In each of the following, find the 6 6 matrix A [aij] that satisfies the given condition: (a) (b) (c)

MATH 2450 WEEK 8 Ch 12 Double Integral in Rectangular Region ʃʃRf(x,y) dA R: a ≤ x ≤ b and c ≤ y ≤ d 2D: XY plane 3D: XYZ dimension, Projection on the XY plane Integral 2D b u ʃaf(x) dx = lim u ->∞∑ i =1f(i) ∆x Linearity Rule ʃʃR(c1*f(x,y) + 2 *g(x,y)) dA c1* ʃʃRf(x,y) dA + c 2ʃʃRg(x,y) dA Dominance Rule When f(x,y) ≥ g(x,y) for all (x,y) in R Then ʃʃ R(x,y) dA ≥ ʃʃ R(x,y) dA Subdivision Rule R = R 1 R 2 ʃʃRf(x,y) dA = ʃʃR1 f(x,y) dA + ʃʃR2 f(x,y) dA if f(x,y) ≥ 0 for all (x,y) in R ʃR f(x,y) dA = volume above where R is in between z = f(x,y) and z =0 EX. ʃʃR(2-y) dA R is in the rectangle with vertices of (0,0) , (3,0) , (3,2) , (0,2) Note: the equation z = 2-y is a plane parallel to the x-axis Area of the triangle is equal to (length*width)/2 = (2*2)/2 = 2 Volume of the prism is Area of the triangle * height(you could also consider this to be depth) Height = 3 Volume = 3*2 = 6 Integrated integral on R ʃa( ʃ c(x,y) dy ) dx b outer integral :ʃ ax inner integral: ʃ cy The above integral has a different process than ʃc( ʃ a(x,y) dx ) dy Fubini’s Theorem However, Fubini’s Theorem says that the two integrals will produce the same answer ʃʃRf(x,y) dA = ʃ (aʃ f(xcy) dy ) dx = ʃ ( ʃ fcx,y)adx ) dy EX. ʃʃ R2-y) dA From a prior example we know the answer is 6 According to Fubini’s Theorem ʃ0( ʃ 02-y) dy ) dx = ʃ ( ʃ0(2-y)0dx ) dy ʃ0( ʃ 02-y) dy ) dx ʃ (2y-(y /2)) | dx 0 0 3 ʃ0(2*