Geometric and Mechanical Applications of Integrals

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1 5 Geometric nd Mechnicl Applictions of Integrls 5.1 Computing Are Using Crtesin Coordintes Suppose curve is given by n eqution y = f(x), x b, where f : [, b] R is continuous function such tht f(x) for ll x [, b]. Then, the re under the curve, i.e., the re of the region bounded by the grph of f, the x-xis, nd the ordintes t x = nd x = b, is µ(p ) k f(ξ i ) x i = Suppose the curve is given in prmetric form: y. x = ϕ(t), y = ψ(t),, α t β, such tht = ϕ(α), b = ψ(β). Then the re under the curve tkes the form β α ψ(t)ϕ (t)dt. If f tkes both positive nd negtive vlues, but chnges sign only t finite number of points, then the re bounded by the curve, the x-xis, nd the ordintes t x = nd x = b, is given by f(x). Suppose f : [, b] R nd g : [, b] R re continuous functions such tht f(x) g(x) for ll x [, b]. Then the re of the region bounded by the grphs of f nd g, nd the ordintes t x = nd x = b is given by µ(p ) k [g(ξ i ) f(ξ i )] x i = 146 [g(x) f(x)].

2 Computing Are Using Polr Coordintes Suppose curve is given in polr coordintes s ρ = ϕ(θ), α θ β, where ϕ : [α, β] R is continuous function. Then the re of the region bounded by the grph of ϕ nd the rys θ = α nd θ = β is is given by Thus, µ(p ) k 1 [ϕ(ξ i) θ i ]ϕ(ξ i ) = Are := 1 k µ(p ) j=1 β α 1 [ϕ(ξ i)] θ i = 1 β ρ dθ. α ρ dθ. (5.1.1) Exmple 5.1 We find the re bounded by the cures defined by y = x, y = x, x : Note tht the points of intersection of the curves re t x = nd x = 1. Also, x x for x 1. Hence, the required re is 1 ( x x ) [ ] 1 x 3/ = 3/ x3 = Exmple 5. We find the re bounded by the ellipse the prmetriztion Then the required re is 4 y = 4 x x = cos t, y = b sin t, t π. π/ (b sin t)( sint) dt = b π/ + y = 1. Let us use b (1 cos t) dt = πb. Next, let us consider the polr form of the ellipse. For this, consider the polr form of points (x, y) on the ellipse, i.e., x = ρ cos θ, y = ρ sin θ, where (ρ, θ) stisfies x + y = 1, i.e. b ρ cos θ + ρ sin θ b = 1, i.e., ρ = ( cos θ ) 1 + sin θ b. Thus, the re cn lso be computed (Exercise) using the formul (5.1.1).

3 148 Geometric nd Mechnicl Applictions of Integrls M.T. Nir Exmple 5.3 We find the re bounded by one rch of the cycloid x = (t sin t), y = (1 cos t). One rch of the cycloid is obtined by vrying t over the intervl [, π]. Thus, the required re is π y = π y(t)x (t) dt = π (1 cos t) dt = 3π. Exmple 5.4 We find the re bounded by circle of rdius. Without loss of generlity ssume tht the centre of the circle is the origin. Then, the circle cn be represented in polr coordintes s ρ =. Hence the required re is 1 π ρ dθ = π. Exmple 5.5 We find the re bounded by the lemniscte The required re is ρ = cos θ. [ ] 1 π/4 π/4 ρ dθ = cos θdθ =. π/4 π/4 5. Computing Arc Length Suppose curve C in the plin R is given prmetriclly by C : γ(t) := (x(t), y(t)), t b, where x(t) nd y(t) re continuous functions of t [, b]. Note tht, s t moves from to b, the point γ(t) := (x(t), y(t)) moves long the curve C from the point γ() = (x(), y()) to γ(b) = (x(b), y(b)). In order to compute the length of C, we first consider polygonl pproximtion of it. Wht we men by tht is the following:

