A parametric curve in R 3 \mathbb{R}^3 R 3 C 1 C^1 C 1 r : [ a , b ] → R 3 \mathbf{r} : [a, b] \to \mathbb{R}^3 r : [ a , b ] → R 3 r ( t ) = ( x ( t ) , y ( t ) , z ( t ) ) \mathbf{r}(t) = (x(t),\, y(t),\, z(t)) r ( t ) = ( x ( t ) , y ( t ) , z ( t ))
Definition. The arc length of r \mathbf{r} r [ a , b ] [a, b] [ a , b ]
L = ∫ a b ∥ r " ( t ) ∥ d t = ∫ a b ( d x d t ) 2 + ( d y d t ) 2 + ( d z d t ) 2 d t L = \int_a^b \lVert \mathbf{r}"(t) \rVert\, dt = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\, dt L = ∫ a b ∥ r " ( t )∥ d t = ∫ a b ( d t d x ) 2 + ( d t d y ) 2 + ( d t d z ) 2 d t
Proposition 5.1. The arc length function s ( t ) = ∫ a t ∥ r ′ ( τ ) ∥ d τ s(t) = \int_a^t \lVert \mathbf{r}'(\tau) \rVert\, d\tau s ( t ) = ∫ a t ∥ r ′ ( τ )∥ d τ d s d t = ∥ r ′ ( t ) ∥ \frac{ds}{dt} = \lVert \mathbf{r}'(t) \rVert d t d s = ∥ r ′ ( t )∥ ∥ d r d s ∥ = 1 \lVert \frac{d\mathbf{r}}{ds} \rVert = 1 ∥ d s d r ∥ = 1
Proof. By the Fundamental Theorem of Calculus, d s d t = ∥ r ′ ( t ) ∥ \frac{ds}{dt} = \lVert \mathbf{r}'(t) \rVert d t d s = ∥ r ′ ( t )∥ s s s r ( s ) = r ( t ( s ) ) \mathbf{r}(s) = \mathbf{r}(t(s)) r ( s ) = r ( t ( s )) d r d s = r ′ ( t ) ⋅ d t d s \frac{d\mathbf{r}}{ds} = \mathbf{r}'(t) \cdot \frac{dt}{ds} d s d r = r ′ ( t ) ⋅ d s d t ∥ d r d s ∥ = ∥ r ′ ( t ) ∥ ⋅ ∣ d t d s ∣ = 1 \lVert \frac{d\mathbf{r}}{ds} \rVert = \lVert \mathbf{r}'(t) \rVert \cdot \left\lvert \frac{dt}{ds} \right\rvert = 1 ∥ d s d r ∥ = ∥ r ′ ( t )∥ ⋅ d s d t = 1 ■ \blacksquare ■
Problem. Find the arc length of the curve r ( t ) = ( e t cos t , e t sin t , e t ) \mathbf{r}(t) = (e^t \cos t,\, e^t \sin t,\, e^t) r ( t ) = ( e t cos t , e t sin t , e t ) 0 ≤ t ≤ ln 2 0 \leq t \leq \ln 2 0 ≤ t ≤ ln 2
Solution r ′ ( t ) = ( e t cos t − e t sin t , e t sin t + e t cos t , e t ) \mathbf{r}'(t) = (e^t \cos t - e^t \sin t,\, e^t \sin t + e^t \cos t,\, e^t) r ′ ( t ) = ( e t cos t − e t sin t , e t sin t + e t cos t , e t )
∥ r ′ ( t ) ∥ 2 = e 2 t ( cos t − sin t ) 2 + e 2 t ( sin t + cos t ) 2 + e 2 t \lVert \mathbf{r}'(t) \rVert^2 = e^{2t}(\cos t - \sin t)^2 + e^{2t}(\sin t + \cos t)^2 + e^{2t} ∥ r ′ ( t ) ∥ 2 = e 2 t ( cos t − sin t ) 2 + e 2 t ( sin t + cos t ) 2 + e 2 t
= e 2 t [ ( cos 2 t − 2 sin t cos t + sin 2 t ) + ( sin 2 t + 2 sin t cos t + cos 2 t ) + 1 ] = e^{2t}[(\cos^2 t - 2\sin t \cos t + \sin^2 t) + (\sin^2 t + 2\sin t \cos t + \cos^2 t) + 1] = e 2 t [( cos 2 t − 2 sin t cos t + sin 2 t ) + ( sin 2 t + 2 sin t cos t + cos 2 t ) + 1 ]
= e 2 t [ 1 + 1 + 1 ] = 3 e 2 t = e^{2t}[1 + 1 + 1] = 3e^{2t} = e 2 t [ 1 + 1 + 1 ] = 3 e 2 t
∥ r ′ ( t ) ∥ = 3 e t \lVert \mathbf{r}'(t) \rVert = \sqrt{3}\, e^t ∥ r ′ ( t )∥ = 3 e t
