Evaluate the surface integral of vector A over the plane x+y+z=1 in the first octant. Using spherical coordinates, evaluate the integral of (x2+y2+z2)dxdydz where V is the sphere x2+y2+z2=a2 in the first octant.

Physics
Evaluate the surface integral of vector A over the plane x+y+z=1 in the first octant. Using spherical coordinates, evaluate the integral of (x2+y2+z2)dxdydz where V is the sphere x2+y2+z2=a2 in the first octant.

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Answer

16π29\dfrac{16\pi\sqrt{2}}{9}

Triple integral using spherical coordinates

Step 1: The region is x2+y2+z2≤4x^2 + y^2 + z^2 \leq 4, x≥0x \geq 0, y≥0y \geq 0, z≥0z \geq 0. In spherical coordinates, 0≤ρ≤20 \leq \rho \leq 2, 0≤θ≤π/20 \leq \theta \leq \pi/2, 0≤ϕ≤π/20 \leq \phi \leq \pi/2. Integrand (x2+y2+z2)3/4=ρ3/2(x^2 + y^2 + z^2)^{3/4} = \rho^{3/2}. dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.

I=∫0π/2∫0π/2∫02ρ3/2ρ2sin⁡ϕ dρ dϕ dθ=∫0π/2∫0π/2∫02ρ7/2sin⁡ϕ dρ dϕ dθ.I = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho^{3/2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho^{7/2} \sin\phi \, d\rho \, d\phi \, d\theta.

Step 2: ∫02ρ7/2 dρ=[29ρ9/2]02=29⋅29/2=29⋅24+1/2=29⋅162=3229\int_0^2 \rho^{7/2} \, d\rho = \left[ \frac{2}{9} \rho^{9/2} \right]_0^2 = \frac{2}{9} \cdot 2^{9/2} = \frac{2}{9} \cdot 2^{4 + 1/2} = \frac{2}{9} \cdot 16\sqrt{2} = \frac{32\sqrt{2}}{9}.

Step 3: ∫0π/2sin⁡ϕ dϕ=[−cos⁡ϕ]0π/2=1\int_0^{\pi/2} \sin\phi \, d\phi = [-\cos\phi]_0^{\pi/2} = 1.

Step 4: ∫0π/2dθ=π/2\int_0^{\pi/2} d\theta = \pi/2.

Step 5: I=3229⋅1⋅π2=16π29I = \frac{32\sqrt{2}}{9} \cdot 1 \cdot \frac{\pi}{2} = \frac{16\pi\sqrt{2}}{9}.

\dfrac{16\pi\sqrt{2}{9}}

Eigenvalues of A=[121−112]A = \begin{bmatrix} \frac{1}{2} & 1 \\ -1 & \frac{1}{2} \end{bmatrix}

Step 1: A−λI=[12−λ1−112−λ]A - \lambda I = \begin{bmatrix} \frac{1}{2}-\lambda & 1 \\ -1 & \frac{1}{2}-\lambda \end{bmatrix}.

det⁡(A−λI)=(12−λ)2−(1)(−1)=(12−λ)2+1=0\det(A - \lambda I) = \left(\frac{1}{2} - \lambda\right)^2 - (1)(-1) = \left(\frac{1}{2} - \lambda\right)^2 + 1 = 0.

Step 2: (12−λ)2=−1\left(\frac{1}{2} - \lambda\right)^2 = -1, 12−λ=±i\frac{1}{2} - \lambda = \pm i, λ=12∓i\lambda = \frac{1}{2} \mp i.

\lambda = \dfrac{1{2} \pm i}

Linear system dxdt=−5y\frac{dx}{dt} = -5y, dydt=−5x\frac{dy}{dt} = -5x

(i) Step 1: Let X(t)=[xy]\mathbf{X}(t) = \begin{bmatrix} x \\ y \end{bmatrix}. Then dXdt=[0−5−50]X\frac{dX}{dt} = \begin{bmatrix} 0 & -5 \\ -5 & 0 \end{bmatrix} \mathbf{X}.

(ii) Step 1: A=[0−5−50]A = \begin{bmatrix} 0 & -5 \\ -5 & 0 \end{bmatrix}. Characteristic equation: det⁡(A−λI)=λ2−25=0\det(A - \lambda I) = \lambda^2 - 25 = 0, λ=±5\lambda = \pm 5.

Step 2: For λ=5\lambda=5, (A−5I)v=0(A - 5I)v = 0 gives v1=[1−1]\mathbf{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}.

For λ=−5\lambda=-5, v2=[11]\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.

Step 3: P=[11−11]P = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}, det⁡P=2\det P = 2, P−1=12[1−111]P^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}.

Step 4: Φ(t)=P[e5t00e−5t]P−1=[cosh⁡(5t)−sinh⁡(5t)−sinh⁡(5t)cosh⁡(5t)]\Phi(t) = P \begin{bmatrix} e^{5t} & 0 \\ 0 & e^{-5t} \end{bmatrix} P^{-1} = \begin{bmatrix} \cosh(5t) & -\sinh(5t) \\ -\sinh(5t) & \cosh(5t) \end{bmatrix}.

