Evaluate _S A * n d S, where A=(x+y) i+(y+z) j+x y z k and S is the surface of the plane x+y+z=1 in the first octant.

Mathematics
Evaluate _S A * n d S, where A=(x+y) i+(y+z) j+x y z k and S is the surface of the plane x+y+z=1 in the first octant.

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Part (a)

Step 1: The surface SS is ϕ=π/4\phi = \pi/4, 0≤ρ≤10 \leq \rho \leq 1, 0≤θ≤π/20 \leq \theta \leq \pi/2 (portion of cone inside unit sphere, first octant).

r(ρ,θ)=(ρsin⁡ϕcos⁡θ,ρsin⁡ϕsin⁡θ,ρcos⁡ϕ),ϕ=π/4.r(\rho, \theta) = (\rho \sin\phi \cos\theta, \rho \sin\phi \sin\theta, \rho \cos\phi), \quad \phi = \pi/4.

Surface element: dS=ρsin⁡ϕ dρ dθdS = \rho \sin\phi \, d\rho \, d\theta.

Step 2: x−y=ρsin⁡ϕ(cos⁡θ−sin⁡θ)x - y = \rho \sin\phi (\cos\theta - \sin\theta).

∬S(x−y) dS=∫0π/2∫01ρsin⁡ϕ(cos⁡θ−sin⁡θ)⋅ρsin⁡ϕ dρ dθ=sin⁡2ϕ∫0π/2(cos⁡θ−sin⁡θ) dθ∫01ρ2 dρ.\iint_S (x - y) \, dS = \int_0^{\pi/2} \int_0^1 \rho \sin\phi (\cos\theta - \sin\theta) \cdot \rho \sin\phi \, d\rho \, d\theta = \sin^2\phi \int_0^{\pi/2} (\cos\theta - \sin\theta) \, d\theta \int_0^1 \rho^2 \, d\rho.

Step 3: sin⁡(π/4)=22\sin(\pi/4) = \frac{\sqrt{2}}{2}, so sin⁡2ϕ=12\sin^2\phi = \frac{1}{2}.

∫01ρ2 dρ=[ρ33]01=13.\int_0^1 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^1 = \frac{1}{3}.

Step 4: ∫0π/2cos⁡θ dθ=[sin⁡θ]0π/2=1−0=1\int_0^{\pi/2} \cos\theta \, d\theta = \left[ \sin\theta \right]_0^{\pi/2} = 1 - 0 = 1,

∫0π/2sin⁡θ dθ=[−cos⁡θ]0π/2=0−(−1)=1\int_0^{\pi/2} \sin\theta \, d\theta = \left[ -\cos\theta \right]_0^{\pi/2} = 0 - (-1) = 1,

so ∫0π/2(cos⁡θ−sin⁡θ) dθ=1−1=0\int_0^{\pi/2} (\cos\theta - \sin\theta) \, d\theta = 1 - 1 = 0.

Step 5: Integral =12⋅0⋅13=0= \frac{1}{2} \cdot 0 \cdot \frac{1}{3} = \boxed{0}.

Part (b)

Step 1: x2+y2=ρsin⁡ϕ\sqrt{x^2 + y^2} = \rho \sin\phi, dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.

Region VV: 0≤θ≤π/20 \leq \theta \leq \pi/2, 0≤ϕ≤π/40 \leq \phi \leq \pi/4, 0≤ρ≤10 \leq \rho \leq 1.

∭Vx2+y2 dV=∫0π/2dθ∫0π/4sin⁡2ϕ dϕ∫01ρ3 dρ.\iiint_V \sqrt{x^2 + y^2} \, dV = \int_0^{\pi/2} d\theta \int_0^{\pi/4} \sin^2\phi \, d\phi \int_0^1 \rho^3 \, d\rho.

Step 2: ∫0π/2dθ=π/2\int_0^{\pi/2} d\theta = \pi/2,

∫01ρ3 dρ=[ρ44]01=1/4\int_0^1 \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_0^1 = 1/4.

Step 3: sin⁡2ϕ=1−cos⁡2ϕ2\sin^2\phi = \frac{1 - \cos 2\phi}{2},

∫0π/4sin⁡2ϕ dϕ=∫0π/41−cos⁡2ϕ2 dϕ=12[ϕ−sin⁡2ϕ2]0π/4=12(π4−sin⁡(π/2)2)=12(π4−12)=π8−14.\int_0^{\pi/4} \sin^2\phi \, d\phi = \int_0^{\pi/4} \frac{1 - \cos 2\phi}{2} \, d\phi = \frac{1}{2} \left[ \phi - \frac{\sin 2\phi}{2} \right]_0^{\pi/4} = \frac{1}{2} \left( \frac{\pi}{4} - \frac{\sin(\pi/2)}{2} \right) = \frac{1}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{8} - \frac{1}{4}.

Step 4: Total = \frac{\pi}{2} \cdot \left( \frac{\pi}{8} - \frac{1}{4} \right) \cdot \frac{1}{4} = \frac{\pi}{8} \left( \frac{\pi}{8} - \frac{1}{4} \right) = \frac{\pi^2}{64} - \frac{\pi}{32} = \dfrac{\pi^2{64} - \dfrac{\pi}{32}}.

Part (c)

Step 1: A=[1001]A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Characteristic equation: det⁡(A−λI)=(1−λ)2=0\det(A - \lambda I) = (1 - \lambda)^2 = 0, so λ=1\lambda = 1 (multiplicity 2).

