Question

the curve f(x)=x^2 from x=0 to x=4 is revoloved around the axis x=5. Determine the volume of the solis of revolution using the method of curcular cross sections

Answer 1

Need more info. The curve is one boundary. Is the region to revolve bounded by y=0 and x=4, or by y=0 and x=0, or what?

Just revolving a curve will produce a surface, but with no corresponding area, there's no volume.

Answer 2

That is all the information I was given, so that is why I am stuck on it.

Answer 3

Well, let's take a stab at it. The simplest region would be the region under the curve from x=0 to x=4, bonded by the line x=4. Kind of a little curved triangle.

If we rotate that around the line x=5 we'll have a kind of volcano-shaped solid with a hole down the center.

To find the volume using circular cross-sections, think of a stack of thin washers, getting smaller as they stack up.

Each washer has i hole inside of radius 1 (distance from x=4 to x=5).

The outside radius is 5-x.

Each washer has a thickness dy.

Now, y = x^2 so x = sqrt(y)

The area of each washer is pi*(5-x)^2 - pi*1^2 = pi(25 - 10x + x^2 - 1)
= pi(y - 10*sqrt(y) + 24)

The volume is Integral(pi*(y - 10y^1/2 + 24) dy [0,16]

That will be

pi (1/2 y^2 - 20/3 y^3/2 + 24y)[0,16]
= pi(128 - 1280/3 + 384)
= 256pi/3