MATH2111 Higher Several Variable Calculus Integration

MATH2 Higher Several Variable Calculus Integration Dr. Jonathan Kress School of Mathematics and Statistics University of New South Wales Semester, 26 [updated: April 3, 26] JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 / 58 iemann Integral (one variable) A set of points P {x, x, x 2,..., x n }, where a x < x < x 2 < < x n b, is called a partition of [a, b]. For a function f : [a, b], the upper and lower iemann sums of f with respect to P are S P (f ) n f k x k and S P (f ) k n f k x k k where f k and f k are the inmum and supremum of f on [x k, x k ] and x k x k x k. For a bounded function f : [a, b], if there exists a unique number I such that S P (f ) I S P (f ) for every partition P of [a, b], then f is iemann integrable on [a, b] and I [a,b] f b a f (x)dx. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 2 / 58

Ê Ê Integration First consider f :, where [a, b] [c, d] is a rectangle in 2. Let P {a x, x, x 2,..., x n b} be a partition of [a, b] and P 2 {c y, y, y 2,..., y m d} ¾ Ê ¾ a partition of [c, d]. ¼ Then is the union of the mn subrectangles ¼ ¾ jk [x j, x j ] [y k, y k ]. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 3 / 58 Integration The upper and lower iemann sums of f with respect to these partitions are S P,P 2 (f ) j,k f jk x j y k and S P,P 2 (f ) j,k f jk x j y k. ¾ Ê ¾ ¼ The sums are over all pairs (j, k) with j n and k m. ¼ ¾ The numbers f jk and f jk are the inmum and supremum of f on jk and x j and y k its width and height. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 4 / 58

Integration Denition (iemann integral) For a bounded function f :, if there exists a unique number I such that S P,P 2 (f ) I S P,P 2 (f ) for every pair of partitions P, P 2 of, then f is iemann integrable on and I f f (x, y) da. I is called the iemann integral of f over. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 5 / 58 iemann Integral Properties: If f and g are integrable on, () Linearity: αf + βg α f + β g, α, β. (2) Positivity (monotonicity): If f (x) g(x) for all x then f g. (3) f f. (4) If 2 and (interior ) (interior 2 ) then f f + f. 2 JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 6 / 58

Lower sum Þ Þ (, ) Ê A lower sum of f over. S P,P 2 (f ) j,k f jk x j y k. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 7 / 58 Upper sum Þ Þ (, ) Ê An upper sum of f over. S P,P 2 (f ) j,k f jk x j y k. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 8 / 58

iemann integral interpretation For a function of one variable, the iemann integral is interpreted as the (signed) area bounded by the graph y f (x) and the x-axis over the interval [a, b]. For a function of two variables, f is the (signed) volume bounded by the graph z f (x, y) and the xy-plane over the rectangle. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 9 / 58 Fubini's theorem on rectangles Theorem (Fubini's Theorem (version )) Let f : be continuous on a rectangular domain [a, b] [c, d]. Then b d a c f (x, y)dydx d b c a f (x, y)dxdy f. Note that means b a b d a c f (x, y)dydx ( ) d f (x, y)dy dx. c The integral inside the brackets is, for each x, a one variable integral. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 / 58

Fubini's theorem on rectangles Example: Calculate the integral of f : 2, where f (x, y) xy 3, over the rectanglur region [, 3] [, 2]. Fubini's theorem for rectangular regions says gives two ways of calculating this integral. Let's integrate rst with respect to y. f 3 2 3 5 4 35 8. 5 4 x dx [ ] 3 2 x 2 xy 3 dy dx 3 [ 4 xy 4 ] 2 dx ¾ JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 / 58 Fubini's theorem on rectangles We can also integrate rst with respect to x. f 9 2 2 3 2 9 2 y 3 dy [ 4 y 4 35 8. xy 3 dx dy ] 2 2 [ 2 x 2 y 3 ] 3 dy ¾ In this case we also have 2 3 f xy 3 dxdy 2 3 ( 2 ) ( 3 ) y 3 xdx dy y 3 dy x dx. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 2 / 58

