MULTIPLE INTEGRATION NOTES AND EXERCISES by Antony Foster Overview In this discussionwe consider the integralof a function of two variables z = f(x,y) over a regionD in the xy-plane (2-space) and the integral of a function of three variables w = F(x,y,z) over a solid region E in 3-space. These integrals are called multiple integrals and are defined as the limit of approximating Riemann sums, much like the single-variable integrals you learned about in earlier calculus courses. We can use multiple integrals to calculate quantities that vary over two or three dimensions, such as the total mass or the angular momentum of an object of varying density and the volumes of solids with general curved boundaries. DOUBLE INTEGRATION In your earlier studies of calculus we defined the definite integral of a continuous function y = f(x) over a closed interval[a,b]asalimitofRiemannsums. Inthis discussionweextendthisideatodefine theintegralofacontinuous function of two variables z = f(x,y) over a bounded regionD in the plane. In both cases the integrals are limits of approximatingRiemannsums. TheRiemannsumsfortheintegralofasingle-variablefunctiony = (x)areobtained by: 1. Partitioning a finite interval [a,b] into thin subintervals, [x , x ], [x , x ], [x , x ],...,[x , x ], ...,[x , x ] where a = x and x = b 0 1 1 2 2 3 i 1 i n 1 n 0 n − − multiplyingthe widthofeachsubinterval,∆x ,bythevalueoff atapointx insidethatsubinterval,andthen i ∗i 2. Forming a Riemann Sum by adding together all the products of the form f(x )∆x . ∗i i A similar method of partitioning, multiplying, and summing is used to construct double integrals. However, this time we pack a planar region D with small rectangles, rather than small subintervals. We then take the product of each small rectangles area with the value of f(x,y) at a point inside that rectangle, and finally sum together all these products. When f(x,y) is continuous, these sums converge to a single number as each of the small rectangles shrinks in both width and height. The limit is the double integral of f(x,y) over D. As with single integrals, we can evaluate multiple integrals via antiderivatives, which frees us from the formidable task of calculating a double integraldirectlyfromitsdefinitionasalimitofRiemannsums. Themajorpracticalproblemthatarisesinevaluating multiple integrals lies in determining the limits of integration. While the integrals of your earlier studies of calculus were evaluated over an interval, which is determined by its two endpoints, multiple integrals are evaluated over a regionintheplaneorinspace. Thisgivesrisetolimitsofintegrationwhichofteninvolvevariables,notjustconstants. Describing the regions of integration is the main new issue that arises in the calculation of multiple integrals. DOUBLE INTEGRALS OVER RECTANGULAR REGIONS OF THE xy-PLANE We begin our investigation of double integrals by considering the simplest type of planar region, a rectangle. We consider a function z = f(x,y) defined on a rectangular region, D, defined as D = (x,y): a 6 x 6 b, c 6 y 6 d . { } IDEA: We subdivide theplaneregionD intosmallrectanglesusinganetworkoflinesparalleltothe x- andy-axes. The lines divide D into n rectangular pieces, where the number of such pieces n gets large as the width and height of each piece gets small. These rectangles form a partition of D. A small rectangular piece of width, ∆x and height ∆y has area ∆A = ∆x∆y. If we number the small pieces partitioning D in some order, then their areas are given by numbers ∆A , ∆A ,...,∆A where ∆A is the area of the kth small rectangle. 