THE DOUBLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; POLAR COORDINATES

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1 HE DOUBLE INEGRAL AS HE LIMI OF RIEMANN SUMS; POLAR COORDINAES HE DOUBLE INEGRAL AS HE LIMI OF RIEMANN SUMS; POLAR COORDINAES he Double Integral as the Limit of Riemann Sums In the one-variable case we can write the integral as the limit of Riemann sums: b n f(x) dx = lim f(xi ) x i. a max x i i= he same approach works with double integrals. o explain it we need to explain what we mean by the diameter of a set. Suppose that S is a bounded closed set (on the line, in the plane, or in three-space). For any two points P and Q of S we can measure their separation, d(p, Q). he greatest separation between points of S is called the diameter of S: diam S = max d(p, Q). P,Q S For a circle, a circular disc, a sphere, or a ball, this sense of diameter agrees with the usual one. Now let s start with a basic region and decompose it into a finite number of basic subregions,, N. If f is continuous on, then f is continuous on each i. Now from each i, we pick an arbitrary point (x i, y i ) and form the Riemann sum N f(x i, yi )(area of i ) i= he double integral over can be obtained as the limit of such sums; namely, given any ɛ >, there exists δ > such that, if the diameters of the are all i less than δ then N f(x i, yi )(area of i ) f(x, y) dxdy < ɛ i= no matter how the (x i, y i ) are chosen within the. We express this by writing N f(x, y) dxdy = lim f(x i, yi )(area of i ). diam i i= Evaluating Double Integrals Using Polar Coordinates Here we explain how to calculate double integrals f(x, y) dxdy using polar coordinates (r, θ). hroughout we take r and restrict θ to [, π]. March 7,

2 HE DOUBLE INEGRAL AS HE LIMI OF RIEMANN SUMS; POLAR COORDINAES We will work with the type of region consisting of all points (x, y) that have polar coordinates (r, θ) in the set : α θ β, ρ(θ) r P (θ). According to the formula in one-variable case the area of is given by area of = β α [P (θ) ρ (θ)] dθ. We can write this as a double integral over : area of = r drdθ. Proof. Simply note that and therefore P (θ) [P (θ) ρ (θ)] = rdr ρ(θ) area of = β P (θ) α ρ(θ) rdrdθ = r drdθ. Now let s suppose that f is some function continuous at each point (x, y) of. hen the composition F (r, θ) = f(r cos θ, r sin θ) is continuous at each point (r, θ) of. We will show that f(x, y) dxdy = f(r cos θ, r sin θ)r drdθ. Proof. Our first step is to place a grid on by using a finite number of rays θ = θ j and a finite number of continuous curves r = ρ k (θ). his grid decomposes into a finite number of little regions,, N with polar coordinates in sets,, N. Note that the area of each i = r drdθ. i Writing F (r, θ) for f(r cos θ, r sin θ), we have N F (r, θ)r drdθ = i= i F (r, θ)r drdθ (additivity) = N i= F (ri, θi ) i r drdθ with some (r i, θ i ) i (Mean-value heorem) Evaluating Double Integrals Using Polar Coordinates March 7,

3 HE DOUBLE INEGRAL AS HE LIMI OF RIEMANN SUMS; POLAR COORDINAES 3 N N = F (ri, θi )(area of i ) = f(xi, yi )(area of i ). i= i= his last expression is a Riemann sum for the double integral f(x, y) dxdy and, as such, differs from that integral by less than any preassigned positive ɛ provided only that the diameters of all the i are sufficiently small. his we can guarantee by making our grid sufficiently fine. Problem. Use polar coordinates to evaluate xy dxdy where is the portion of the unit disc that lies in the first quadrant. Solution. Here : θ π, r. herefore R. xy dxdy = (r cos θ)(r sin θ)r drdθ = π/ r 3 cos θ sin θ drdθ = 8. Problem. Use polar coordinates to calculate the volume of a sphere of radius Solution. In rectangular coordinates V = R (x + y ) dxdy where is the disc of radius R centered at the origin. Here herefore : θ π, r R. V = R (x + y ) rdrdθ π R = R r rdrdθ = 4 3 πr3. Problem 3. Calculate the volume of the solid bounded above by the lower nappe of the cone z = x + y and bounded below by the disc : (x ) + y. Solution V = = dxdy (x + y ) dxdy (x + y ) dxdy. Evaluating Double Integrals Using Polar Coordinates March 7,

