17. Find the moments of inertia I x, I y, I 0 for the lamina of. 4. D x, y 0 x a, 0 y b ; CAS. 20. D is enclosed by the cardioid r 1 cos ; x, y 3

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1 SCTION 2.5 TRIPL INTGRALS XRCISS. lectric charge is distributed over the rectangle, 2 so that the charge densit at, is, 2 2 (measured in coulombs per square meter). Find the total charge on the rectangle. 2. lectric charge is distributed over the disk so that the charge densit at, is, 2 2 (measured in coulombs per square meter). Find the total charge on the disk. Find the mass and center of mass of the lamina that occupies the region and has the given densit function.., 2, ;, 2 4., a, b;, c 5. is the triangular region with vertices,, 2,,, ;, 6. is the triangular region with vertices,,,, 4, ;, 7. is bounded b e,,, and ;, 8. is bounded b s,, and ; 9. is bounded b the parabola 2 and the line 2;,., cos, 2;,,. A lamina occupies the part of the disk 2 2 in the first quadrant. Find its center of mass if the densit at an point is proportional to its distance from the -ais. 2. Find the center of mass of the lamina in ercise if the densit at an point is proportional to the square of its distance from the origin. CAS. Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length a if the densit at an point is proportional to the square of the distance from the verte opposite the hpotenuse. 4. A lamina occupies the region inside the circle but outside the circle 2 2. Find the center of mass if the densit at an point is inversel proportional to its distance from the origin. 5. Find the moments of inertia I, I, I for the lamina of ercise Find the moments of inertia I, I, I for the lamina of ercise Find the moments of inertia I, I, I for the lamina of ercise Consider a square fan blade with sides of length 2 and the lower left corner placed at the origin. If the densit of the blade is,., is it more difficult to rotate the blade about the -ais or the -ais? 9 2 Use a computer algebra sstem to find the mass, center of mass, and moments of inertia of the lamina that occupies the region and has the given densit function. 9., sin, ; 2. is enclosed b the cardioid r cos ;, s A lamina with constant densit, occupies a square with vertices,, a,, a, a, and, a. Find the moments of inertia and and the radii of gration and. I I 22. A lamina with constant densit, occupies the region under the curve sin from to. Find the moments of inertia I and I and the radii of gration and., 2.5 TRIPL INTGRALS Just as we defined single integrals for functions of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables. Let s first deal with the simplest case where f is defined on a rectangular bo:,, a b, c d, r s The first step is to divide into sub-boes. We do this b dividing the interval a, b into l subintervals i, i with lengths i i i, dividing c, d into m subintervals with lengths j j j, and dividing r, s into n subintervals with lengths k k k. The planes through the endpoints of these subintervals paral-

2 694 CHAPTR 2 MULTIPL INTGRALS lel to the coordinate planes divide the bo into lmn sub-boes ijk i, i j, j k, k which are shown in Figure. The sub-bo ijk has volume V ijk i j k. Then we form the triple Riemann sum 2 l m n i j k f ijk *, * ijk, ijk * V ijk ijk Î k where the sample point ijk *, ijk *, ijk * is in ijk. analog with the definition of a double integral (2..5), we define the triple integral as the limit of the triple Riemann sums in (2) as the sub-boes shrink. Î j Î i FINITION The triple integral of f over the bo is f,, dv lim ma i, j, k l m n l i j k f ijk *, ijk *, ijk * V ijk FIGUR if this limit eists. Again, the triple integral alwas eists if f is continuous. We can choose the sample point to be an point in the sub-bo, but if we choose it to be the point i, j, k, and if we choose sub-boes with the same dimensions. so that V ijk V, we get a simplerlooking epression for the triple integral: f,, dv lim l, m, n l l i m n j k f i, j, k V Just as for double integrals, the practical method for evaluating triple integrals is to epress them as iterated integrals as follows. 4 FUINI S THORM FOR TRIPL INTGRALS If f is continuous on the rectangular bo a, b c, d r, s, then f,, dv s r d c b f,, d d d a The iterated integral on the right side of Fubini s Theorem means that we integrate first with respect to (keeping and fied), then we integrate with respect to (keeping fied), and finall we integrate with respect to. There are five other possible orders in which we can integrate, all of which give the same value. For instance, if we integrate with respect to, then, and then, we have f,, dv b a s r d c f,, d d d

