Review: Lagrange multiplier method: Max or min of a function f x, y subject to the constraint g x, y k : At a critical point x00, y , the gradient of f is parallel to the gradient of g. f x0, y 0 g x 0 , y 0 for some 1) write the constraint as gk 0 (subtract the constant ) 2 set up the function F x, y , f x , y g x , y 3 solve FFFF 0 xy 0, 0, and 0 4 find all such points that satisfy these equations and evaluate f at these points 5) The largest/smallest value of f will be the max/min of fg subject to the constraint . Max or min of f( x , y , z ) subject to the constraint g( x , y , z ) k : At a critical poit, fg for some constant Use above method for Fxyz , , , fxyz , , gxyz , , Suppose that the temperature at a point x, y on a metal plate is T x, y 4 x22 4 xy y . Example: An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant? f44 x22 xy y g x22 y 25 F x, y , 4 x2 4 xy y 2 x 2 y 2 25 Fx 8 x 4 y 2 x 0 8x 2 x 4 y F x22 y 25 Fy 4 x 2 y 2 y 0 2y 2 y 4 x solve: xy4 2 , yx12 and xy2225 2y 4y x yx12 1 (4 ) 4 (what if y 0?) 4 4 or 2 5 4 4 ( 5)=0 0 or 5 If 0 then yx 2 hence x2+4 x 2 5 x 2 25 or x2 5 and y2 20 critical points: 5,2 5 and 5, 2 5 If 5 then xy 2 hence 4y2 +y 2 5y 2 25 or y22 5 and x 20 TT 5,2 5 5, 2 5 0 2 5, 5 and 2 5, 5 TT2 5, 5 2 5, 5 125 highest temparature: 125, lowest temparature: 0 The same principle works for a function of 3 variables. Find min or max of f( x , y , z ) subject to the constraint g( x , y , z ) k : At a critical poit, fg for some constant Example: Find the point on the cone z2 x 2 y 2 which is closest to the point (3,4,0). Minimize f( x , y , z ) ( x 3)2 ( y 4) 2 z 2 subject to the constraint g(x , y , z ) x2 y 2 z 2 0 Find critical points of F( x , y , z ) ( x 3)2 ( y 4) 2 z 2 ( x 2 y 2 z 2 ) Fx 2( x 3) 2 x 0 or (xx 3) Fyy 2(y 4) 2 0 or (y 4) y Fz 2 z 2 z 0 or z z solve (x 3) x , (y 4) y and z z and z2 x 2 y 2 a) 1 3 (x 3) x , or 2 x 3, x and (y 4) y or 2 y 4, or y 2 2 2 9 25 5 3 5 3 5 zz 4 or critical points ,2, and ,2, 4 4 2 2 2 2 2 b) z 0 xy 0 from the constraint does not solve the other equations A picture: Find the point on the cone z2 x 2 y 2 which is closest to the point (3,4,0). Minimize f( x , y , z ) ( x 3)2 ( y 4) 2 z 2 3 5 3 5 ff,2, ,2, 2 2 2 2 2 35 2 3 (2 4) 22 9 25 36 4 4 13 4 4 4 3 5 3 5 distance = 13 critical points ,2, and ,2, 2 2 2 2 There are 2 points on the cone that are closest! 15.1 Double Integrals over rectangles Single variable integral : area Recall: under the graph of the function f and above the x-axis is found by using the area of “infinitely” many rectangles. 1) Divide the interval [a,b ] into n small intervals. * 2) Choose some point xk in the n-th interval. * 3) Add the area f( xkk ) x of the rectangles. 4) Take the limit as n . n n b * * A f xkk x Alim f xkk x A f() x dx n i1 i1 a Double variable integral : volume under the graph of the function (surface) and above the xy-plane found by using the volume of “infinitely” many rectangular prisms. Compute the volume under the graph of f and over the rectangle [a,b]x[c,d] 1) Divide the square [a,b ] [c,d] into n small squares. 