Fubini's theorem on rectangles Fubini's theorem is essentially the same as the method of slicing for calculating volumes from rst year. However, proving the iterated integrals give the same number for the volume as the denition involves some subtlety. The lower iemann sum can be written as a double sum. S P,P 2 (f ) m n f j,k x j y k f j,k x j y k j,k and it is tempting to consider the sum in the brackets as a one variable iemann lower sum of f (x, y) for a xed value of y. However, f j,k is not necessarily an inmum for f (x, y) as a function of x for any particular xed value of y. One way around this problem is to use the continuity of f to show that n j f j,k x j can be made to be within ε of the one variable lower sum of f (x, y k ) by requiring the spacing in the partition P 2 to be suciently small. A more eligant approach can be found on pages 657 of the Internet Supplement to Marsden and Tromba 5th Edition. k j JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 3 / 58 iemann integral over more general regions Theorem (Integrability of bounded functions) Let f : be a bounded function on the rectangle and suppose that the set of points where f is discontinuous lies on a nite union of graphs of continuous functions. Then f is integrable over. The proof of this theorem is exercises 4, 5 and 6 from the Internet Supplement to Marsden and Tromba. The essence of the proof is that the contribution to the upper and lower iemann sums from subrectangles containing the the lines of discontinuity can be made arbitrarily small by taking the width of subintervals in the partitions to be small enough. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 4 / 58

iemann integral over more general regions Theorem (Fubini's Theorem (version 2)) Let f : be bounded on a rectangular domain [a, b] [c, d] with the discontinuities of f conned to a nite union of graphs of continuous functions. If d the integral f (x, y)dy exists for each x [a, b], then c f b a c ( ) d f (x, y)dy dx. Similarly, if the integral b a f (x, y)dx exists for each y [c, d], then f d c a ( ) b f (x, y)dx dy. Since f is not continuous there is no guarantee that d c or that b f (x, y)dx exists for each y. f (x, y)dy exists for each x a JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 5 / 58 iemann integral over more general regions An elementary region is a region of the type illustrated below. A y-simple region. An x-simple region. φ ¾ () φ () ψ () ¾ ψ ¾ () D {(x, y) : x [a, b] and φ (x) y φ 2 (x)} where φ and φ 2 are continuous functions from [a, b] to. D 2 {(x, y) : y [c, d] and ψ (y) x ψ 2 (y)} where ψ and ψ 2 are continuous functions from [c, d] to. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 6 / 58

iemann integral over more general regions Suppose D is an elementary region, a rectangle containing D and f a function from D to. First extend f to a function dened on all of by { f f (x, y) if (x, y) D (x, y) if (x, y) D and (x, y). Then f is integrable on D if f is integrable on and f f. D If f is continuous except perhaps on a set of points made from the graphs of continuous functions, then f also has this property and so the second version of Fubini's theorem gives us a way to calculate the integral of f over D as an iterated integral. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 7 / 58 Iterated integrals for elementary regions Suppose D is a y-simple region contained in [a, b] [c, d] and bounded by x a, x b, y φ (x) and y φ 2 (x), and f : D is such that f satises the conditions of Fubini's Theorem (version 2). Then b d f f f (x, y)dydx, D but since f (x, y) for y < φ (x) or y > φ 2 (x), a c d c f (x, y)dy φ2 (x) φ (x) f (x, y)dy φ2 (x) φ (x) f (x, y)dy. and hence f D a b φ2 (x) φ (x) f (x, y)dydx. A similar result hold for integrals of x-simple regions like D 2. d ψ2 (y) f f (x, y)dxdy. D2 c ψ (y) JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 8 / 58

Notes The properties of the integral over rectangular regions also apply to more general regions. The area of a region D of 2 is given by. (That is, the integral of the D function f (x, y) over D.) Everything that we have done for integrals in 2 extends to n. The volume of a region D of 3 is given by. (That is, the integral of the D function f (x, y, z) over D.) JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 9 / 58 Fubini's theorem (a) Find the integral of f (x, y) x 2 y over the triangular region with vertices (, ), (, ) and (, ). (b) For each of the following two integrals, nd the region over which the integral is calculated and reverse the order of integration. (i) I 2 2x f (x, y)dydx. (ii) I 2 x 2 f (x, y)dydx. x JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 2 / 58

Fubini's theorem examples Example (a): Find the integral of f (x, y) x 2 y over the triangular region with vertices (, ), (, ) and (, ). Fubini's theorem gives two ways of calculating this integral. Let's integrate with respect to y rst. f x x 2 y dy dx 2 x 2 ( x) 2 dx x 2 2x 3 + x 4 dx 2 [ 3 x 3 2 x 4 + ] 5 x 5 6. [ 2 x 2 y 2 ] x dx ¼ JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 2 / 58 Fubini's theorem examples Now check this by integrating with respect to x rst. y f x 2 y dx dy 3 3 6. [ 3 x 3 y ] y dy 3 ( y)3 y dy y 3y 2 + 3y 3 y 4 dy [ 2 y 2 y 3 + 3 4 y 4 5 y 5 ] ¼ JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 22 / 58