3 SCTION 2.5 TRIPL INTGRALS 695 V XAMPL valuate the triple integral, where is the rectangular bo given b SOLUTION We could use an of the si possible orders of integration. If we choose to integrate with respect to, then, and then, we obtain 2 dv d d d 2 d d Now we define the triple integral over a general bounded region in threedimensional space (a solid) b much the same procedure that we used for double integrals (2.2.2). We enclose in a bo of the tpe given b quation. Then we define a function F so that it agrees with f on but is for points in that are outside. definition, f,, dv 2 dv,,, 2, d d 2 d d 4 F,, dv FIGUR 2 A tpe solid region =u (, ) =u (, ) This integral eists if f is continuous and the boundar of is reasonabl smooth. The triple integral has essentiall the same properties as the double integral (Properties 6 9 in Section 2.2). We restrict our attention to continuous functions f and to certain simple tpes of regions. A solid region is said to be of tpe if it lies between the graphs of two continuous functions of and, that is, 5,,,, u, u 2, where is the projection of onto the -plane as shown in Figure 2. Notice that the upper boundar of the solid is the surface with equation u 2,, while the lower boundar is the surface u,. the same sort of argument that led to (2.2.), it can be shown that if is a tpe region given b quation 5, then =u (, ) 6 f,, dv u 2, f,, u d da, b a =g () FIGUR A tpe solid region =u (, ) =g () The meaning of the inner integral on the right side of quation 6 is that and are held fied, and therefore u, and u 2, are regarded as constants, while f,, is integrated with respect to. In particular, if the projection of onto the -plane is a tpe I plane region (as in Figure ), then,, a b, t t 2, u, u 2,

4 696 CHAPTR 2 MULTIPL INTGRALS =u (, ) and quation 6 becomes c =u (, ) =h () d =h () 7 f,, dv b 2 t a t u 2, u, f,, d d d If, on the other hand, is a tpe II plane region (as in Figure 4), then,, c d, h h 2, u, u 2, FIGUR 4 Another tpe solid region and quation 6 becomes 8 f,, dv d c 2 h h u 2, u, f,, d d d (,, ) =-- (,, ) (,, ) = FIGUR 5 =- = FIGUR 6 XAMPL 2 valuate, where is the solid tetrahedron bounded b the four planes,,, and. SOLUTION When we set up a triple integral it s wise to draw two diagrams: one of the solid region (see Figure 5) and one of its projection on the -plane (see Figure 6). The lower boundar of the tetrahedron is the plane and the upper boundar is the plane (or ), so we use u, and u 2, in Formula 7. Notice that the planes and intersect in the line (or ) in the -plane. So the projection of is the triangular region shown in Figure 6, and we have 9 This description of as a tpe region enables us to evaluate the integral as follows: dv d d d 2 d d 2,,,, 2 dv 2 d d 2 d 6 d FIGUR 7 A tpe 2 region =u (, ) =u (, ) A solid region is of tpe 2 if it is of the form where, this time, is the projection of onto the -plane (see Figure 7). The back surface is u,, the front surface is u 2,, and we have,,,, u, u 2, f,, dv u 2, f,, u d da,

5 SCTION 2.5 TRIPL INTGRALS 697 Finall, a tpe region is of the form FIGUR 8 A tpe region =u (, ) =u (, ) where is the projection of onto the -plane, u, is the left surface, and u 2, is the right surface (see Figure 8). For this tpe of region we have,,,, u, u 2, f,, dv u 2, f,, u d da, In each of quations and there ma be two possible epressions for the integral depending on whether is a tpe I or tpe II plane region (and corresponding to quations 7 and 8). s 2 2 dv V XAMPL valuate, where is the region bounded b the paraboloid 2 2 and the plane 4. SOLUTION The solid is shown in Figure 9. If we regard it as a tpe region, then we need to consider its projection onto the -plane, which is the parabolic region in Figure. (The trace of 2 2 in the plane is the parabola 2.) = +@ =4 Visual 2.5 illustrates how solid regions (including the one in Figure 9) project onto coordinate planes. 4 = FIGUR 9 Region of integration FIGUR Projection on -plane +@=4 From 2 2 we obtain s 2, so the lower boundar surface of is s 2 and the upper surface is s 2. Therefore, the description of as a tpe region is,, 2 2, 2 4, s 2 s 2 and so we obtain _2 2 FIGUR Projection on -plane s 2 2 dv 2 s s 2 s 2 2 d d d Although this epression is correct, it is etremel difficult to evaluate. So let s instead consider as a tpe region. As such, its projection onto the -plane is the disk shown in Figure. Then the left boundar of is the paraboloid 2 2 and the right boundar is the plane 4, so taking u, 2 2 and u 2, 4 in quation, we

6 698 CHAPTR 2 MULTIPL INTGRALS have s 2 2 dv s 2 2 d da s 2 2 da Although this integral could be written as The most difficult step in evaluating a triple integral is setting up an epression for the region of integration (such as quation 9 in ample 2). Remember that the limits of integration in the inner integral contain at most two variables, the limits of integration in the middle integral contain at most one variable, and the limits of integration in the outer integral must be constants. it s easier to convert to polar coordinates in the -plane: r cos, r sin. This gives s 2 2 dv 2 2 s4 2 s s 2 2 d d s 2 2 da 2 d 2 4r 2 r 4 dr 2 4r 2 r 5 4 r 2 rrdrd APPLICATIONS OF TRIPL INTGRALS Recall that if f, then the single integral f d represents the area under the curve f from a to b, and if f,, then the double integral f, da represents the volume under the surface f, and above. The corresponding interpretation of a triple integral f,, dv, where f,,, is not ver useful because it would be the hpervolume of a four-dimensional object and, of course, that is ver difficult to visualie. (Remember that is just the domain of the function f ; the graph of f lies in four-dimensional space.) Nonetheless, the triple integral f,, dv can be interpreted in different was in different phsical situations, depending on the phsical interpretations of,, and f,,. Let s begin with the special case where f,, for all points in. Then the triple integral does represent the volume of : b a 2 V dv For eample, ou can see this in the case of a tpe region b putting in Formula 6: f,, dv u 2, u d da u 2, u, da, and from Section 2.2 we know this represents the volume that lies between the surfaces u, and u 2,. XAMPL 4 Use a triple integral to find the volume of the tetrahedron T bounded b the planes 2 2, 2,, and.

