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? f44 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: xy4  2 , yx12  and xy22 25 2y 4y x  yx12    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 TT2 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  critical points  ,2,  and  ,2,   2 2   2 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

distance = 13

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 Alim f xkk x A f() x dx  n   i1 i1 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

Vollim f xk , y k  A k where  A k  0 as n   n  i1

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 z 8  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 x1 2x 2  8x xy dy 0  3 x0 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 x3 1 3 1 2 dxdy x4 y dxdy x4 y dy  3      02 xy 4  02 0 2 x2 1 1 22 3  4y  2  4 y dy      2 0

111 1 13 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 0xy  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 y0  u  1, y  1  u  x2  1

2 2 2 1 x 1 x 1 x 1 2 y ydy du 1 du 1ux2 1 ln(x  1) 22 dy   2   ln(u )   xy 1   2xu 2  22u1 0 1 u 1 2xu1 22xx y 2 1 dAln( x2  1) dx  use u ' v uv uv ' 2 2  2  R x y120 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

Vollim f xk , y k A k n  i1 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 440 12 4 52

7 3 1 11 52 3 55    12 52 3 52 156 156