L = ∫ 0 ln 2 3 e t d t = 3 [ e t ] 0 ln 2 = 3 ( 2 − 1 ) = 3 L = \int_0^{\ln 2} \sqrt{3}\, e^t\, dt = \sqrt{3}\, [e^t]_0^{\ln 2} = \sqrt{3}(2 - 1) = \sqrt{3} L = ∫ 0 l n 2 3 e t d t = 3 [ e t ] 0 l n 2 = 3 ( 2 − 1 ) = 3
■ \blacksquare ■
Problem. Find the arc length of the helix r ( t ) = ( cos t , sin t , t ) \mathbf{r}(t) = (\cos t,\, \sin t,\, t) r ( t ) = ( cos t , sin t , t ) 0 ≤ t ≤ 4 π 0 \leq t \leq 4\pi 0 ≤ t ≤ 4 π
Solution r ′ ( t ) = ( − sin t , cos t , 1 ) \mathbf{r}'(t) = (-\sin t,\, \cos t,\, 1) r ′ ( t ) = ( − sin t , cos t , 1 ) ∥ r ′ ( t ) ∥ = sin 2 t + cos 2 t + 1 = 2 \lVert \mathbf{r}'(t) \rVert = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2} ∥ r ′ ( t )∥ = sin 2 t + cos 2 t + 1 = 2
L = ∫ 0 4 π 2 d t = 4 2 π L = \int_0^{4\pi} \sqrt{2}\, dt = 4\sqrt{2}\,\pi L = ∫ 0 4 π 2 d t = 4 2 π
■ \blacksquare ■
Definition. Let r ( s ) \mathbf{r}(s) r ( s ) ∥ r ′ ( s ) ∥ = 1 \lVert \mathbf{r}'(s) \rVert = 1 ∥ r ′ ( s )∥ = 1
Unit tangent vector: T ( s ) = r ′ ( s ) \mathbf{T}(s) = \mathbf{r}'(s) T ( s ) = r ′ ( s ) Curvature: κ ( s ) = ∥ T ′ ( s ) ∥ = ∥ r ′ ′ ( s ) ∥ \kappa(s) = \lVert \mathbf{T}'(s) \rVert = \lVert \mathbf{r}''(s) \rVert κ ( s ) = ∥ T ′ ( s )∥ = ∥ r ′′ ( s )∥ Principal normal: N ( s ) = T ′ ( s ) ∥ T ′ ( s ) ∥ \mathbf{N}(s) = \frac{\mathbf{T}'(s)}{\lVert \mathbf{T}'(s) \rVert} N ( s ) = ∥ T ′ ( s )∥ T ′ ( s ) κ ≠ 0 \kappa \neq 0 κ = 0 Binormal: B ( s ) = T ( s ) × N ( s ) \mathbf{B}(s) = \mathbf{T}(s) \times \mathbf{N}(s) B ( s ) = T ( s ) × N ( s ) Torsion: τ ( s ) = − B ′ ( s ) ⋅ N ( s ) \tau(s) = -\mathbf{B}'(s) \cdot \mathbf{N}(s) τ ( s ) = − B ′ ( s ) ⋅ N ( s ) The vectors T \mathbf{T} T N \mathbf{N} N B \mathbf{B} B Frenet—Serret frame , an orthonormal Basis that moves with the curve.
Theorem 5.2 (Frenet—Serret Formulas).
T ′ = κ N , N ′ = − κ T + τ B , B ′ = − τ N \mathbf{T}' = \kappa\, \mathbf{N}, \quad \mathbf{N}' = -\kappa\, \mathbf{T} + \tau\, \mathbf{B}, \quad \mathbf{B}' = -\tau\, \mathbf{N} T ′ = κ N , N ′ = − κ T + τ B , B ′ = − τ N
Proof. Since T \mathbf{T} T T ⋅ T = 1 \mathbf{T} \cdot \mathbf{T} = 1 T ⋅ T = 1 T ′ ⋅ T = 0 \mathbf{T}' \cdot \mathbf{T} = 0 T ′ ⋅ T = 0 T ′ \mathbf{T}' T ′ T \mathbf{T} T T ′ \mathbf{T}' T ′ N \mathbf{N} N κ ≠ 0 \kappa \neq 0 κ = 0 T ′ = κ N \mathbf{T}' = \kappa\,\mathbf{N} T ′ = κ N
Similarly, B = T × N \mathbf{B} = \mathbf{T} \times \mathbf{N} B = T × N B ′ ⋅ B = 0 \mathbf{B}' \cdot \mathbf{B} = 0 B ′ ⋅ B = 0 B ⋅ T = 0 \mathbf{B} \cdot \mathbf{T} = 0 B ⋅ T = 0 B ′ ⋅ T + B ⋅ T ′ = 0 \mathbf{B}' \cdot \mathbf{T} + \mathbf{B} \cdot \mathbf{T}' = 0 B ′ ⋅ T + B ⋅ T ′ = 0 B ′ ⋅ T = − B ⋅ κ N = 0 \mathbf{B}' \cdot \mathbf{T} = -\mathbf{B} \cdot \kappa\,\mathbf{N} = 0 B ′ ⋅ T = − B ⋅ κ N = 0 B ′ \mathbf{B}' B ′ N \mathbf{N} N B ′ = − τ N \mathbf{B}' = -\tau\,\mathbf{N} B ′ = − τ N
