I prefer to use the method of cylindrical shells for a problem like this. This method can be thought of as the summation circumferences multiplied by the height. It is like spinning an area around the axis of choice. It looks like this;

∫2πxf(x)dx...note that shell method spins around the **opposite** axis as the variable. The disk or washer methods spins around the **same** axis as the variable so you would have to use f(y) functions for those.

First, we need f(x) for the boundary or side of the object.

Note the end points of the right side of the frustrum. (3,30),(8,0)

If you are taking calculus, I suppose that you know how to make a line equation from this.

For the sake of clarity anyway;

(y1 - 30) = (dy/dx)(x1 - 3), dy/dx from the points above gives -30/5 = -6

(y1 - 30) = -6x1 + 18, we will call this ys (side)

So, ys = 48 - 6x, but notice that we are only going to use this function when x goes from 3 to 8

Before x passes 3 we simply have a cylinder with a radius of 3 and height of 30.

Here y = 30 for x from 0 to 3, we will call this yc (center)

So using the shell method from way above for both functions with their proper limits of integration gives us;

∫2πx(30)dx] from 0 to 3 + ∫2πx(48 - 6x)dx] from 3 to 8 =

2π {∫30x dx]0,3 + ∫48x - 6x² dx]3,8}

2π {(15x²)]0,3 + (24x² - 2x³)]3,8}

2π [(135) + (1536 - 1024 - 216 + 54)]

2π (485) =

V = 970π cm³

edit*

I had to correct a silly error concerning multiplication!