Fubini's theorem examples Example (b) (i): Find the region over which the integral I 2 2x f (x, y) dydx is calculated and reverse the order of integration. The integration is rst with respect to y. I 2 2x f (x, y) dy dx ¾ For each value of x, the lower limit of integration is y and the upper limit of integration is y 2 2x. The resulting function is integrated for x from to. ¾ ¾ The region of integration is the triangle with vertices (, ), (, 2) and (, ). ¼ JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 23 / 58 Fubini's theorem examples To reverse the order of integration, take slices of constant y and integrate rst with respect to x. ¾ For each value of y, the lower limit of integration is x and the upper limit of integration is x 2 y. The resulting function is integrated with respect to y from to 2. ¾ ¾ ¾ ¼ I 2 2 y f (x, y) dx dy. So, I 2 2x f (x, y) dydx 2 2 y f (x, y) dxdy. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 24 / 58

Fubini's theorem examples Example (b) (ii): Find the region over which the integral I 2 x 2 x is calculated and reverse the order of integration. The integration is rst with respect to y. I 2 x 2 x f (x, y) dy dx f (x, y) dydx ¾ ¾ For each value of x, the lower limit of integration is y x and the upper limit of integration is y x 2. The resulting function is integrated for x from to. The region of integration 2 is the triangle with vertices (, ), (, ) and (, 2) with a semi-circular cap on top. ¾ JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 25 / 58 Fubini's theorem examples To reverse the order of integration and integrate with respect to x rst, it is convenient to split the region into two pieces. I 2A I 2B y 2 2 y 2 f (x, y) dx dy. y+ f (x, y) dx dy. ¾ ¾ ¾ ¾ ¾ + So, I 2 x 2 x f (x, y) dydx ¾ y 2 y 2 f (x, y) dxdy + y+ 2 f (x, y) dxdy. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 26 / 58

Uniform continuity Denition A function f : n m is uniformly continuous on if for all x, y and for all positive ɛ there exists δ such that ( ) d(x, y) < δ d f(x), f(y) < ɛ. In the denition of continuity, δ may depend on x, but here, given ɛ, the same δ must work for all x. Theorem A continuous function on a compact set is uniformly continuous on. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 27 / 58 Leibniz' rule Theorem (Leibniz' rule for dierentiation under the integral sign) Consider function f : 2 that is continuous on and f continuous on. If x is uniformly for some ɛ > and if {(x, y) : x x < ɛ, a y b}, F (x) b a f (x, y)dy, x x ɛ then F (x ) d dx b a b a f (x, y)dy f x (x, y)dy. xx JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 28 / 58

JKress-proof-of-Leibniz-rule.pdf JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 29 / 58 Leibniz' rule examples Use the fact that for a, b >, dx 2 ( ) a + b b ax + b a to nd (a) (b) xdx (ax + b) 3/2 dx (ax + b) 3/2 JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 3 / 58

Leibniz' rule examples (a) [c, d] [, ] for < c < d is compact and f : with f (a, x) ax + b and f x (a, x) a 2(ax + b) 3/2 for b > are bounded and continuous and hence are both uniformly continuous on. ( ) dx ( 2 ( ) ) a + b b a ax + b a a ( ) dx 2 ( ) a + b b a ax + b a 2 + a a + b ( ) x 2 (ax + b) 3/2 dx 2 ( ) a + b b a 2 + a a + b x (ax + b) 3/2 dx 4 ( ) 2 a + b a 2 b a a +. b JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 3 / 58 Leibniz' rule examples (b) [c, d] [, ] for < c < d is compact and f : with f (b, x) ax + b and f b (b, x) 2(ax + b) 3/2 for a > are bounded and continuous and hence are both uniformly continuous on. ( ) dx ( 2 b ax + b b a ( ) dx b ax + b a ( ) 2 (ax + b) 3/2 dx a (ax + b) 3/2 dx 2 a ( ) ) a + b b ( ) a + b b ( ) a + b b ( ). a + b b JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 32 / 58

