Math 21a Handout on Triple IntegralsThe purpose of this handout is to provide a few more examples of triple integrals. Inparticular, I provide one example in the usual x-y-z coordinates, one in cylindrical coordinates andone in spherical coordinates.Example 1 : Here is the problem: Integrate the function f(x, y, z) = z over the tetrahedral pyramidin space where• 0 ≤ x.• 0 ≤ y.• 0 ≤ z.• x + y + z ≤ 1.(1)The integral in question isI ≡ zdz dAxyR01−−∫∫∫ ,(2)where R is the ‘shadow’ region in the x-y plane; the region where• 0 ≤ x,• 0 ≤ y,• x + y ≤ 1.(3)Indeed, a vertical line (where x and y are constant) will hit the pyramid only if x and y are non-negative and x + y ≤ 1. Otherwise, one of the conditions in (1) is violated when z ≥ 0. This lineenters our pyramid from below where z = 0 and it then exits with z value where x + y + z = 1,which is to say where z = 1 - x - y. This information provides the lower bound to the z-integral, 0,and also the upper bound, 1 - x - y.The integral over the z coordinate in (2) gives 2−1 (1 - x - y)2, and soI = 2−1()120101−−−∫∫x y dy dxx .(4)With regard to the upper and lower bounds to the x and y integrals in (4), remark that a vertical line(where x is constant) hits the shadow region R only if 0 ≤ x ≤ 1. Otherwise, one of the conditionsin (3) will be violated when y is non-negative. (These bounds on x give the lower and upperbounds for the x integral in (4).) This line enters the shadow region where x = 0 and exits with yvalue where x + y = 1. In this way, the lower bound to the y integral in (4) is found to be 0 andthe upper bound is found to be 1 - x.In any event, the y integral in (4) gives 3−1 (1 - x)3 which leavesI = 6−1 ()1301−∫x dx = (24)−1 .(5)Example 2 : The problem in this example is to integrate the function z over the region where• 0 ≤ z,• x2 + y2 ≤ 1,• z ≤ 1 - x2 - y2 .(6)To accomplish this task, note that the integration region as just described involves only thecoordinates x and y through the combination x2 + y2, and the function to be integrated doesn’tinvolve these coordinates at all. In particular, since x and y only appear here through x2 + y2, itmakes sense to solve the problem using the cylindrical coordinates (r, θ, z). And, in terms ofthese coordinates, the integral in question has the formI = zdz rdrdrr0112−≤∫∫∫θ .(7)Here, the shadow region is seen to be the disk in the x-y plane where r ≤ 1 since a vertical linewhich hits the x-y plane outside of this disk violates the middle point in (6). Meanwhile, the lowerand upper bounds for the z-integral in (7) come about via the observation that a vertical line whichhas r ≤ 1 enters the region where z = 0 (because of the first point in (6)) and exits where z = 1 - r2(because of the third point in (6)).In any event, the z-integral in (7) gives 2−1 (1 - r2)2 which makesI = 2−1 ()1220201−π∫∫rdrdrθ(8)Here, the bounds of 0 and 1 for the r integral come from the fact that the shadow region is the diskin the x-y plane where r ≤ 1. This fact also explains why the 0 and 2π bounds for the θ integral.The θ integral in (8) gives 2π, which implies thatI = π ()12201−∫r rdr .(9)This last integral can be done by substituting u = r2. The answer is I = π/6.Example 3 : The problem here is to integrate the function f(x, y, z) = z over the upper half of theball; this being the region where• 0 ≤ z.• x2 + y2 + z2 ≤ 1.(10)Although this problem can be worked in cylindrical coordinates, spherical coordinateswork as well. For the latter, the iterated integral isI = ( cos ) sin/ρφρ φρθφ2010202∫∫∫ππddd .(11)Concerning (11), some explanation is in order: First, the term ρ cosφ in the parenthesis is just theexpression for z in spherical coordinates. Second, the integration bounds of 0 and 1 for ρ, 0 and2π for θ and 0 and π/2 for φ insure that the integration takes place only over the upper half of theball. In particular, the function z is non-negative only where φ ≤ π/2 and so the π/2 upper boundon the φ integral restricts the integration as required.The ρ integration in (11) can be done first with the result being 1/4. The θ integration thenprovides a factor of 2π. Finally, the φ integration provides a factor of 1/2. (Use the substitution u= cos φ while noticing that du = -sin φ dφ.) Thus, I =
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