How does one actually find or set up the limits for a triple integral given cert

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Triple integrals can be used to compute volumes of more general shapes, as well as masses and moments of more general solids (especially those of varying density) and average values of three variable functions. Let f(x; y; z) be a function dened on a closed, bounded region G in space. We can dene the integral of f over G. First draw a box containing G. Then partition the box into n rectangular cells. The kth cell has dimensions ?xk, ?yk, and ?zk, and volume Vk = ?xk?yk?zk. Pick any point in the kth cell, and denote it by (?xk; ?yk; ?zk). Then form the sum Sn = Sf(xk; yk; zk)?Vk We are interested in what happens as the partition gets increasingly ne. Let kPk be the largest value among all of the ?xk, ?yk, and ?zk. Then we let ||P|| -> 0. If we achieve a single limiting value no matter how the partition and points (xk; yk; zk) are chosen, then f is integrable over G. As long as G is obtained by joining nitely many smooth surfaces along nitely many smooth curves, then f is integrable over G. If f is integrable over G, then the sums approach a limit as ||P|| -> 0, which we call the triple integral of f over G: lim Sn = ???f(x,y,z)dV

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