4 HE DOUBLE INEGRAL AS HE LIMI OF RIEMANN SUMS; POLAR COORDINAES 4 he first integral is (area of ) = π. We evaluate the second integral by changing to polar coordinates. he equation (x ) + y = simplifies to x + y = x. In polar coordinates this becomes r = r cos θ, which simplifies to r = cos θ. he disc is the set of all points with polar coordinates in the set herefore : π θ π, r cos θ. (x + y ) dxdy = We then have V = π 3 9. Problem 4. Evaluate where is the triangle bounded by r drdθ = π/ π/ dxdy ( + x + y ) 3/ y =, x = y, x =. cos θ r drdθ = 3 9. Solution. he vertical side of the triangle is part of the line x =. In polar coordinates this is r cos θ =, which can be written r = sec θ. herefore ( + x + y ) dxdy = r drdθ 3/ ( + r ) 3/ where he double integral over reduces to For θ [, π/4] π/4 sec θ : θ π/4, r sec θ. [ r π/4 ( + r ) drdθ = 3/ = ( π/4 + sec θ = herefore the integral can be written ( π/4 + r ) + sec dθ. θ cos θ cos θ + = ] sec θ cos θ sin θ. ) [ ( )] π/4 cos θ sin θ sin dθ = θ sin = π θ 4 π 6 = π. he function f(x) = e x has no elementary antiderivative. Nevertheless, by taking a circuitous route and then using polar coordinates, we can show that e x dx = π. Evaluating Double Integrals Using Polar Coordinates March 7, dθ

5 SOME APPLICAIONS OF DOUBLE INEGRAION 5 Proof. he circular disc D b : x +y b is the set of all (x, y) with polar coordinates (r, θ) in the set : θ π, r b. herefore D b e (x+y) dxdy = e r r drdθ = = π π b ( e b)dθ = π( e b ). e r rdrdθ Let S a be the square a x a, a y a. Since D a S a D a and e (x +y ) is positive, e (x+y) dxdy e (x+y) dxdy e (x+y) dxdy. It follows that D a S a π( e a ) S a D a e (x+y) dxdy π( e 4a ). As a, π( e a ) π and π( e 4a ) π. herefore But herefore lim a S a e (x+y) dxdy = = ( a = e x dx a S a e (x+y) dxdy = π. a a a a ) ( a a a a a a e x e y dxdy e (x +y ) dxdy ) ( a e y dy = e dx) x. a a ( ) / lim e x dx = lim a a a e (x +y) dxdy = π. S a his integral comes up frequently in probability theory and plays an important role in what is called statistical mechanics. SOME APPLICAIONS OF DOUBLE INEGRA- ION he Mass of a Plate Suppose that a thin distribution of matter, what we call a plate, is laid out in the xy-plane in the form of a basic region. If the mass density of the plate (the mass per unit area) is constant, then the total mass M of the plate is simply the density λ times the area of the plate: M = λ the area of. March 7,

6 SOME APPLICAIONS OF DOUBLE INEGRAION 6 If the density varies continuously from point to point, say λ = λ(x, y), then the mass of the plate is the average density of the plate times the area of the plate: M = average density the area of. his is an integral: M = λ(x, y) dxdy. he Center of Mass of a Plate he center of mass of a rod x M is a density-weighted average taken over the interval occupied by the rod: x M M = b a xλ(x) dx. he center of mass of a plate (x M, y M ) is determined by two density-weighted averages, each taken over the region occupied by the plate: x M M = xλ(x, y) dxdy, y M M = yλ(x, y) dxdy. Problem. A plate is in the form of a half-disc of radius a. Find the mass of the plate and the center of mass given that the mass density of the plate varies directly as the distance from the center of the straight edge of the plate. Solution. Place the plate over the x-axis. he mass density can then be written λ(x, y) = k x + y. Here M = x M M = hus x M =. π a ( π ) ( a ) k x + y dxdy = (kr)rdrdθ = k dθ r dr y M M = ( ) x k x + y = k(π)( 3 a3 ) = 3 ka3 π. ( ) y k x + y dxdy =. (the integrand is odd w.r.t. x) dxdy = π a (r sin θ)(kr)rdrdθ ( π ) ( a ) = k sin θ dθ r 3 dr = k()( 4 a4 ) = ka4. Since M = 3 ka3 π, we have y M = ( ka4) / ( 3 ka3 π ) = 3a/π. he Center of Mass of a Plate March 7,