1 2 n k ToformaRiemannsumoverD,wechooseapoint(x ,y )inthekthsmallrectangle,multiplythenumberf(x ,y ) k k k k by the area ∆A and add together the products f(x ,y )∆A : k k k k n S = f(x ,y )∆A n k k k k=1 X Depending on how we pick (x ,y ) in the kth small rectangle, we may get different values for S . k k n We are interested in what happens to these Riemann sums as the widths and heights of all the small rectangles in the partition of D approach zero. The norm of a partition P, written P is the largest width or height of any rectangle in the partition. For example, if P = 0.1, then all the recktanglkes in the partition of D have width at most 0.1 and height at most 0.1. Sometimkes thke Riemann sums converge as the norm of P goes to zero, written P 0. The resulting limit is then written as k k → n lim f(x ,y )∆A . k k k P 0 k k→ k=1 X As P 0 and the rectangles get narrowand short, their number n increases,so we can also write this limit as k k → n lim f(x ,y )∆A . k k k n →∞k=1 X with the understanding that ∆A 0 as n and P 0. There are many choices involved in a limit of k → → ∞ k k → this kind. The collectionof smallrectanglesis determined by the gridof verticalandhorizontallines that determine a rectangular partition of P. In each of the resulting small rectangles there is a choice of an arbitrary point at which f is evaluated. These choices together determine a single Riemannsum. To forma limit, we repeatthe whole process againand again, choosing partitions whose rectangle widths and heights both go to zero and whose number goes to infinity. When a limit of the sums exists, giving the same limiting value no matter what choices are made, then the function f(x,y) is saidto be integrable overD and the limit is calledthe double integral of f(x,y) over D, written as f(x,y)dA or f(x,y)dxdy. ZDZ ZDZ It can be shown that if f(x,y) is a continuous function throughout D, then f is integrable, as in the single-variable casediscussedinyourearlierstudiesofcalculus. Manydiscontinuousfunctionsarealsointegrable,includingfunctions which are discontinuous only on a finite number of points or smooth curves. The proof of these facts can be found in more advanced textbooks about calculus. Interpreting Double Integrals as Volumes of Solids in 3-space When f(x,y) is a positive function over a plane region D in the xy-plane, we may interpret the positive number f(x,y)dA as the volume of the 3-dimensional solid region E over the xy-plane bounded below by D and D bRoRunded above by the graph of the surface given as z = f(x,y). Each term f(x ,y ) ∆A (a positive number) in k k k the sum S = n f(x ,y ) ∆A is the volume of a vertical rectangular box that approximates the volume of n k=1 k k k the portion of the solid region E in space that stands directly above the base ∆A . The sum S thus approximates k n what we want toPcall the total volume of the solid E. Definition Volume as a double Integral If f(x,y) > 0 for all (x,y) D, then we define the Volume of {the solid E in space whose base} is D and bounded above by t∈he surface z = f(x,y) as V(E), and n V(E) = lim S = f(x,y)dA n n →∞Xk=1 ZDZ where ∆A 0 as n . As youmight expect, this more generalmethod of calculating volume agreeswith the k → → ∞ Solids of Revolution methods you encountered in your earlier studies of calculus, but we do not prove this here. Fubini’s Theorem for Calculating Double Integrals Suppose that we wish to calculate the volume, V, under the plane z = f(x,y) = 4 x y over the rectangular region D = (x,y) : 0 6 x 6 2, 0 6 y 6 1 in the xy-plane. If we apply the me−thod−of slicing you learned in { } working with volumes in your earlier studies of calculus, with slices perpendicular to the x-axis, then the volume is x=2 A(x)dx (1) Zx=0 where A(x) is the cross-sectionalarea at x. For each value of x, we may calculate A(x) as the integral y=1 A(x) = (4 x y)dy (2) − − Zy=0 which is the area under the curve in the plane of the cross-section at x. In calculating A(x), x is held fixed and the integration takes place with respect to y. Combining Equations (1) and (2), we see that the volume of the entire solid is x=2 x=2 y=1 Volume = A(x)dx = (4 x y)dy dx − − Zx=0 Zx=0 (cid:18)Zy=0 (cid:19) x=2 y2 y=1 = 4y xy dx − − 2 Zx=0 (cid:20) (cid:21)y=0 x=2 7 = x dx 2 − Zx=0 (cid:18) (cid:19) 7 x2 x=2 = x = 5 cubic units 2 − 2 (cid:20) (cid:21)x=0 To obtain the cross-sectionalarea A(x), we hold x fixed and integrate with respect to y. If we just wanted to write a formula for the