2) Choose some point (x ,y ) in the k-th square. kk Notation: V f( x , y ) dA 3) Add the volumes f( xk ,y k ) A k of the cylinders. R 4) Take the limit as n . n Vollim f xk , y k A k where A k 0 as n n i1 n Fubini' s Theorem for Rectangular Regions Let f be continuous over the rectangle R x, y a x b , c y d, then db f( x , y ) dA f ( x , y ) dx dy R c a Integrate w.r.t. x first Work inside out treating y as a constant b A y f( x , y ) dx cross sectional area a "Add" up the cross sectional areas: d Volume A( y ) dy c We can also do this the other way around: d b b d f(,) x y dA f (,) x y dx dy f(,) x y dy dx R c a a c Integrate w.r.t. y first treating x as a constant d A x f( x , y ) dy cross sectional area c "Add" up the cross sectional areas: b Volume A( x ) dx a We usually simply write: d b b d f(,) x y dA f (,) x y dx dy f(,) x y dy dx R c a a c Example : Find the volume V of the solid lying under the graph of the elliptic paraboloid z8 2 x22 y and above the rectangle R x, y 0 x 1,0 y 2 . V 82 x22 y dA R 21 8 2x22 y dxdy 00 2 3 x1 2x 2 8x xy dy 0 3 x0 2 2 3 2 2 2 22 2 22 y 44 8 8 y dy y dy y 3 0 0 3 330 33 Ay 12 13 1 Example : Find the value of the double interal dxdy 3 02 xy 4 13 13 1 x3 1 3 1 2 dxdy x4 y dxdy x4 y dy 3 02 xy 4 02 0 2 x2 1 1 22 3 4y 2 4 y dy 2 0 111 1 13 4yy 1 2 4 1 1 1 1 2 1 4 1 4 2 4 3 4yy 4 2 4 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 3 4yy 2 4 0 8 7 6 3 2 8 7 6 3 2 1 6 7 14 21 16 1 8 42 8 42 56 y Example : Find the value of the double integral dA 22 R xy 1 where R is the rectangle 0xy 2, 0 1 We have to decide if we integrate first with respect to xy or with respect to . 1 If we do x first, we have to use the indefinite integral dx arctan x 1 x2 21 y y dA dydx maybe y first: 22 22 R xy 1 00xy 1 substitute: u x2 y 2 1, du 2 x 2 ydy y0 u 1, y 1 u x2 1 2 2 2 1 x 1 x 1 x 1 2 y ydy du 1 du 1ux2 1 ln(x 1) 22 dy 2 ln(u ) xy 1 2xu 2 22u1 0 1 u 1 2xu1 22xx y 2 1 dAln( x2 1) dx use u' v uv uv ' 2 2 2 R x y120 x 1 2 2 u, v ln( x 1) 122 1 2x ln(5) 1 ln(x2 1) dx dx 2x 0022 2x0 2 x x 1 4 x 1 12x uv'22 , ' ln(5) 2 ln(5) 21xx arctan(x ) arctan(2) 4 0 4 f x, y is called separable if it can be writtten as f x, y g x h y . If fR is separable and integrable over a rectangle , then bd f x, y dA g x h y dA g x dx h y dy RR ac 5 7 e x Example : Evaluate dydx 1 e y 55 7 77ee 5 2 x 1 e x 5 49 1 dydx dy xdx ln y lnee ln e 22 11eeyy 2 1 48 51 96 2 15.2 Double Integrals over general regions For a double integral f x, y dA R over a more general region R in the x y plane we again divide R into many small rectangles We will miss smaller and smaller regions as the subdivision gets finer. We then add up the volume of the rectangular boxes n Vollim f xk , y k A k n i1 Compute the volume under fR and over a region bounded by the two curves y g12( x ) and y g( x ) by slicing: gx2 Cross sectional area A( x ) f x , y dy gx1 b Volume A( x ) dx a b gx2 f x,, y dA f x y dydx R a g1 x A picture in the x-y plane: “sweep” out the region b gx2 f x, y dA f x, y dydx R a g1 x Example : Compute the double intregral 4xy y3 dA R where R is the region bounded by y x and y x3 4xy y3 dydx R yx 1 x 1 4 1 2 12 3 2 y 27xx 4xy y dydx 2xy dx 22x x dx 4 3 44 0 x3 0 yx 0 2 1 7x 1 3 8 13 2 12 4 7xx7 7x x x 7 1 1 2x dx 12 4 52 0 440 12 4 52 7 3 1 11 52 3 55 12 52 3 52 156 156.