\Phi(t)=\begin{bmatrix\cosh5t&-\sinh5t\-\sinh5t&\cosh5t\end{bmatrix}}

Green's theorem: ∮C(x−y) dx+(x+y) dy\oint_C (x-y)\,dx + (x+y)\,dy, CC bounds x=0x=0, y=0y=0, y=10−x2y=10-x^2

Step 1: P=x−yP=x-y, Q=x+yQ=x+y. ∂Q∂x−∂P∂y=1−(−1)=2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2.

Step 2: ∮C=∬D2 dA=2⋅Area(D)\oint_C = \iint_D 2\, dA = 2 \cdot Area(D).

Step 3: DD: 0≤x≤100 \leq x \leq \sqrt{10}, 0≤y≤10−x20 \leq y \leq 10-x^2.

Area = ∫010(10−x2) dx=[10x−x33]010=1010−103/23=1010(1−13)=20103\int_0^{\sqrt{10}} (10 - x^2)\, dx = \left[10x - \frac{x^3}{3}\right]_0^{\sqrt{10}} = 10\sqrt{10} - \frac{10^{3/2}}{3} = 10\sqrt{10} \left(1 - \frac{1}{3}\right) = \frac{20\sqrt{10}}{3}.

Step 4: ∮C=2⋅20103=40103\oint_C = 2 \cdot \frac{20\sqrt{10}}{3} = \frac{40\sqrt{10}}{3}.

\dfrac{40\sqrt{10}{3}}

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Quick Answer
  1. Triple integral using spherical coordinates Step 1: The region is x^2 + y^2 + z^2 ≤ 4, x ≥ 0, y ≥ 0, z ≥ 0.
  2. In spherical coordinates, 0 ≤ ≤ 2, 0 ≤ ≤ /2, 0 ≤ ≤ /2.
  3. Integrand (x^2 + y^2 + z^2)^3/4 = ^3/2.
  4. dV = ^2 \, d \, d \, d.
Evaluate the surface integral of vector A over the plane x+y+z=1 in the first octant. Using spherical coordinates, evaluate the integral of (x2+y2+z2)dxdydz where V is the sphere x2+y2+z2=a2 in the first octant.
Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

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Triple integral using spherical coordinates Step 1: The region is x^2 + y^2 + z^2 ≤ 4, x ≥ 0, y ≥ 0, z ≥ 0. In spherical coordinates, 0 ≤ ≤ 2, 0 ≤ ≤ /2, 0 ≤ ≤ /2. Integrand (x^2 + y^2 + z^2)^3/4 = ^3/2. dV = ^2 \, d \, d \, d. I = _0^/2 _0^/2 _0^2 ^3/2 ^2 \, d \, d \, d = _0^/2 _0^/2 _0^2 ^7/2 \, d \, d \, d. Step 2: _0^2 ^7/2 \, d = [ (2)/(9) ^9/2 ]_0^2 = (2)/(9) · 2^9/2 = (2)/(9) · 2^4 + 1/2 = (2)/(9) · 16sqrt(2) = 32sqrt(2)9. Step 3: _0^/2 \, d = [-]_0^/2 = 1. Step 4: _0^/2 d = /2. Step 5: I = 32sqrt(2)9 · 1 · ()/(2) = 16(2)9. 16(2)9 Eigenvalues of A = (1)/(2) & 1 \\ -1 & (1)/(2) Step 1: A - I = (1)/(2)- & 1 \\ -1 & (1)/(2)- . (A - I) = ((1)/(2) - )^2 - (1)(-1) = ((1)/(2) - )^2 + 1 = 0. Step 2: ((1)/(2) - )^2 = -1, (1)/(2) - = ± i, = (1)/(2) i. = (1)/(2) ± i Linear system (dx)/(dt) = -5y, (dy)/(dt) = -5x (i) Step 1: Let X(t) = x \\ y . Then dXdt = 0 & -5 \\ -5 & 0 X. (ii) Step 1: A = 0 & -5 \\ -5 & 0 . Characteristic equation: (A - I) = ^2 - 25 = 0, = ± 5. Step 2: For =5, (A - 5I)v = 0 gives v_1 = 1 \\ -1 . For =-5, v_2 = 1 \\ 1 . Step 3: P = 1 & 1 \\ -1 & 1 , P = 2, P^-1 = (1)/(2) 1 & -1 \\ 1 & 1 . Step 4: (t) = P e^5t & 0 \\ 0 & e^-5t P^-1 = (5t) & -(5t) \\ -(5t) & (5t) . (t)=5t&-5t\\-5t&5t Green's theorem: _C (x-y)\,dx + (x+y)\,dy, C bounds x=0, y=0, y=10-x^2 Step 1: P=x-y, Q=x+y. ( Q)/( x) - ( P)/( y) = 1 - (-1) = 2. Step 2: _C = _D 2\, dA = 2 · Area(D). Step 3: D: 0 ≤ x ≤ sqrt(10), 0 ≤ y ≤ 10-x^2. Area = _0^sqrt(10) (10 - x^2)\, dx = [10x - (x^3)/(3)]_0^sqrt(10) = 10sqrt(10) - 10^3/23 = 10sqrt(10) (1 - (1)/(3)) = 20sqrt(10)3. Step 4: _C = 2 · 20sqrt(10)3 = 40sqrt(10)3. 40sqrt(10)3