4 Computing Arc Length 149 Corresponding to prtition P : = t < t 1 < < t k = b of [, b], consider the length of the polygonl line obtined by joining the points γ(t ), γ(t 1 ),..., γ(t k ), i.e., the quntity l P (C) := k (x(ti ) x(t i 1 )) + (y(t i ) y(t i 1 )). Using the bove quntity, we define the length of C s follows: Definition 5.1 the length of the curve C is defined s l(c) := sup P l P (C), where the supremum is tken over ll prtitions P of [, b]. In order to compute the quntity l(c), we ssume tht the curve is smooth in the sense tht x(t) nd y(t) re differentible in (, b) nd their derivtives x (t) nd y (t) re continuous. Under this ssumption (by Men Vlue Theorem), there exist ξ i, η i (t i 1, t i ) such tht x(t i ) x(t i 1 ) = x (ξ i )(t i t i 1 ), y(t i ) y(t i 1 ) = y (η i )(t i t i 1 ) for i = 1,,..., k so tht the quntity l P (C) tkes the form l P (C) = k x (ξ i ) + y (η i ) (t i t i 1 ). By our ssumptions on the functions x(t) nd y(t), the function f(t) := x (t) + y (t), t b, is integrble over [, b]. Hence, if we tke sequence (P n ) of prtitions on [, b], sy P n : = t,n < t,n < < t,kn = b such tht µ(p n ) s n, then S(P n, f, T n ) := k n x (c i,n ) + y (c i,n ) (t i,n t i 1,n ) x (t) + y (t) dt s n, where T n := {c i,n } n is ny set of tgs on P n. Corresponding to the prtition P n l Pn (C) tkes the form l Pn (C) = k n x (ξ i,n ) + y (η i,n ) (t i,n t i 1,n ).

5 15 Geometric nd Mechnicl Applictions of Integrls M.T. Nir By the continuity of the functions x (t) nd y (t), we obtin S(P n, f, T n ) l Pn (C) s n. Now, using the property P Q = l P (C) l C (Q), we cn conclude tht so tht l(c) = sup l P (C) = l P P n n (C), l(c) = x (t) + y (t) dt Using crtesin coordintes If the curve C is given by n eqution y = f(x), x b, where f is continuous function on [, b], then we my write Then we hve l(c) = C : γ(t) := (t, f(t)), t b. ( 1 + f (t), i.e., l(c) = 1 + ( dy ). Remrk 5.1 Curves in prmetric form re ssumed to be piecewise smooth, i.e., hving unique tngents except possibly t finite number of points. Note tht if curve is given in prmetric form s x = φ(t), y = ψ(t) with α t b, then it hs unique tngent t (x, y ) if φ (t ), ψ (t ) exists, where t is such tht (x = φ(t ), y = ψ(t ), nd φ (t ) + ψ (t ) ). 5.. Using Polr Coordintes Suppose curve is given in polr coordintes s ρ = ϕ(θ), α θ β, where ϕ : [α, β] R is continuous function. Since x = ρ cos θ, y = ρ sin θ, α θ β, we hve A = β α ( ) + dθ ( ) dy dθ. dθ

6 Computing Arc Length 151 Note tht Hence, it follows tht dθ = ρ cos θ + ρ( sin θ), dy dθ = ρ sin θ + ρ cos θ. A = = β α β α ( ) + dθ ρ + ( ) dy dθ dθ ( ) dρ dθ. dθ 5..3 Illustrtive exmples Exmple 5.6 We find the circumference of circle of rdius. Without loss of generlity ssume tht the centre of the circle is the origin,i.e., the circle is given by x + y =. The required length is L := ( ) dy, y = x. Thus, L := 4 x = π. Exmple 5.7 Now we find the length of the circle when it is represented by the equtions The required length is x = cos θ, y = sin θ, θ π. L := 4 = 4 π/ π/ ( ) ( ) dy + dθ dθ dθ sin θ + cos θ dθ = π. Exmple 5.8 Let us find the length of the ellipse x = cos θ, y = b sin θ, θ π.