7 SOME APPLICAIONS OF DOUBLE INEGRAION 7 Centroids If the plate is homogeneous, then the mass density λ is constantly M/A where A is the area of the base region. In this case the center of mass of the plate falls on the centroid of the base region. he centroid ( x, ȳ) depends only on the geometry of : xm = x(m/a) dxdy = (M/A) x dxdy ȳm = y(m/a) dxdy = (M/A) Dividing by M and multiplying through by A we have xa = x dxdy, ȳa = y dxdy. y dxdy. hus x is the average x-coordinate on and ȳ is the average y-coordinate on. he mass of the plate does not enter into this at all. Problem. Find the centroid of the region Solution. xa = x dxdy = ȳa = y dxdy = : a x b, φ (x) y φ (x). b φ (x) a φ (x) b φ (x) a φ (x) x dydx = y dydx = Kinetic Energy and Moment of Inertia b a b a x[φ (x) φ (x)] dx; ([φ (x)] [φ (x)] ) dx. A particle of mass m at a distance r from a given line rotates about that line with angular speed ω. he speed v of the particle is then rω and the kinetic energy is given by the formula KE = mv = mr ω. Imagine now a rigid body composed of a finite number of point masses m i located at distances r i from some fixed line. If the rigid body rotates about that line with angular speed ω, then all the point masses rotate about that same line with that same angular speed ω. he kinetic energy of the body can be obtained by adding up the kinetic energies of all the individual particles: KE = i m ir i ω = ( ) m i ri ω. i he expression in parentheses is called the moment of inertia (or rotational inertia) of the body and is denoted by the letter I: I = i m i r i. Centroids March 7,

8 SOME APPLICAIONS OF DOUBLE INEGRAION 8 For a rigid body in straight-line motion KE = Mv where v is the speed of the body. For a rigid body in rotational motion KE = Iω where ω is the angular speed of the body. he Moment of Inertia of a Plate Suppose that a plate in the shape of a basic region rotates about an axis. moment of inertia of the plate about that axis is given by the formula I = λ(x, y)[r(x, y)] dxdy where λ = λ(x, y) is the mass density function and r(x, y) is the distance from the axis to the point (x, y). Derivation. Decompose the plate into N pieces in the form of basic regions,, N. From each i, choose a point (x i, y i ) and view all the mass of the plate as concentrated there. he moment of inertia of this piece is then approximately [λ(x i, y i )(area of i )][r(x i, y i )] = λ(x i, y i )[r(x i, y i )] (area of i ) he sum of these approximations, N λ(x i, yi )[r(x i, yi )] (area of i ) i= is a Riemann sum for the double integral λ(x, y)[r(x, y)] dxdy. As the maximum diameter of the i tends to zero, the Riemann sum tends to this integral. Problem 3. A rectangular plate of mass M, length L, and width W with lower left-corner situated at (, ) rotates about the y-axis. Find the moment of inertia of the plate about that axis (a) given that the plate has uniform mass density. (b) given that the mass density of the plate varies directly as the square of the distance from the rightmost side. Solution. Let the plate be placed in the rectangle R with corners (, ), (L, ), (L, W ), (, W ). (a) Here λ(x, y) = M/LW and r(x, y) = x. hus I = R M LW x dxdy = M LW W L = M L LW W x dx = 3 ML. x dxdy he Moment of Inertia of a Plate March 7, he

