I=R⌠⌠⌠f(x,y,z)dv

Question

Set up an iterated integral to evaluate I=_R⌠⌠⌠f(x,y,z) dv, where R is the region in the first octant bounded by the surface z=x+y^2, the cylinders y=2sqrt(x), x=2sqrt(y), and the xy-plane

 

 

The ⌠ symbol is an integral symbol.

Thank you.

Answer 1

The intersection, S, of region R with the xy-plane in the first quadrant is bounded below by y = x²/4 and above by y = 2√x. We see that the curves intersect when x = 4 and when x = 0.   For each x from x = 0 to x = 4, y varies from y = x²/4 to y = 2√x.  And, for each point (x,y,z) in R where (x,y) lies in S, z varies from z = 0 to z = x+y^2.

 

So, I = ∫_{(0 to 4)} ∫_{(x²/4 to 2√x)} ∫_{(0 to x+y²)} f(x,y,z) dz dy dx