7 15 Geometric nd Mechnicl Applictions of Integrls M.T. Nir The required length is b L := 4 = 4 = 4 = 4 = 4 π/ π/ π/ π/ π/ ( ) ( ) dy + dθ dθ dθ sin θ + b cos θ dθ (1 cos θ) + b cos θ dθ ( b ) cos θ dθ 1 β cos θ dθ, where β =. The bove integrl is not expressible in stndrd form unless β = 1, i.e., unless b = in which cse the ellipse is the circle. But, the integrl cn be pproximtely computed numericlly. Exmple 5.9 We find the length of the stroid: x = cos 3 t, y = sin 3 t. The required length is L := 4 = 4 = 1 π/ π/ π/ ( ) ( ) dy + dt dt dt 9 cos 4 t sin t + 9 sin 4 t cos 3 t dt cos t sin dt = 6. Exmple 5.1 We find the length of the crdioid ρ = (1 + cos θ). The required length is L := π ρ + ρ dθ. Since ρ = (1 + cos θ), ρ = sin θ, we hve L = π π 1 + cos θ dθ = 4 cos θ dθ = Computing Volume of Solid Suppose tht three dimensionl object, solid, lies between two prllel plnes x = nd x = b. Let α(x) be the re of the cross section of the solid t the point x,

8 Computing Volume of Solid of Revolution 153 with cross section being prllel to the yz-plne. We ssume tht the function α(x), x [, b] is continuous. Now, consider prtition P : = x < x 1 <... < x k = b of the intervl [, b]. Then the volume of the solid is given by µ(p ) k α(ξ i ) x i = α(x). Exmple 5.11 Let us compute the volume of the solid enclosed by the ellipsoid x + y b + z c = 1. For fixed x [, ], the boundry of the cross section t x is given by the eqution i.e., y b + z c = 1 x, y φ(x) + z = 1, ψ(x) where φ(x) = b 1 x, ψ(x) = c 1 x. Hence, nd the required volume is V := ) α(x) = πφ(x)ψ(x) = πbc (1 x, α(x) = πbc ) (1 x = 4 3 πbc. In prticulr, volume of the solid bounded by the sphere x + y + z = is 4 3 π Computing Volume of Solid of Revolution Suppose solid is obtined by revolving curve y = f(x), x b, with x-xis s xis of revolution. We would like to find the volume of the solid. In this cse the re of cross section t x is given by α(x) = πy = π[f(x)], x b. Hence, the volume of the solid of revolution is V := α(x) = π y.

9 154 Geometric nd Mechnicl Applictions of Integrls M.T. Nir Exmple 5.1 Let us compute the volume of the solid of revolution of the curve y = x bout x-xis for x. The required volume is V := π y = π x 4 = 5 5. Exmple 5.13 We compute the volume of the solid of revolution of the ctenry y = (e x/ + e x/) bout x-xis for x b. The required volume is V := π y = π (e x/ + e x/) π = 4 4 We see tht V = π3 8 ( ) e x/ + e x/ +. ( e b/ e b/) + π b. 5.5 Computing Are of Surfce of Revolution Suppose solid is obtined by revolving curve y = f(x), x b, with x-xis s xis of revolution. We would like to find the re of the surfce of the solid. The required re is A := µ(p ) j=1 k πf(ξ i ) s i, where P : = x < x 1 <... < x k = b is prtition of the intervl [, b], nd Thus A = µ(p ) s i := 1 + [f (ξ i )] x i, i = 1,..., k. k πf(ξ i ) 1 + [f (ξ i )] x i = π y 1 + ( ) dy. Exmple 5.14 We find the surfce of revolution of the prbol y = px, x for p >. The required re is ( ) dy A = π y 1 + = π px 1 + p x = π p p + x = π p [ (x + p) 3/ 1 ] = π p [ ( + p) 3/ p 3/]. 3 3

10 Centre of Grvity Centre of Grvity Suppose A 1, A,..., A n re mteril prticles on the plne t coordintes (x 1, y 1 ), (x, y ),..., (x n, y n ) nd msses m 1, m,... m n respectively. Then the centre of grvity of the system of these prticles is t the point A = (x C, y C ), where x C := n x n im i n m, y C := y im i n i m. i Now we ttempt to define the centre of grvity of mteril line nd mteril plnr region enclosed by certin curves Centre of grvity of mteril line in the plne Suppose curve L is given by the eqution y = f(x), x b. We ssume tht this curve is mteril line. Suppose the density of the mteril t the point X = (x, y) is γ(x). This density is defined s follows: Suppose M(X, r) is the mss of n rc of the line contining the point X with length r. Then the density of the mteril t the point x is defined by M(X, r) γ(x) :=. r r Now, in order to find the centre of grvity of L, we first consider prtition P : = x < x 1 <... < x k, nd tke points ξ i = [x i 1, x i ], i = 1,..., n. Then we tke the the centre of grvity of the system of mteril points (ξ 1, f(ξ 1 )), (ξ, f(ξ )),..., (ξ k, f(ξ k )) s x C (P ) = n ξ iγ i s n i n γ, y C (P ) := f(ξ i)γ i s i n i s i γ. i s i Here, s i is the length of the rcs joining (x i 1, y i 1 ) to (x i, y i ), nd γ i is the density t the point (ξ i, f(ξ i ). Here y i = f(x i ). Note tht γ i s i is the pproximte mss of the rc joining (x i 1, y i 1 ) to (x i, y i ). Now, the centre of grvity of L is t (x C, y C ), where n x C = ξ iγ i s n i n µ(p ) γ, y C := f(ξ i)γ i s i n i s i µ(p ) γ. i s i

