Hoping to someone can help me with #1. I scanned it in so I don't flail around in Gimp for the next half hour trying to make the diagrams.

http://dl.dropbox.com/u/1938505/math.pdf

For a I did the integral going from -1 to 1 of 2(pi*((1-x^2)^1/2)*dx)

I multiplied by 2 since (1-x^2)^1/2 is the equation of the semi circle and since it volume I don't think subtracting is the best option here.

Thanks for the help.

y(x) = √(1-x²)

Since you're looking for the area as a function of x, in a A(x) = π*y(x)² = π(1-x²), since the cross section's a circle.

In b, half of the base is y(x). Thus the A(x) = (2*y(x))² = 4(1-x²).

In c, half of the diagonal is y(x). The diagonal's then 2y(x). From Pythagoras theorem, it then follows that one side is 2y(x)/√(2) = √(2)*y(x). Thus A(x) = (√(2)*y(x))² = 2*y(x)² = 2(1-x²).

In d, since half of the base of the triangle is y(x), A(x) = (√(3)/4)*(2y(x))² = √(3)y(x)² = √(3)(1-x²)

Thanks that clears up some things. Mods can lock this now if they want.