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
916π2
Triple integral using spherical coordinates
Step 1: The region is x2+y2+z2≤4, x≥0, y≥0, z≥0. In spherical coordinates, 0≤ρ≤2, 0≤θ≤π/2, 0≤ϕ≤π/2. Integrand (x2+y2+z2)3/4=ρ3/2. dV=ρ2sinϕdρdϕdθ.
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.
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.
ScanSolve AI Answer
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