9 SOME APPLICAIONS OF DOUBLE INEGRAION 9 (b) In this case λ(x, y) = k(l x) but we still have r(x, y) = x. herefore I = R W L k(l x) x dxdy = k (L x) x dxdy = 3 kl5 W. We can eliminate the constant of proportionality k by noting that M = R W L k(l x) dxdy = k (L Lx + x ) dxdy = 3 kl3 W. herefore k = 3M/L 3 W and I = ML. Radius of Gyration If the mass M of an object is all concentrated at a distance r from a given axis, then the moment of inertia about that axis is given by the product Mr. Suppose now that we have a plate of mass M (actually any object of mass M will do here), and suppose that l is some line. he object has some moment of inertia I about l. Its radius of gyration about l is the distance K for which I = MK. Namely, the radius of gyration about I is the distance from l at which all the mass of the object would have to be concentrated to effect the same moment of inertia. he formula for radius of gyration is usually written K = I/M. Problem 4. A homogeneous circular plate of mass M and radius R rotates about an axle that passes through the center of the plate and is perpendicular to the plate. Calculate the moment of inertia and the radius of gyration. Solution. ake the axle as the z-axis and let the plate rest on the circular region : x + y R. he density of the plate is M/A = M/πR and r(x, y) = x + y. Hence M I = πr (x + y ) dxdy = M π R r 3 drdθ = πr MR. he radius of gyration K is I/M = R/. he circular plate of radius R has the same moment of inertia about the central axle as a circular wire of the same mass with radius R/. he circular wire is a more efficient carrier of moment of inertia than the circular plate. he Parallel Axis heorem Suppose we have an object of mass M and a line l M that passes through the center of mass of the object. he object has some moment of inertia about that line; call it I M. Radius of Gyration March 7,

10 RIPLE INEGRALS If l is any line parallel to l M, then the object has a certain moment of inertia about l; call that I. he parallel axis theorem states that I = I M + d M where d is the distance between the axes. We will prove the theorem under somewhat restrictive assumptions. Assume that the object is a plate of mass M in the shape of a basic region, and assume that l M is perpendicular to the plate. Call l the z-axis. Call the plane of the plate the xy-plane. Denoting the points of by (x, y) we have I I M = λ(x, y)(x + y ) dxdy λ(x, y)[(x x M ) + (y y M ) ] dxdy = λ(x, y)[x M x + y M y (xm + ym)] dxdy = x M xλ(x, y) dxdy + y M (x M + ym) λ(x, y) dxdy yλ(x, y) dxdy = x MM + y MM (x M + y M)M = (x M + y M)M = d M. An obvious consequence of the parallel axis theorem is that I M I for all lines l parallel to l M. o minimize moment of inertia we must pass our axis through the center of mass. RIPLE INEGRALS Once we are familiar with double integrals it is not hard to understand triple integrals f(x, y) dxdy, f(x, y, z) dxdydz. Basically the only difference is this: instead of working with functions of two variables continuous on a plane region, we will be working with functions of three variables continuous on some portion of three-space. March 7,

11 RIPLE INEGRALS he riple Integral over a Box For double integration we began with a rectangle R : a x a, b y b. For triple integration we begin with a box Π : a x a, b y b, c z c. o partition this box, we first partition the edges. aking a partition P : a = x < x < x < < x m = a of [a, a ], a partition P : b = y < y < y < < y n = b of [b, b ] and a partition P 3 : c = z < z < z < < z q = c of [c, c ] we form the Cartesian product P = P P P 3 = {(x i, y j, z k ) : x i P, y j P, z k P 3 } and call this a partition of Π. P breaks up Π into m n q nonoverlapping boxes Π ijk : x i x x i, y j y y j, z k z z k. aking and M ijk as the maximum value of f on Π ijk m ijk as the minimum value of f on Π ijk we form the upper sum m n q m n q U f (P ) = M ijk (volume of Π ijk ) = M ijk (x i x i )(y j y j )(z k z k ) i= j= k= i= j= k= and the lower sum m n q m n q L f (P ) = m ijk (volume of Π ijk ) = m ijk (x i x i )(y j y j )(z k z k ) i= j= k= i= j= k= As in the case of functions of one and two variables, it turns out that, with f continuous on Π, there is one and only one number I that satisfies the inequality L f (P ) I U f (P ) for all partitions P of Π. DEFINIION HE RIPLE INEGRAL OVER A BOX Π he unique number I that satisfies the inequality L f (P ) I U f (P ) for all partitions P of Π. is called the triple integral of f over Π and is denoted by f(x, y, z) dxdydz. Π he riple Integral over a Box March 7,

