Calculus Integral Problem HELP revolving function about axis find Volume: set up integral?

in general you have a choice when doing these revolution problems to use disks / washers or shells.

Disks -- works particularly well when revolving about the x axis.

f(x) is the radius of the disk. dx is the thickness. The volume is π f(x)^2 dx

washers --

It is the same as disks, but there is a hole in the disk. Find the volume of the disk subtract the volume of the hole.

Shells -- works best when revolving around the y axis.

We take a cylindar of height f(x) and raduis (x+dx) and subtract a cylindar of height f(x) and raduis f(x) leaving behind a shell of thickness dx.

The volume of this shell = π f(x) (x+dx)^2 - π f(x) (x)^2 = 2 π f(x) x dx

But you are not locked into shells when rotating around the y axis and disk when rotating around the x axis. It could turn out that you want to use disks when rotating around the y axis. In which case, your disks would be.

π f^-1(y)^2 dy

To the problems at hand:

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

Let R be the region between the graph of f and the x axis on the interval [0,4]. Set up an appropriate integral to find the volume V of the solid region obtained by revolving R about the x axis and then find V.

V = π∫ f(x)^2 dx

= π∫ sqrt (16 - x^2)^2 | x = [0,4]

= π ∫16 - x^2 dx

2. Let f(x) = sec(5x)

Let R be the region between the graph of f and the x axis on the interval [0,pi/20]. Set up an appropriate integral to find the volume V of the solid region obtained by revolving R about the x axis and then find the volume (exact value).

again, use disks...

V = π∫ f(x)^2 dx

= π∫ sec^2 (5x) | x = [0,pi/20]

= π (1/5) tan 5x | x = [0,pi/20]

π/5