Integrate f(x, y, z) = x over the region in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) above z = y2 and below z = 8 − 2x2 − y2.
Find the boundary of D in the xy-plane. The upper and lower surfaces intersect where they have the same z-values. Find the intersection of the two surfaces in terms of x and y ≥ 0.
y2 = 8 − 2x2 − y2
x2 + y2 = h4XZagboIgc=
Thus the boundary of D is a 3Fxr8xmp24lukJ+1AGBVkCFMSN3GWMbf8e5lWEhHnCmTFEDt in the first quadrant.
The region D can be expressed as either vertically or horizontally simple, depending on which variable we express the boundary in.
Solve the boundary, x2 + y2 = 4, for y in terms of x, thus making DM2jxtJGVDpVt/Kx8Ux1JJ4ofYkdaWFI9ZSKzvQ== simple.
0 ≤ x ≤ XvVM00l89Is=
Write the triple integral as an iterated integral.
Evaluate the inner integral.
At this point, we will choose to write the integrand as 8x − 2x3 − 2xy2 = 2x(4 − x2) − 2xy2. Integrate the middle integral.
Evaluate f(x, y, z) = x over the defined region by integrating the last of the iterated integral. Round your answer to two decimal places.
= SFgqQUkJGdg=