12 RIPLE INEGRALS he riple Integral over a More General Solid We start with a three-dimensional, bounded, open, connected set and adjoin to it the boundary. We now have a three-dimensional, bounded, closed, connected set. We assume that is a basic solid; that is, we assume that the boundary of consists of a finite number of continuous surfaces z = α(x, y), y = β(x, z), x = γ(y, z). Now let s suppose that f is some function continuous on. o define the triple integral of f over we first encase in a rectangular box Π. We then extend f to all of Π by defining f to be zero outside of. his extended function f is bounded on Π, and it is continuous on all of Π except possibly at the boundary of. In spite of these possible discontinuities, f is still integrable over Π; that is, there still exists a unique number I such that L f (P ) I U f (P ) for all partitions P of Π. he number I is by definition the triple integral Π f(x, y, z) dxdydz. We define the triple integral over by setting f(x, y, z) dxdydz = Π f(x, y, z) dxdydz. Volume as a riple Integral he simplest triple integral of interest is the triple integral of the function that is constantly one on. his gives the volume of : volume of = dxdydz. Proof. Set f(x, y, z) = for all (x, y, z) in. Encase in a box Π. Define f to be zero outside of. An arbitrary partition P of Π breaks up into little boxes Π ijk. Note that L f (P ) = the sum of the volumes of all the Π ijk that are contained in U f (P ) = the sum of the volumes of all the Π ijk that intersect. It follows that L f (P ) the volume of U f (P ). he arbitrariness of P gives the formula. he riple Integral over a More General Solid March 7,

13 RIPLE INEGRALS 3 Some Properties of the riple Integral Below we give without proof the salient elementary properties of the triple integral. hey are all analogous to what you saw in the one- and two-variable cases. he referred to is a basic solid. he functions f and g are assumed to be continuous on. I. he triple integral is linear: = α II. It preserves order: [αf(x, y, z) + βg(x, y, z)] dxdydz f(x, y, z) dxdydz + β g(x, y, z) dxdydz. if f on, then f(x, y, z) dxdydz ; if f g on, then f(x, y, z) dxdydz g(x, y, z) dxdydz. III. It is additive: if is broken up into a finite number of basic solids,, n, then f(x, y, z) dxdydz = f(x, y, z) dxdydz + + n f(x, y, z) dxdydz IV. It satisfies a mean-value condition: namely, there is a point (x, y, z ) in for which f(x, y, z) dxdydz = f(x, y, z )(the volume of ). We call f(x, y, z ) the average value of f on. he notion of average given above enables us to write f(x, y, z) dxdydz = (the average value of f on )(the volume of ). We can also take weighted averages: if f and g are continuous and g is nonnegative on, then there is a point (x, y, z ) in for which f(x, y, z)g(x, y, z) dxdydz = f(x, y, z ) g(x, y, z) dxdydz. We call f(x, y, z ) the g-weighted average of f on. he formulas for mass, center of mass, and moments of inertia derived in the previous section for two-dimensional plates are easily extended to three-dimensional objects. Suppose that is an object in the form of a basic solid. If has constant mass density λ (here density is mass per unit volume), then the mass of is the density λ times the volume of : M = λv. Some Properties of the riple Integral March 7,

14 RIPLE INEGRALS 4 If the mass density varies continuously over, say λ = λ(x, y, z), then the mass of is the average density of times the volume of : M = λ(x, y, z) dxdydz. he coordinates of the center of mass (x M, y M, z M ) are density-weighted averages: x M M = xλ(x, y, z) dxdydz, y M M = z M M = yλ(x, y, z) dxdydz, zλ(x, y, z) dxdydz. If the object is homogeneous (constant mass density M/V ), then the center of mass of depends only on the geometry of and falls on the centroid ( x, ȳ, z) of the space occupied by. he density is irrelevant. he coordinates of the centroid are simple averages over : xv = x dxdydz, ȳv = y dxdydz, zv = z dxdydz. he moment of inertia of about a line is given by the formula I = λ(x, y, z)[r(x, y, z)] dxdydz. Here λ(x, y, z) is the mass density of at (x, y, z) and r(x, y, z) is the distance of (x, y, z) from the line in question. he moments of inertia about the x, y, z axes are again denoted by I x, I y, I z. Some Properties of the riple Integral March 7,