11 156 Geometric nd Mechnicl Applictions of Integrls M.T. Nir Assuming tht the function γ(x) := γ(x, f(x)) is continuous on [, b], we see tht ( b xγ(x, y) 1 + dy ( b ) yγ(x, y) 1 + dy ) x C = ), y C = b γ(x, y) ). b 1 + γ(x, y) 1 + ( dy ( dy Exmple. We find the centre of grvity of the semi-circlulr rc x + y =, y, ssuming tht the density of the mteril is constnt. In this cse, so tht it follows tht Hence, since γ(x, y) is constnt, y = f(x) := x, 1 + ( ) dy = y 1 + x C = y C = x. 1 + ( dy ) ( dy ) = π Centre of grvity of mteril plnr region Next we consider the centre of grvity of mteril plnr region Ω bounded by two curves y = f(x), y = g(x), with f(x) g(x) x b. Suppose tht the density of the mteril t the point X is γ(x). This density is defined s follows: Suppose M(X, r) is the mss of the circulr region S(X, r) Ω with centre t x nd rdius r >, nd α(x, r) is the re of the sme circulr region. Then the density of the mteril t the point x is defined by γ(x) := r M(X, r) α(x, r). Now, in order to find the centre of grvity of Ω, we first look t the following specil cse: Suppose Ω is rectngle given by 1 x b 1, y b. Then we cn infer tht the centre of grvity of such rectngle is locted t the point ( 1 + b 1, + b ). Tking the bove obervtion into ccount, we consider prtition P = {x i } n i= of the intervl [, b], nd consider the rectngulr strips: R i : x i 1 x x i, f(ξ i ) y g(ξ i ), i = 1,..., n,

12 Centre of Grvity 157 where ξ i = x i 1+x i, i = 1,..., n. If γ is the (constnt) density of the mteril, then the mss of the rectngulr strip R i is m i = γ[g(ξ i ) f(ξ i )] x i, i = 1,..., n. Assuming tht the mss of the rectngulr strip R i is concentrted t its mid-point: ( X i : ξ i, f(ξ ) i) + g(ξ i ), we consider the centre of grvity of the system of mteril points t X i s x C,P := n ξ n im i n m, y C,P := n i m i f(ξ i )+g(ξ i ) m i. Now the centre of grvity of Ω is defined s i.e., x C = x C,P, y C = y C,P, µ(p ) µ(p ) x C = = µ(p ) n ξ iγ[g(ξ i ) f(ξ i )] x i n γ[g(ξ i) f(ξ i )] x i x[g(x) f(x)] [g(x) f(x)] y C = = 1 µ(p ) n 1 [f(ξ i) + g(ξ i )]γ[g(ξ i ) f(ξ i )] x i n γ[g(ξ i) f(ξ i )] x i [f(x) + g(x)][g(x) f(x)] [g(x) f(x)] Exmple 5.15 We find the coordintes of the centre of grvity of segment of prbol y = x cut off by the stright line x =. In this cse f(x) = x, g(x) = x, x. Hence the coordintes of the centre of grvity re x C = y C = 1 x[g(x) f(x)] = x x [g(x) f(x)] x = 3 5. [f(x + g(x)][g(x) f(x)] [g(x) f(x)] =.