Leibniz' rule examples Given that for a >, nd (a) xe ax sin(kx) dx e ax sin(kx) dx (a) Assuming that Leibniz' rule is applicable, (b) k a 2 + k 2 xe ax cos(kx) dx. ( ) e ax sin(kx) dx ( ) k a a a 2 + k 2 ( e ax sin(kx) ) dx 2ak a (a 2 + k 2 ) 2 xe ax sin(kx)dx 2ak (a 2 + k 2 ) 2 xe ax sin(kx)dx 2ak (a 2 + k 2 ) 2. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 33 / 58 Leibniz' rule examples e ax sin(kx) dx k a 2 + k 2 (b) Assuming that Leibniz' rule is applicable, k ( ) e ax sin(kx) dx k ( e ax sin(kx) dx ) k ( k a 2 + k 2 ) a 2 + k 2 2k 2 (a 2 + k 2 ) 2 xe ax cos(kx)dx a2 k 2 (a 2 + k 2 ) 2. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 34 / 58

Leibniz' rule with variable limits To nd d dt v(t) u(t) f (x, t)dx we need to use the Fundamental Theorem of Calculus d du v u f (x, w)dx f (u, w) and Now let u u(t), v v(t) and w t and The chain rule says d dt F (u, v, w) F u F (u, v, w) du dt + F v f (u, w) du dt v u d dv dv dt + F v u f (x, w)dx. dw w dt + f (v, w)dv dt + w f (u, t) du dv + f (v, t) dt dt + f (x, w)dx f (v, w). v v u f (x, w)dx f (x, t)dx t u JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 35 / 58 Leibniz' rule with variable limits Example: d dt t t cos(tx) x dx cos(t t) t d ( t ) + cos(t.t) dt dt t dt + cos(t t) t 2 + cos(t t t cos(t t) 2t cos(t t) 2t 3 cos(t t) 2t + cos(t2 ) t + cos(t2 ) t + 2 cos t2 t 2 ) + t [ cos(tx) + t + cos(t2 ) t t t t x sin(tx) dx t x ] t t cos(t t) t ( ) cos(tx) dx x JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 36 / 58

Change of variables JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 37 / 58 Change of variables JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 38 / 58

Change of variables Consider the integral of x over the region shown on the right, (, ¾) I x +x x x dydx + 2 3 x x x dydx. (¾, ) Note that is bounded by the lines x + y, x + y 3, x y, x y. Perhaps it might be easier to use u x + y and v x y as coordinates. In the uv-plane, the corresponding region has boundaries Ú Ù u, u 3, v and v. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 39 / 58 Change of variables If then u x + y and v x y x (u + v). 2 Ú Ù Can we simply write the integral as I 3 (u + v) dudv? 2 No! The map ( ) u v ( scales areas by a factor of ) ( ) x y ( ) det 2. dudv 2dxdy dxdy 2 dudv. So, I 3 2 (u + v) 2 dudv. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 4 / 58

Change of variables For a more general change of variables by a dierentiable function f : ( ) ( ) ( ) ( ) ( ) ( ) u x x x x x x f T f + Jf v y y y y y y ( x ) ( x )( ) ( ¼, ¼ ) Ì Ù Ú Ù A f B f C f D f y ( x y ( x y ( x y + Jf ) + Jf ) + Jf ) + Jf y ( x y ( x y ( x y )( ) x )( ) x y )( ) y Area A B C D Ú Area A B C D JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 4 / 58 Area ABCD detjf. Change of variables Theorem (Change of variables) Suppose F : n n is C, det(j x F ) for x and F is one-to-one. Then, if f is integrable on F () f (f F ) detjf Alternative notation: f (x, y) dxdy f (x(u, v), y(u, v)) (x, y) (u, v) dudv where (x, y) (u, v) x det(jf ) det u y u x v y v JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 42 / 58

Change of variables Find the area of the region bounded by x 2 y 2, x 2 y 2 4 y x 2, y x 4. Let u x 2 y 2 and v y x. Ú ¾ (x, y) (u, v) Ù x u y u x v y v Area dxdy (x, y) (u, v) dudv. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 43 / 58 Change of variables ( ) ( ) ( ) ( x u F F x x 2 y 2 ) y y v y x Note that F is dierentiable for x. So u u ( ) J(F x y 2x 2y ) v v y. x x y 2 x ( ) det J(F ) 2x ( x ( 2y) yx ) 2 2 2 y 2 x 2 2 2v 2. But we want det(jf ) ) ( ) det ((JF ) det J(F ) 2 2v 2, that is, (x, y) (u, v) ( ) (u, v) (x, y) 2 2v 2. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 44 / 58

