disk method cross section?

Question

using the disk method to find the volume of the solid bounded by x=y^2 and the line x=4 rotated about the x axis, what is the area of the largest cross section?

Answer 1

the largest cross section would be the circle at x=4 parallel to the z-axis with radius y. y=sqrt(x)=sqrt(4)=2. A circle with radius 2 has area 4pi.

P.S. why would you need to know the area of the largest cross section?

Answer 2

Rotating the figure about the x-axis gives you a three-dimensional parabola that's chopped off 4 units from its vertex. Since x = 4 at this point, and x = y^2, y must be +/- 2.

So, the cross-sectional slice of the parabola where x = 4 is simply a circle with radius 2. It therefore has an area of

pi*r^2
= pi*2^2
= 4*pi square units

Answer 3

volume=
integral with limits 0 and 4 of sq. root x
(this would be much easier to show you if i could use the symbols...i'll fix it if i can find them)
(i found it...the computer tells me to write it like this: ∫_0^4▒√x, if you have Microsoft word 2007, you can change this back to professional form)
largest x section
the point at which the equaiton has a maximum
so set the derivative equal to zero, that's the x value, put it into the equation for y, y is the radius of the x section which is a circle (area = pi * r^2)