This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Part (a) Step 1: The surface S is = /4, 0 ≤ ≤ 1, 0 ≤ ≤ /2 (portion of cone inside unit sphere, first octant). r(, ) = ( , , ), = /4. Surface element: dS = \, d \, d. Step 2: x - y = ( - ). _S (x - y) \, dS = _0^/2 _0^1 ( - ) · \, d \, d = ^2 _0^/2 ( - ) \, d _0^1 ^2 \, d. Step 3: (/4) = sqrt(2)2, so ^2 = (1)/(2). _0^1 ^2 \, d = [ (^3)/(3) ]_0^1 = (1)/(3). Step 4: _0^/2 \, d = [ ]_0^/2 = 1 - 0 = 1, _0^/2 \, d = [ - ]_0^/2 = 0 - (-1) = 1, so _0^/2 ( - ) \, d = 1 - 1 = 0. Step 5: Integral = (1)/(2) · 0 · (1)/(3) = 0. Part (b) Step 1: sqrt(x^2 + y^2) = , dV = ^2 \, d \, d \, d. Region V: 0 ≤ ≤ /2, 0 ≤ ≤ /4, 0 ≤ ≤ 1. _V sqrt(x^2 + y^2) \, dV = _0^/2 d _0^/4 ^2 \, d _0^1 ^3 \, d. Step 2: _0^/2 d = /2, _0^1 ^3 \, d = [ (^4)/(4) ]_0^1 = 1/4. Step 3: ^2 = (1 - 2)/(2), _0^/4 ^2 \, d = _0^/4 (1 - 2)/(2) \, d = (1)/(2) [ - ( 2)/(2) ]_0^/4 = (1)/(2) ( ()/(4) - ((/2))/(2) ) = (1)/(2) ( ()/(4) - (1)/(2) ) = ()/(8) - (1)/(4). Step 4: Total = ()/(2) · ( ()/(8) - (1)/(4) ) · (1)/(4) = ()/(8) ( ()/(8) - (1)/(4) ) = (^2)/(64) - ()/(32) = (^2)/(64) - ()/(32). Part (c) Step 1: A = 1 & 0 \\ 0 & 1 . Characteristic equation: (A - I) = (1 - )^2 = 0, so = 1 (multiplicity 2). Step 2: Eigenvectors: (A - I)v = 0 gives zero matrix equation, eigenspace is all R^2. Basis: v_1 = 1 \\ 0 , v_2 = 0 \\ 1 . =1 (mult. 2),\ eigenspace\ R^2 Part (d) Step 1: V = -1 & 0 \\ 0 & 1 . (V - I) = (-1 - )(1 - ) = 0, so _1 = -1, _2 = 1. Step 2: For _1 = -1: (V + I)v = 0 & 0 \\ 0 & 2 v = 0, so v_1 = 1 \\ 0 . For _2 = 1: (V - I)v = -2 & 0 \\ 0 & 0 v = 0, so v_2 = 0 \\ 1 . _1=-1\ ( 1 \\ 0 ),\ _2=1\ ( 0 \\ 1 ) Part (e): System (dx)/(dt) = x^2 - y^2, (dy)/(dt) = -y^2 in vector form: (d)/(dt) x \\ y = x^2 - y^2 \\ -y^2 . x = x^2-y^2 \\ -y^2 Part (f): Assuming linear system x = V x with V from (d). Fundamental matrix (transition) (t) = e^V t = e^-t & 0 \\ 0 & e^t (diagonal). Step 1: Since V diagonal, e^V t = e^-t & 0 \\ 0 & e^t . (t)= e^-t & 0 \\ 0 & e^t Part (g): Solution x(t) = (t) x(0). x(t) = e^-t & 0 \\ 0 & e^t x(0) \\ y(0) = x(0) e^-t \\ y(0) e^t . x(t) \\ y(t) = x(0)e^-t \\ y(0)e^t Part (h): Green's theorem for _C P\,dx + Q\,dy, P = x^2 - y^2, Q = x^2 + y^2, C: upper semicircle x^2 + y^2 = 1, y ≥ 0 + x-axis from (1,0) to (-1,0) (counterclockwise). _R ( ( Q)/( x) - ( P)/( y) ) dA = _R (2x - (-2y)) \, dA = _R 2(x + y) \, dA, R: x^2 + y^2 ≤ 1, y ≥ 0. Step 1: Polar: x = r , y = r , dA = r \, dr \, d, : 0 , r: 0 1. 2 _0^ _0^1 (r + r ) r \, dr \, d = 2 _0^ ( + ) d _0^1 r^2 \, dr. Step 2: _0^1 r^2 \, dr = 1/3, _0^ \, d = []_0^ = 0, _0^ \, d = [-]_0^ = 1 - (-1) = 2, so 2 · (0 + 2) · (1)/(3) = 2 · 2 · (1)/(3) = (4)/(3). (4)/(3)