Change of variables JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 45 / 58 Change of variables Integrate x over the part of the unit disc that lies in the rst quadrant. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 46 / 58

Change of variables JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 47 / 58 substitution-example-.canvas JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 48 / 58

Change of variables JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 49 / 58 substitution-example-2.canvas JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 5 / 58

Normal distribution The standard normal distribution p(x) e x2 /2 is a probability distribution 2π on. But how do we know that If I I 2 e x2 dx, then e x2 dx π 2 π 2 dθ e y2 dy e x2 e y2 dxdy e (x2 +y 2 ) dxdy e r2 rdrdθ e r2 rdr p(x)dx? So, [ π 2 π 4. I 2 θ2 ( ] π 2 [ ] 2 e r2 ( 2 )) e e x2 dx π 2. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 5 / 58 Cylindrical and spherical polar coordinates Cylindrical polar coordinates Spherical polar coordinates Þ Þ Ö Þ φ ρ θ θ x r cos θ y r sin θ z z (x, y, z) (r, θ, z) r x ρ sin φ cos θ y ρ sin φ sin θ z ρ cos φ (x, y, z) (ρ, θ, φ) ρ2 sin φ JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 52 / 58

Spherical and polar coordinates Example (Exam 28): Using cylindrical or spherical polar coordinates, write down iterated integrals that would give the volume of the region bounded below by z 3x 2 + 3y 2 and above by z 4 x 2 y 2. substitution-example-3.canvas GNUPLOT commands: gnuplot> set parametric gnuplot> set view equal gnuplot> set isosamples 2 gnuplot> splot [-pi:pi][:] 2*sin(v*pi/6)*cos(u), \ 2*sin(v*pi/6)*sin(u), \ 2*cos(v*pi/6), \ 2*v*sin(pi/6)*cos(u), \ 2*v*sin(pi/6)*sin(u), \ 2*v*cos(pi/6) JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 53 / 58 Mass, centre of mass, centroid The balance point of masses on a line is the point x about which the torque is. Ñ Ñ ¾ ¾ Ñ m (x x) + m 2 (x 2 x) + m 3 (x 3 x) 3 3 3 m i (x i x) m i x i x m i. i x 3 m i x i i. 3 m i i+ i i+ The total mass is and M 3 i+ p k m k M m i is a discrete probability distribution with x E(X ). JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 54 / 58

Mass, centre of mass, centroid Consider a continuous mass distribution ρ :. ρ() We can nd the approximate centre of mass using an upper or lower sum of ρ with respect to a partition P. ρ() x n ρ i x i x i i, x n ρ i x i b a b ρ(x) x dx. ρ(x) dx a i JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 55 / 58 M b a and ρ(x) M ρ(x) dx is the total mass is a probability density. Mass, centre of mass, centroid For a lamina occupying the region 2 with density ρ :, the total mass is M ρ(x, y) dxdy The coordinates of the centre of mass are x xρ(x, y) dxdy M ȳ yρ(x, y) dxdy M For a solid body occupying the region 3 with density ρ :, the total mass is M ρ(x, y, z) dxdydz The coordinates of the centre of mass are x xρ(x, y, z) dxdydz M ȳ yρ(x, y, z) dxdydz M z zρ(x, y, z) dxdydz M For a lamina or solid body of constant density, the centre of mass is called the centroid and denoted (x c, y c, z c ). JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 56 / 58

Mass, centre of mass, centroid Example: Find the centre of mass of the trianglular lamina with vertices at (, ), (, ) and (, ) with density ρ(x, y) y. M y y 3. y ] y [ yx y dx dy y dy y( y) y(y ) dy The symmetry of and ρ gives x. yρ y 6. ȳ M y ] y [ y 2 x y y dx dy y dy y 2 ( y) y 2 (y ) dy yρ 3 6 2. JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 57 / 58 Mass, centre of mass, centroid Example: Find the centroid of the region bounded by r θ and the x-axis. To calculate the centroid, take ρ. We have already calculated the total mass (area) to be A π 3 /6. x π θ π r cos θ rdr dθ [ 3 r 3 cos θ ] θ dθ JM Kress (UNSW Maths & Stats) MATH2 Integration Semester, 26 58 / 58 π 3 θ3 cos θ dθ 4 π 2 (int by parts 3 times) x c 6(4 π2 ) π 3 A similar calculation gives y π θ.358. r sin θ rdr dθ π3 6π. 3 y c 2(π2 6) π 2.7845.