Step 2: Eigenvectors: (A−I)v=0(A - I)v = \mathbf{0} gives zero matrix equation, eigenspace is all R2\mathbb{R}^2.

Basis: v1=[10]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, v2=[01]\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.

\lambda=1 (mult. 2),\ eigenspace\ \mathbb{R^2}

Part (d)

Step 1: V=[−1001]V = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}.

det⁡(V−λI)=(−1−λ)(1−λ)=0\det(V - \lambda I) = (-1 - \lambda)(1 - \lambda) = 0, so λ1=−1\lambda_1 = -1, λ2=1\lambda_2 = 1.

Step 2: For λ1=−1\lambda_1 = -1: (V+I)v=[0002]v=0(V + I)v = \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix} v = \mathbf{0}, so v1=[10]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.

For λ2=1\lambda_2 = 1: (V−I)v=[−2000]v=0(V - I)v = \begin{bmatrix} -2 & 0 \\ 0 & 0 \end{bmatrix} v = \mathbf{0}, so v2=[01]\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.

\lambda_1=-1\ ( \begin{bmatrix 1 \ 0 \end{bmatrix} ),\ \lambda_2=1\ ( \begin{bmatrix} 0 \ 1 \end{bmatrix} )}

Part (e): System dxdt=x2−y2\dfrac{dx}{dt} = x^2 - y^2, dydt=−y2\dfrac{dy}{dt} = -y^2 in vector form:

ddt[xy]=[x2−y2−y2].\dfrac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x^2 - y^2 \\ -y^2 \end{bmatrix}.

\dot{\mathbf{x} = \begin{bmatrix} x^{2}-y^{2} \ -y^{2} \end{bmatrix}}

Part (f): Assuming linear system x˙=Vx\dot{x} = V \mathbf{x} with VV from (d).

Fundamental matrix (transition) Φ(t)=eVt=[e−t00et]\Phi(t) = e^{V t} = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{t} \end{bmatrix} (diagonal).

Step 1: Since VV diagonal, eVt=[e−t00et]e^{V t} = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{t} \end{bmatrix}.

\Phi(t)=\begin{bmatrix e^{-t} & 0 \ 0 & e^{t} \end{bmatrix}}

Part (g): Solution x(t)=Φ(t)x(0)\mathbf{x}(t) = \Phi(t) x(0).

x(t)=[e−t00et][x(0)y(0)]=[x(0)e−ty(0)et].x(t) = \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{t} \end{bmatrix} \begin{bmatrix} x(0) \\ y(0) \end{bmatrix} = \begin{bmatrix} x(0) e^{-t} \\ y(0) e^{t} \end{bmatrix}.

\begin{bmatrix x(t) \ y(t) \end{bmatrix} = \begin{bmatrix} x(0)e^{-t} \ y(0)e^{t} \end{bmatrix}}

Part (h): Green's theorem for ∮CP dx+Q dy\oint_C P\,dx + Q\,dy, P=x2−y2P = x^2 - y^2, Q=x2+y2Q = x^2 + y^2, CC: upper semicircle x2+y2=1x^2 + y^2 = 1, y≥0y \geq 0 + x-axis from (1,0) to (-1,0) (counterclockwise).

∬R(∂Q∂x−∂P∂y)dA=∬R(2x−(−2y)) dA=∬R2(x+y) dA,\iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = \iint_R (2x - (-2y)) \, dA = \iint_R 2(x + y) \, dA,

RR: x2+y2≤1x^2 + y^2 \leq 1, y≥0y \geq 0.

Step 1: Polar: x=rcos⁡θx = r \cos\theta, y=rsin⁡θy = r \sin\theta, dA=r dr dθdA = r \, dr \, d\theta, θ:0→π\theta: 0 \to \pi, r:0→1r: 0 \to 1.

2∫0π∫01(rcos⁡θ+rsin⁡θ)r dr dθ=2∫0π(cos⁡θ+sin⁡θ)dθ∫01r2 dr.2 \int_0^\pi \int_0^1 (r \cos\theta + r \sin\theta) r \, dr \, d\theta = 2 \int_0^\pi (\cos\theta + \sin\theta) d\theta \int_0^1 r^2 \, dr.

Step 2: ∫01r2 dr=1/3\int_0^1 r^2 \, dr = 1/3,

∫0πcos⁡θ dθ=[sin⁡θ]0π=0\int_0^\pi \cos\theta \, d\theta = [\sin\theta]_0^\pi = 0,

∫0πsin⁡θ dθ=[−cos⁡θ]0π=1−(−1)=2\int_0^\pi \sin\theta \, d\theta = [-\cos\theta]_0^\pi = 1 - (-1) = 2,

so 2⋅(0+2)⋅13=2⋅2⋅13=432 \cdot (0 + 2) \cdot \frac{1}{3} = 2 \cdot 2 \cdot \frac{1}{3} = \frac{4}{3}.

\dfrac{4{3}}

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Quick Answer
  1. Part (a) Step 1: The surface S is = /4, 0 ≤ ≤ 1, 0 ≤ ≤ /2 (portion of cone inside unit sphere, first octant).
  2. r(, ) = ( , , ), = /4.
  3. Surface element: dS = \, d \, d.
  4. _S (x - y) \, dS = _0^/2 _0^1 ( - ) · \, d \, d = ^2 _0^/2 ( - ) \, d _0^1 ^2 \, d.
Evaluate _S A * n d S, where A=(x+y) i+(y+z) j+x y z k and S is the surface of the plane x+y+z=1 in the first octant.
Mathematics

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.

ScanSolve AI Answer
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)