13 158 Geometric nd Mechnicl Applictions of Integrls M.T. Nir 5.7 Moment of Inerti Suppose there re n mteril points in the plne. Let their msses be m 1, m,... m n respectively. Suppose tht these points re t distnces d 1,..., d n from fixed point O. Then the moment of inerti of the system of these points with respect to the point O is defined by the quntity: I O := d i m i. If O is the origin, nd (x 1, y 1 ), (x, y ),..., (x n, y n ) re the points, then I O := (x i + yi )m i Moment of inerti of mteril line in the plne Suppose curve L is given by the eqution y = f(x), x b. We ssume tht this curve is mteril line. Suppose the density of the mteril t the point X = (x, y) is γ(x). Now, in order to find the moment of inerti of L, we first consider prtition P : = x < x 1 <... < x k, nd tke points ξ i = [x i 1, x i ], i = 1,..., n. Then we consider the moment of inerti of the system of mteril points t (ξ 1, η i ), i = 1,..., n. Here, η i = f(ξ i ), i = 1,..., n. Thus, I O,P := I O,P := (ξi + ηi )m i. (ξi + ηi )γ i s i. Here, s i is the length of the rcs joining (x i 1, y i 1 ) to (x i, y i ), nd γ i is the density t the point (ξ i, η i ). Note tht γ i s i is the pproximte mss of the rc joining (x i 1, f(x i 1 )) to (x i, f(x i )). Now, ssuming tht the functions f(x) nd γ(x) := γ(x, f(x)) re continuous on [, b], the moment of inertil of L with respect to O is I O = µ(p ) I O,P = µ(p ) = (ξi + ηi )γ i s i (x + y )γ(x, y) 1 + ( ) dy.

14 Moment of Inerti Moment of inerti of circulr rc with respect to the centre Suppose the given curve is circulr rc: ρ =, α θ β. Following the rguments in the bove prgrph, we compute the moment of inerti using polr coordintes: The moment of inerti, in this, cse is given by I O := d i m i, µ(p ) where d i =, m i = γ i θ i, for i = 1,..., n, so tht β I O = γ i [ θ i ] = 3 γ(θ)dθ. µ(p ) Here, γ(θ) is the point density. If γ(θ) = γ, constnt, then β I O = 3 γ(θ)dθ = (β α)γ 3. In prticulr, M.I of the circle ρ =, θ π, is α I O = πγ Moment of inerti of mteril sector in the plne The region is R : ρ, α θ β with constnt density γ. To find the M.I. of R, we prtition it by rys nd circulr rcs: P : α = θ < θ 1 < θ <... < θ n = β, Q : = ρ < ρ 1 < ρ <... < ρ m =. Consider the elementry region obtined by the bove prtition: R ij : ρ j 1 ρ ρ j, θ i 1 θ θ i. Assume tht the mss of this region R ij is concentrted t the point (ˆρ j, ˆθ i ), where ˆρ j [ρ j 1, ρ j ], ˆθ i [θ i 1, θ i ]. Then the MI of the mteril point t (ˆρ j, ˆθ i ) is m ij d ij where m ij is the mss of the region R ij which is pproximtely equl to [ˆρ j θ i ρ j ]γ, nd d ij = ˆρ j. Thus the MI of the sub-sector θ i 1 θ θ i is defined by µ(q) j=1 m ij d ij = µ(q) j=1 = µ(q) j=1 α [ˆρ j θ i ρ j ]γ ˆρ j ( ) γ ˆρ 3 j ρ j θi ( ) = γ ρ 3 dρ θ i = γ4 4 θ i.

15 16 Geometric nd Mechnicl Applictions of Integrls M.T. Nir From this, it follows tht, the moment of inerti of the sector α θ β is µ(p ) m γ 4 4 θ i = In prticulr, moment of inerti of circulr disc is πγ 4 where M = π γ is the mss of the disc. = M, (β α)γ4. 4 Exercise 5.1 If M is the mss of right circulr homogeneous cylinder with bse rdius, then show tht its moment of inerti is M. 5.8 Additionl Exercises 1. Find the re of the portion of the circle x + y = 1 which lies inside the prbol y = 1 x. [Hind: Are enclosed by the circle in the second nd third qudrnt nd the re enclosed by the prbol in the first nd fourth qudrnt. The required re is π x. Ans: π ]. Find the re common to the crdioid ρ = (1 + cos θ) nd the circle ρ = 3. [Hind: The points of intersections of the given curves re given by 1+cos θ = 3, i.e., for θ = ± π 3. Hence the required re is [ 1 π/3 ( ) 3 dθ + 1 ] π (1 + cos θ) dθ. π/3 Ans: 7 4 π ] 3. For, b >, find the re included between the prbols y = 4(x + ) nd y = 4b(b x). [Hind: Points of intersection of the curves is given by (x + ) = b(b x), i.e., x = b +b = b ; y = b. The required re is [ ] 4(x + ) + 4b(b x). Ans: 8 3 b( + b).] b 4. Find the re of the loop of the curve r cos θ = sin 3θ [Hint:r = for θ = nd θ = π/3, nd r is mximum for θ = π/6. The re is π/3 r dθ. ]