volume, without carrying out any of the integrations, we could write 2 1 Volume = (4 x y)dy dx − − Z0 Z0 The expression on the right, called an iterated or repeated integral, says that the volume is obtained by integrating 4 x y with respect to y from y = 0 to y = 1, holding x − − fixed, and then integrating the resulting expression in x with respect to x from x = 0 to x = 2. The limits of integration 0 and 1 are associated with y, so they are placed on the integral closest to dy. The other limits of integration, 0 and 2, are associated with the variable x, so they are placed on the outside integral symbol that is paired with dx. What would have happened if we had calculated the volume by slicing with planes perpendicular to the y-axis? As a function of y, the typical cross-sectionalarea is x=2 x2 x=2 A(y) = (4 x y)dx = 4x xy = 6 2y. (4) − − − 2 − − Zx=0 (cid:20) (cid:21)x=0 The volume of the entire solid is therefore y=1 y=1 Volume = A(y)dy = (6 2y)dy = 6y y2 y=1 = 5 cubic units, − − y=0 Zy=0 Zy=0 (cid:2) (cid:3) in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing 1 2 Volume = (4 x y)dxdy. − − Z0 Z0 The expression on the right says we can find the volume by integrating with respect to x from x = 0 to x = 2 as in Equation (4) and then integrating the result with respect to y from y =0 to y = 1. In this iterated integral, the order of integration is first x and then y, the reverse of the order in Equation (3). Whatdoesthetwovolumecalculationsabovewith iterated integralshavetodowith thedouble integral (4 x y)dA − − ZDZ over the rectangle D = (x,y) : 0 6 x 6 2, 0 6 y 6 1 ? The answer is that both iterated integrals give { } the value of the double integral. This is what we would reasonably expect, since the double integral measures the volume of the same region as the two iterated integrals. A theorem published in 1907 by Guido Fubini says that: The double integral of any continuous function f(x,y) over a rectangle D can be calculated as an iterated integral in either order of integration. Guido Fubini proved his theorem in greater generality, but this is what it says in our setting: Theorem 1. Fubini’s Theorem (Weak Form) If f(x,y) is continuous throughout the rectangular region D = (x,y): a 6 x 6 b, c 6 y 6 d , then { } d b b d f(x,y)dA = f(x,y)dxdy = f(x,y)dy dx. ZDZ Zc Za Za Zc Fubini’s Theorem says that: Double integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time. Fubini’s Theorem also says that: We may calculate the double integral by integrating in either order. Whenwecalculateavolumebyslicing,wemayuse,eitherplanesperpendiculartothex-axisorplanesperpendicular to the y-axis. Example 1. Calculate f(x,y)dA for f(x,y) = 1 6x2y and D = (x,y): 0 6 x 6 2, 1 6 y 6 1 . − { − } D RR Solution. By Fubini’s Theorem 1 2 f(x,y)dA = (1 6x2y)dxdy − ZDZ Z−1Z0 1 = x 2x3y x=2dy − x=0 Z−11(cid:2) (cid:3) = (2 16y)dy − Z−1 = 2y 8y2 y=1 = 4 − y= 1 − (cid:2) (cid:3) Reversing the order of integration gives the same answer: 2 1 1 (1 6x2y)dy dx = y 3x2y2 y=1 dx − − y= 1 Z0 Z−1 Z−21(cid:2) (cid:3) − = (1 3x2) ( 1 3x2) dx − − − − Z0 2(cid:2) (cid:3) = 2dy = 4 Z0 ⊔⊓ Double Integrals over bounded Non Rectangular Regions To define the double integral of a function f(x,y) over a bounded, nonrectangular region D, we again begin by covering D with a grid of small rectangular cells whose union contains all points of D. This time, however, we cannot exactly fill D with a finite number of rectangles lying inside D, since its boundary is curved, and some of the small rectangles in the grid lie partly outside D. A partition of D is formed by taking the rectangles that lie completely inside it, not using any that are either partly or completely outside. For commonly arising regions,more and more of D is included as the norm of a partition (the largest width or height of any rectangle used) approaches zero. Once we have a partition of D, we number the rectangles in some order from 1 to n and let ∆A be the area of the k k-th rectangle. We then choose a point (x ,y ) in the k-th