Find the volume enclosed by the two surfaces

find the volume enclosed by the cylinder x^2+y^2=16 that is to the right of the plane y=1, above the plane z=0, and below the plane z=10, using triple integrals.

Here is a plot of the x-projection of this solid:
http://www.wolframalpha.com/input/?i=x^2…

From this, we see that this region is x-simple and we have the following bounds for x and y:
-√15 ≤ x ≤ √15, 1 ≤ y ≤ √(16 - x^2).

Since we want the volume above z = 0 and below z = 10, we also have the following bounds for z:
0 ≤ z ≤ 10.

So, we can describe the solid as:
W = {(x, y, z) | -√15 ≤ x ≤ √15, 1 ≤ y ≤ √(16 - x^2), 0 ≤ z ≤ 10}.

Therefore, the required volume is:
V = ∫∫∫W dV = ∫∫∫ dz dy dx (from z=0 to 10) (from y=1 to √(16 - x^2)) (from x=-√15 to 15)
= 10 ∫∫ dy dx (from y=1 to √(16 - x^2)) (from x=-√15 to 15)
= 10 ∫ [√(16 - x^2) - 1] dx (from x=-√15 to 15)
= 10[(1/2)x√(16 - x^2) + 8arcsin(x/4) - x] (evaluated from x=-√15 to 15)
= 10{(1/2)(√15)(1) + 8arcsin(√15/4) - √15 - [(1/2)(-√15)(1) - 8arcsin(√15/4) + √15]
= 160arcsin(√15/4) - 10√15.

I hope this helps!