pi * int[0 to 4] (6-x)(4-x)dx

2 and 3 don't really require calculus if you can just use the formulas for a hemisphere and truncated cone though.

1. π∫y^2 dx, where x runs from 0 to 4.

    Transform this into: π∫x dx = π(x^2)/2 from x = 0 to 4.  Answer: 8π cubic units.

2. π∫x^2 dy, where y runs from 0 to 4.

    Transform this into: π∫ (16-y^2)  dy, where y runs from 0 to 4.  Answer: 64π cubic units.

3. π∫(6-x)^2 dy , where y runs from 0 to 4.

    Transform this into π∫(6-y)^2 dy, where y runs from 0 to 4.  Answer: 80π/3 cubic units.