rectangle and form the Riemann sum k k n S = f(x ,y )∆A n k k k k=1 X As the norm of the partition forming goes to zero, i.e., P 0, the width and height of each enclosed rectangle k k → goestozeroandtheirnumbergoesto infinity. Iff(x,y)isacontinuousfunction, thentheseRiemannsumsconverge to a limiting value, not dependent on any of the choices we made. This limit is called the double integral of f(x,y) over D: n lim f(x ,y )∆A = f(x,y)dA k k k P 0 k k→ Xk=1 ZDZ The nature of the boundary of D introduces issues not found in integrals over an interval. When D has a curved boundary,thenrectanglesofapartitionlieinsideD butdonotcoverallofD. Inorderforapartitiontoapproximate D well, the parts of D coveredby smallrectangleslying partly outsideD must become negligible as the normof the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made frompolygons, circles,ellipses, and from continuous graphs over an interval, joined end to end. A curve with a fractal type of shape would be problematic, but such curves are not relevant for most applications. A careful discussion of which type of regions D can be used for computing double integrals is left to a more advanced text. Doubleintegralsofcontinuousfunctionsovernonrectangularregionshavethesamealgebraicproperties(summarized further on)as integralsoverrectangularregions. The domainAdditivity Propertysaysthat if D is decomposedinto nonoverlappingregionsD andD withboundariesthatareagainmadeofafinitenumberoflinesegmentsorsmooth 1 2 curves, then f(x,y)dA = f(x,y)dA + f(x,y)dA ZDZ ZDZ1 ZDZ2 If f(x,y) is positive and continuous over D we define the volume of the solid region, E, between D and the surface z = f(x,y) to be f(x,y)dA as before. D RR A Procedure for evaluating Double Integrals over x-simple bounded nonrectangular Regions If D = (x,y) : a 6 x 6 b, g (x) 6 y 6 g (x) is an x-simple region in the xy-plane that is, D is bounded 1 2 { } above and bounded below by the plane curves y = g (x) and y = g (x) andonthe sides by the lines x = a and 2 1 x = b we may again calculate the volume f(x,y)dA by the method of slicing. D RR STEP 1. We first calculate the cross-sectionalarea y=g2(x) A(x) = f(x,y)dy Zy=g1(x) and then STEP 2. integrate A(x) from x = a to x = b to get the volume as an iterated integral: b b g2(x) Volume = V(E) = f(x,y)dA = A(x)dx = f(x,y)dy dx (5) ZDZ Za Za Zg1(x) A Procedure for Evaluating Double Integrals over y-simple bounded nonrectangular Regions Similarly, if D = (x,y) : c 6 x 6 d, h (y) 6 x 6 h (y) is an y-simple region in the xy-plane that is, D 1 2 { } is bounded above and bounded below by the lines y = c and y = d and on the sides by the plane curves x = h (y) and x = h (y), then the volume calculated by slicing as follows: 2 1 Step 1. We first calculate the cross-sectionalarea x=h2(y) A(y) = f(x,y)dx Zx=h1(y) and then Step 2. integrate A(y) from y = c to y = d to get the volume as an iterated integral: d d h2(y) Volume = V(E) = f(x,y)dA = A(y)dy = f(x,y)dxdy (6) ZDZ Zc Zc Zh1(y) The fact that the iterated integrals(5) and(6) both give the volume that we defined to be the double integral of f(x,y) over D is a consequence of the following stronger form of Fubini’s Theorem. Theorem 2. Fubini’s Theorem (Strong Form) Let f(x,y) be a continuous function throughout a region D. 1. If D is defined by a 6 x 6 b, g (x) 6 y 6 g (x) with g (x) and g (x) continuous on [a, b], that 1 2 1 2 is, D is an x-simple region, then b g2(x) f(x,y)dA = f(x,y)dy dx. ZDZ Za Zg1(x) 2. If D is defined by c 6 y 6 d, h (y) 6 x 6 h (y) with h (y) and h (y) continuous on [c, d], that 1 2 1 2 is, D is a y-simple region, then d h2(y) f(x,y)dA = f(x,y)dxdy. ZDZ Zc Zh1(y) Example 2. Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis (i.e., the line y = 0) and the lines y = x and x = 1 and whose top lies in the plane z = f(x,y) = 3 x y. − − Solution. First notice that the region D is bounded below by the curve g (x) = 0 and bounded above the the 1 curve g (x) = x and on the sides by the lines x = 0 and x = 1, i.e, D is an x-simple region. Thus by Fubini’s 2 Theorem (Strong Form): Volume = f(x,y)dA ZDZ b g2(x) = f(x,y)dA Za Zg1(x) 1 x = (3 x y)dy dx − − Z0 Z0 1 y2 y=x = 3y xy dx − − 2 Z0 (cid:20) (cid:21)y=0 1 3 3 1 x=1 = 3x x2 dx = x2 x3 = 1 cubic unit − 2 2 − 2 Z0 (cid:18) (cid:19) (cid:20) (cid:21)x=0 When the order of the integration is reversed, i.e., we consider D as a y-simple region, then the volume is Volume = f(x,y)dA ZDZ d h2(y) = f(x,y)dA Zc Zh1(y) 1 1 = (3 x y)dxdy − − Z0 Zy 1 1 x=1 = 3x x2 xy dy − 2 − Z0 (cid:20) (cid:21)x=y 1 1 1 = 3 y 3y + y2 + y2 dy − 2 − − 2 Z0 (cid:18) (cid:19) 1 5 3 = 4y + y2 dy 2 − 2 Z0 (cid:18) (cid:19) y=1 5 1 = y 2y2 + y3 = 1 cubic unit. 2 − 2 (cid:20) (cid:21)y=0 The two integrals are equal as they should be. ⊔⊓ Although Fubini’s Theorem assures us that: A double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen. Example 3. Calculate f(x,y)dAwhere f(x,y) = sin(x) andD is the triangularregioninthexy-plane bounded x D by the x-axis, the line yR=R x and the line x = 1. Solution. If we first integrate with respect to y and then with respect to x, we find 1 x sin(x) 1 sin(x) y=x 1 dy dx = y dx = sin(x)dx = [ cos(x)]x=1 = 1 cos(1) 0.46. x x − x=0 − ≈ Z0 (cid:20)Z0 (cid:21) Z0 (cid:20) (cid:21)y=0 Z0 If we reversed the order of integration and attempt to calculate 1 1 sin(x) dxdy, x Z0 Zy we run into a problem because the integral sin(x) dx x Z cannot be expressed in terms of elementary functions (there is no simple antiderivative!) ⊔⊓ WhencalculatingdoubleIntegralsoverboundednonrectangularregions,thereisnogeneralruleforpredictingwhich orderofintegrationwillbetheeasiestorthegoodoneincircumstanceslikethese. Iftheorderyouchoosefirst,doesnt work, then try the other. Sometimes neither order will work, and then we need to use numerical approximation methods. A Procedure for Finding the limits of integration for a Double Integrals We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works. When faced with evaluating f(x,y) dA, integrating first with respect to y and then with respect to x, take the D following steps: RR Step 1. Sketch: Sketch the region D of integration and its bounding curves. Step 2. Find the y-limits of integration: Find the y-limits of integration imagining a vertical line, say L passing through the region D in the direction of increasing y. Mark the y values where L enters and leaves D. These are your y-limits of integration and are usually functions of x (instead of constants, unless D is a rectangle). Step 3. Find the x-limits of integration: we find the x-limits of integration by choosing x-limits that include all vertical lines passing through D. Exercise 1. Write out a 3-step procedure similar to that above, when faced with evaluating f(x,y)dA, ZDZ integrating first with respect to x and then with respect to y. Example 4. Reversing the order of integration Reverse the order of integration for the { } double integral 2 2x f(x,y)dA = (4x + 2)dy dx ZDZ Z0 Zx2 using the above 3 Step procedure. Solution. 2 2x 4 √y f(x,y)dA = (4x + 2)dy dx = (4x + 2)dxdy ZDZ Z0 Zx2 Z0 Zy/2 ⊔⊓ Properties of Double Integrals Likesingleintegrals,doubleintegralsofcontinuousfunctionshavealgebraicpropertiesthatareusefulincomputations and applications. If f(x,y) and g(x,y) are continuous function on a plane region D, then 1. Constant Multiple: cf(x,y)dA = c f(x,y)dA for any c. D D RR RR 2. Sum and Difference: (f(x,y) g(x,y))dA = f(x,y)dA g(x,y)dA. ± ± D D D RR RR RR 3. Domination: a) f(x,y)dA > 0 if f(x,y) > 0 on D. D RR b) f(x,y)dA > g(x,y)dA if f(x,y) > g(x,y) on D. D D RR RR f(x,y) = f(x,y)dA + g(x,y)dA 4. Additivity: ZDZ ZDZ1 ZDZ2 if D is the union of two nonoverlapping regions D and D . 