hard calculus question involving triple integral! help!!?

Question

Find volume of the solid bounded by z=1+(X-1)^2 + 4y^2, the planes x=3, y=2, and the coordinate planes.

Answer 1

The volume of a solid is (triple integral) 1 dx dy dz.
The trick is to set up your limits of integration. In this case,
0 < or equal x < or equal 3
0 < or equal y < or equal 2
0 < or equal z < or equal 1+(X-1)^2 + 4y^2
Integrate wrt z first, gives you 1+(X-1)^2 + 4y^2,
now integrate wrt y, gives you 2(x - 1)^2 + 38/3,
now integrate wrt x, gives you 16/3 + 38 - (-2/3) = 44.
Sorry I haven't shown all the steps in detail, but I don't know how to get hold of a mathematical word processor font in these Q & A.

Answer 2

I think you only need a double integral.

∫∫z dy dx where y runs from 0 to 2 and x from 0 to 3.

∫∫z dy dx = ∫∫[1 + (x-1)² + 4y²] dy dx
= ∫∫[1 + x² - 2x + 1 + 4y²] dy dx
= ∫∫[x² - 2x + 2 + 4y²] dy dx
= ∫[x²y - 2xy + 2y + (4/3)y³] (eval from 0 to 2) dx
= ∫[2x² - 4x + 4 + 32/3] dx
= ∫[2x² - 4x + 44/3] dx
= [(2/3)x³ - 2x² + (44/3)x] (eval from 0 to 3)
= [(18 - 18 + 44) - 0] = 44

Answer 3

21 cubic units.
just find xyz.

Answer 4

x=2243557434.5653233686