7 SCTION 2.5 TRIPL INTGRALS 699 (,, 2) SOLUTION The tetrahedron T and its projection on the -plane are shown in Figures 2 and. The lower boundar of T is the plane and the upper boundar is the plane 2 2, that is, 2 2. Therefore, we have =2 +2+=2 VT T dv d d d FIGUR 2 T +2=2 or =- 2 (,, ), 2,, 2 b the same calculation as in ample 4 in Section 2.2. (Notice that it is not necessar to use triple integrals to compute volumes. The simpl give an alternative method for setting up the calculation.) All the applications of double integrals in Section 2.4 can be immediatel etended to triple integrals. For eample, if the densit function of a solid object that occupies the region is,,, in units of mass per unit volume, at an given point,,, then its mass is 2 2 m 2 2 d d,, dv = 2 and its moments about the three coordinate planes are FIGUR 4 M,, dv M,, dv M,, dv The center of mass is located at the point,,, where 5 M m M m M m If the densit is constant, the center of mass of the solid is called the centroid of. The moments of inertia about the three coordinate aes are 6 I 2 2,, dv I I 2 2,, dv 2 2,, dv As in Section 2.4, the total electric charge on a solid object occuping a region and having charge densit,, is Q,, dv

8 7 CHAPTR 2 MULTIPL INTGRALS = V XAMPL 5 Find the center of mass of a solid of constant densit that is bounded b the parabolic clinder 2 and the planes,, and. SOLUTION The solid and its projection onto the -plane are shown in Figure 4. The lower and upper surfaces of are the planes and, so we describe as a tpe region:,,, 2, = = Then, if the densit is,, m dv, the mass is 2 d d d d 2 4 d 2 dd 5 4 d FIGUR 4 ecause of the smmetr of and about the -plane, we can immediatel sa that M and, therefore,. The other moments are M dv 2 d d d 2 2 d d 2 d 2 6 d M dv d d 2 d d d 6 d d d Therefore, the center of mass is,, M m, M m, M m ( 5 7,, 5 4) 2.5 XRCISS. valuate the integral in ample, integrating first with respect to, then, and then. dv 2. valuate the integral, where,,, 2, using three different orders of integration. 6 valuate the iterated integral.. 6 d d d 4. s 2 5. e d d d d d d e 2 d d d

9 SCTION 2.5 TRIPL INTGRALS 7 CAS 7 6 valuate the triple integral. 2 dv 7., where {,, 2, s4 2,} cos 5 dv 8., where,,,, dv, where lies under the plane and above the region in the -plane bounded b the curves s,, and. dv, where is bounded b the planes,,, and dv., where is the solid tetrahedron with vertices,,,,,,, 2,, and,, dv 2., where is the solid tetrahedron with vertices,,,,,,,,, and,, 2 e dv., where is bounded b the parabolic clinder 2 and the planes,, and 2 dv 4., where is bounded b the parabolic clinder 2 and the planes,, and 5. dv, where is bounded b the paraboloid and the plane 4 dv 6., where is bounded b the clinder and the planes,, and in the first octant 7 2 Use a triple integral to find the volume of the given solid. 7. The tetrahedron enclosed b the coordinate planes and the plane The solid bounded b the clinder 2 and the planes, 4, and 9 9. The solid enclosed b the clinder and the planes 5 and 2. The solid enclosed b the paraboloid 2 2 and the plane 6 2. (a) press the volume of the wedge in the first octant that is cut from the clinder 2 2 b the planes and as a triple integral. (b) Use either the Table of Integrals (on Reference Pages 6 ) or a computer algebra sstem to find the eact value of the triple integral in part (a). 22. (a) In the Midpoint Rule for triple integrals we use a triple Riemann sum to approimate a triple integral over a bo, where f,, is evaluated at the center i, j, k of the bo ijk. Use the Midpoint Rule to estimate s dv, where is the cube defined b 4, 4, 4. ivide into eight cubes of equal sie. CAS (b) Use a computer algebra sstem to approimate the integral in part (a) correct to the nearest integer. Compare with the answer to part (a) Use the Midpoint Rule for triple integrals (ercise 22) to estimate the value of the integral. ivide into eight sub-boes of equal sie. 2., where ln dv,, 4, 8, 4 24., where Sketch the solid whose volume is given b the iterated integral sin 2 dv,, 4, 2, press the integral as an iterated integral in si different was, where is the solid bounded b the given surfaces ,, 28.,, 2, 29.,, d d d d d d The figure shows the region of integration for the integral s Rewrite this integral as an equivalent iterated integral in the five other orders. 2 6 =œ f,, dv 2 f,, d d d =-