Triple integral by cylinder and planes?

Question

Set up the iterated (triple) integrals that will solve for the volume of the solid.

----bounded by the cylinder y=x^2 +2 and the planes y=4, z=0 and 3y-4z=0.

Answer 1

Well, y = x^2 + 2 isn't a cylinder. In the xy-plane, this is a parabola, not a circle, so its projection into three-dimensional space is related to a cylindric surface, except that it is also an unbounded region in this form. Anyway, if the region is bounded by y = 4, it is also bounded by x = sqrt(2), because x = sqrt(2) solves y = x^2 + 2 for y = 4. So the region is bounded in x by [0, sqrt(2)], in y by [x^2 + 2, 4], and in z by [0, 3y/4] because 3y - 4z = 0 ==> z = 3y/4.

For a volume, you just take the triple integral of dxdydz over the appropriate bounds, but you rearrange them so that you haven't integrated dx while x remains in the outer part of the expression, dy while y remains, or dz while z remains. Yahoo! Answers is not very well suited to displaying this. But it will be the integral from 0 to sqrt(2) of the integral from x^2 + 2 to 4 of the integral from 0 to 3y/4 of dzdydx. Or int(int(int(dxdydz))) with bounds as given. That way, you can integrate dz using bounds in terms of y, integrate the resulting expression containing y using bounds in terms of x, and finally integrate the resulting expression containing x using bounds that are constants to get a numerical result.

Answer 2

You don't need no stinking calculus or triple integrals for this one. It's just a cylinder of an ellipse of radii 1, 2, and height 4 (total volume 4(2π) = 8π) divided in half by the plane y = z + 2, or 4π. Draw this one out. Edit: Well, if the thumbs-downer INSISTS, a triple integral for figuring out 3D volumes starts with "1", and it's all done with limits, as follows in this case: V = ∫ ∫ ∫ 1 dx dy dz The limits for ∫ dx is -(1/2)√(1-z²) to (1/2)√(1-z²) The limits for ∫ dy is 0 to z+2 The limits for ∫ dz is -2 to 2 Then V = 4π