Question

SOLUTION The solid E is shown in the top figure. If we regard it as a type 1 region, then we need to consider its projection D, onto the xy-plane, which is the parabolic region in the middle figure. (The trace of y = x2 + x in the plane 2 = 0 is the parabola y - 2 .) From y - x+ z we obtain Voor? so the lower boundary surface of Eis --Vy-and the upper boundary surface is za +-√2 - x Therefore the description of E as a type 1 region is X = {(x, y, 2) |-- 5x5 1,x?sys 1,-Vy - xszs Vy - xa} and so we obtain III. 1** *zov - -LIT VY** vx+7 de dy dk. - - * Although this expression is correct, it is extremely difficult to evaluate. So let's instead consider E as a type 3 region. As such, its projection Dy onto the xz plane is the disk x + 2 5 1 shown in the bottom figure. Then the left boundary of Eis the paraboloid y = x + 22 and the right boundary is the plane y = 1, so taking M(x, z) = x2 + 2+ and H2(x,x) = 1 in this equation, we have verzor- Sollen - Solveron on - (1-x2–22)24?). Х Although this integral could be written as IT (1 – x2 - 22) x2 + 2?dzdx √1-x it's easier to convert to polar coordinates in the xz plane: xr cose, 2 = r sin 0. This gives (1 – *2-22)x + 22A This SIIV* *230V 6"/(1-22 - [*] ( Derde do ✓ Jor X 2x یا

Answer 1

The Trace of taxt zo in the plane z=6 is to Pena bola yax from y. nt z Y-22 So lower bound of the Surface E is VY-> a Z = and upperboundary surface ů zvy the description of E as a type I region is and so we obtain E={2,4,2)) - 12 nel; reyal ISS Jazz du June as ſynerg 79- at z dedydu. - vyam Although this is correct, it is difficult to evaluate 1 SS vanzadora Jos for a day) an E 2 s curama que ) D3 da natu I let a zz da Dz Although this integral could be willense Viano (1-1~) /ax zu dz da V Ot y easier to convert

to Polar Coordinates in the az-plane as N=dcoso , Z = 12 Sino D2 of Dor rdr do SSS Jazz du = ll (l_2²_2²) | X + 27 d A y (1) S. do 1 kom - s" di za ( ZA ( 12/25 75