16 Additionl Exercises Find the re of the region bounded by the curves x y 3 = nd x y =. [Hint: Points of intersections of the curves re t x =, 1, 1. The re is 1 (x1/3 x). Ans: 1/ ] 6. Find the re of the region tht lies inside the circle r = cos θ nd outside the crdioid r = (1 cos θ). [Hint: Note tht the circle is the one with centre t (, /) nd rdius /. The curves intersect t θ = ±π/3. The required re is π/3 π/3 (r 1 r )dθ, where r 1 = cos θ, r = (1 cos θ). Ans: 3 3 (3 3 π) ] 7. Find the re of the loop of the curve x = (1 t ), y = t(1 t ) for 1 t 1. [Hint: y = for t { 1,, 1}, nd y negtive for 1 t nd positive for t 1. Also, y = x ( x)/ so tht the curve is symmetric w.r.t. the x-xis. Are is y = 1 y(t)x (t)dt. Ans: 8 /15 ] 8. Find the length of n rch of the cycloid x = (t sin t), y = (1 cos t). [Hint: The curve cuts the x-xis t x = nd x = π for t = nd t = π respectively. Thus the length is π [x (t)] + [y (t)] dt. Ans: 8. ] 9. For >, find the length of the loop of the curve 3 y = x(x ). [Hint: The curve cuts the x-xis t x =, nd the curve is symmetric w.r.t. the x-xis. Thus the required re is ( 1 + dy ). Note tht 6yy = (x )(3x ), so tht 1 + y = (3x+) 1x. Ans: 4 3. ] 1. Find the length of the curve r = 1+cos θ, θ π/. [Hind: l := π/ r + [r ] dθ = π/4 sec 3 θdθ. Ans: + ln( + 1). ] 11. Find the volume of the solid obtined by revolving the curve y = 4 sin x, x π/, bout y-xis. [Hint: writing y = 4 sin x, x π/4 nd y = 4 sin u, π/4 u π/, the required volume is 4 (u x )dy = π π/ π/4 u (8 cos u)du π π/4 x (8 cos x). Also, note tht the curve is symmetric w.r.t. the line x = π/4. Hence, the required volume is given by π π/4 [( π 4 x) x ]dy. Ans: π.] 1. Find the re of the surfce obtined by revolving loop of the curve 9x = y(3 y) bout y-xis. [Hind: x = iff y = or y = 3. The required re is π ( ) 3 x 1 + dy. Ans: 3π. ]

17 16 Geometric nd Mechnicl Applictions of Integrls M.T. Nir 13. Find the re of the surfce obtined by revolving bout x-xis, n rc of the ctenry y = c cosh(x/c) between x = nd x = for >. [Hind: The re is π y 1 + y = π c cosh x c. Ans: πc [ + c sinh c ]. ] 14. The lemniscte ρ = cos θ revolves bout the line θ = π 4. Find the re of the surfce of the solid generted. [Hind: The required surfce is π π/4 π/4 h ρ + ρ dθ, where h := ρ sin ( π 4 θ), ρ = cos 3θ so tht ρ + ρ = cos θ. Ans: 4π. ] 15. Find the volume of the solid generted by the crdioid ρ = (1 + cos θ) bout the initil line. [Ans: 8 3. ]

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 6 (First moments of an arc) A.J.Hobson

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 6 (First moments of an arc) A.J.Hobson JUST THE MATHS UNIT NUMBER 13.6 INTEGRATION APPLICATIONS 6 (First moments of n rc) by A.J.Hobson 13.6.1 Introduction 13.6. First moment of n rc bout the y-xis 13.6.3 First moment of n rc bout the x-xis