1 2 Areas of bounded regions in the Plane If we take f(x,y) = 1 in the definition of the double integralovera regionD in the preceding section, the Riemann sums reduce to n n S = f(x ,y )∆A = ∆A (1) n k k k k k=1 k=1 X X This is simply the sum of the areas of the small rectangles in the partition of D, and approximates what we would like to call the area of the region D. As the normof a partitionof D approacheszero,the height and width of all rectangles in the partition approachzero, and the coverageof D becomes increasingly complete. We define the area of the region D, which we denote by A(D) be the limit n Area of the region D = A(D) = lim ∆A = dA. k P 0 k k→ kX=1 ZDZ Definition area of planar region The area of a closed, bounded region D of the plane is { } A(D) = dA. ZDZ Average Value Theaveragevalueofanintegrablefunctionofonevariableonaclosedintervalisthe integralofthefunctionoverthe intervaldividedbythelengthoftheinterval. Foranintegrablefunctionoftwovariablesdefinedonaboundedregion D intheplane,theaveragevalueisthe integral of f over the region D divided by the area of the region D. This can be visualized by thinking of the function as giving the height at one instant of some water sloshing around in a tank whose vertical walls lie over the boundary of the region. The average height of the water in the tank can be found by letting the water settle down to a constant height. The height is then equal to the volume of water in the tank divided by the area of D. We are led to define the average value of an integrable function f over a region D to be Definition Average value of f(x,y) The average value of f(x,y) on a closed, bounded region D of the { } plane is 1 Avg(f) = f(x,y)dA. A(D) ZDZ Moments, Mass and Centers-of-Mass for Thin Flat Plates (or Laminas) Using multiple integrals we can calculate the mass, Moment and center-of-mass (the balance point) of laminas with varying density. We first consider the problem of finding the center of mass of a thin flat plate or lamina: A disk of Aluminum, say,or a triangularsheet of metal. We assume the distribution of mass in such a plate to be continuous. A materials density function, denoted by δ(x,y) is its mass per unit area. The mass of a thin plate is obtained by integrating the density function δ(x,y) over the region D forming the thin plate. The first moment about an axis is calculated by integrating over D the distance from the axis times the density. The center of mass is found from the first moments. Below we give the double integral formulas for mass, first moments, and center of mass. Mass and first moment formulas for thin plates (also called laminas) covering a region D in the xy-plane. MASS: M = δ(x,y)dA D RR FIRST MOMENTS: M = y δ(x,y)dA, M = xδ(x,y)dA. x y D D RR RR CENTER-OF-MASS: x¯ = My, y¯ = Mx M M Example 5. A thin plate covers a triangular region D of the xy-plane bounded by the x-axis and the lines x = 1 and y = 2x in the first quadrant. The plate’s density at the point (x,y) D is δ(x,y) = 6x + 6y + 6. Find the ∈ plate’s Mass, First Moments and Center-of-Mass about the coordinate axes. Solution. 1. Make a sketch of D = (x,y) 0 6 x 6 1, 0 6 y 6 2x (Not shown). { | } Calculating M = δ(x,y)dA ZDZ by integrating first with respect to y and then with respect to x. 2. Finding the y-limits of integration: Any vertical line L passing through D in the direction of increasing y must enterD at y = 0 and leaves D at y = 2x and so 2x A(x) = (6x + 6y + 6)dy = 6xy +3y2 + 6y y=2x = 24x2 + 12x y=0 Z0 (cid:2) (cid:3) 3. Finding the x-limits of integration: We include all vertical lines cutting through D starting with x = 0 and ending with x = 1. Thus 1 2x 1 M = δ(x,y)dA (6x + 6y + 6)dy dx = (24x2 + 12x)dx ZDZ Z0 (cid:20)Z0 (cid:21) Z0 = 8x3 + 6x2 x=1 = 14 mass Units x=0 (cid:2) (cid:3) First Moment about the x-axis 1 2x 1 M = y δ(x,y)dA = (6xy + 6y2 + 6y)dy dx = 3xy2 + 2y3 + 3y2 y=2x dx x y=0 ZDZ Z0 (cid:20)Z0 (cid:21) Z0 (cid:2) (cid:3) 1 = (28x3 + 12x2)dx = 7x4 + 4x3 x=1 = 11 x=0 Z0 (cid:2) (cid:3) First Moment about the y-axis 1 2x 1 M = xδ(x,y)dA = (6x2 + 6xy + 6x)dy dx = 6x2y + 3xy2 + 6xy y=2x dx y y=0 ZDZ Z0 (cid:20)Z0 (cid:21) Z0 (cid:2) (cid:3) 1 = (24x3 + 12x2)dx = 6x4 + 4x3 x=1 = 10 x=0 Z0 (cid:2) (cid:3) Finally, the coordinates of the Center-Of-Mass are: M 10 M 11 y x x¯ = = and y¯ = = . M 14 M 14 In other words, the triangular plate balances at the point (5/7, 11/14)in D. ⊔⊓