For N ′ \mathbf{N}' N ′ { T , N , B } \{\mathbf{T}, \mathbf{N}, \mathbf{B}\} { T , N , B } N ′ = ( N ′ ⋅ T ) T + ( N ′ ⋅ N ) N + ( N ′ ⋅ B ) B \mathbf{N}' = (\mathbf{N}' \cdot \mathbf{T})\,\mathbf{T} + (\mathbf{N}' \cdot \mathbf{N})\,\mathbf{N} + (\mathbf{N}' \cdot \mathbf{B})\,\mathbf{B} N ′ = ( N ′ ⋅ T ) T + ( N ′ ⋅ N ) N + ( N ′ ⋅ B ) B N ⋅ T = 0 \mathbf{N} \cdot \mathbf{T} = 0 N ⋅ T = 0 N ′ ⋅ T = − N ⋅ T ′ = − κ \mathbf{N}' \cdot \mathbf{T} = -\mathbf{N} \cdot \mathbf{T}' = -\kappa N ′ ⋅ T = − N ⋅ T ′ = − κ N ⋅ N = 1 \mathbf{N} \cdot \mathbf{N} = 1 N ⋅ N = 1 N ′ ⋅ N = 0 \mathbf{N}' \cdot \mathbf{N} = 0 N ′ ⋅ N = 0 N ⋅ B = 0 \mathbf{N} \cdot \mathbf{B} = 0 N ⋅ B = 0 N ′ ⋅ B = − N ⋅ B ′ = τ \mathbf{N}' \cdot \mathbf{B} = -\mathbf{N} \cdot \mathbf{B}' = \tau N ′ ⋅ B = − N ⋅ B ′ = τ N ′ = − κ T + τ B \mathbf{N}' = -\kappa\,\mathbf{T} + \tau\,\mathbf{B} N ′ = − κ T + τ B ■ \blacksquare ■
Intuition. The curvature κ \kappa κ τ \tau τ τ = 0 \tau = 0 τ = 0
For a curve parameterised by an arbitrary parameter t t t
κ = ∥ r ′ ( t ) × r ′ ′ ( t ) ∥ ∥ r ′ ( t ) ∥ 3 \kappa = \frac{\lVert \mathbf{r}'(t) \times \mathbf{r}''(t) \rVert}{\lVert \mathbf{r}'(t) \rVert^3} κ = ∥ r ′ ( t ) ∥ 3 ∥ r ′ ( t ) × r ′′ ( t )∥
τ = [ r ′ ( t ) × r ′ ′ ( t ) ] ⋅ r ′ ′ ′ ( t ) ∥ r ′ ( t ) × r ′ ′ ( t ) ∥ 2 \tau = \frac{[\mathbf{r}'(t) \times \mathbf{r}''(t)] \cdot \mathbf{r}^{\prime\prime\prime}(t)}{\lVert \mathbf{r}'(t) \times \mathbf{r}''(t) \rVert^2} τ = ∥ r ′ ( t ) × r ′′ ( t ) ∥ 2 [ r ′ ( t ) × r ′′ ( t )] ⋅ r ′′′ ( t )
Problem. Find the curvature and torsion of the helix r ( t ) = ( a cos t , a sin t , b t ) \mathbf{r}(t) = (a\cos t,\, a\sin t,\, bt) r ( t ) = ( a cos t , a sin t , b t ) a , b > 0 a, b \gt 0 a , b > 0
Solution r ′ ( t ) = ( − a sin t , a cos t , b ) \mathbf{r}'(t) = (-a\sin t,\, a\cos t,\, b) r ′ ( t ) = ( − a sin t , a cos t , b )
r ′ ′ ( t ) = ( − a cos t , − a sin t , 0 ) \mathbf{r}''(t) = (-a\cos t,\, -a\sin t,\, 0) r ′′ ( t ) = ( − a cos t , − a sin t , 0 )
r ′ ′ ′ ( t ) = ( a sin t , − a cos t , 0 ) \mathbf{r}^{\prime\prime\prime}(t) = (a\sin t,\, -a\cos t,\, 0) r ′′′ ( t ) = ( a sin t , − a cos t , 0 )
∥ r ′ ∥ = a 2 + b 2 \lVert \mathbf{r}' \rVert = \sqrt{a^2 + b^2} ∥ r ′ ∥ = a 2 + b 2
r ′ × r ′ ′ = ∣ i j k − a sin t a cos t b − a cos t − a sin t 0 ∣ = ( a b sin t , − a b cos t , a 2 ) \mathbf{r}' \times \mathbf{r}'' = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -a\sin t & a\cos t & b \\ -a\cos t & -a\sin t & 0 \end{vmatrix} = (ab\sin t,\, -ab\cos t,\, a^2) r ′ × r ′′ = i − a sin t − a cos t j a cos t − a sin t k b 0 = ( ab sin t , − ab cos t , a 2 )
∥ r ′ × r ′ ′ ∥ = a 2 b 2 + a 4 = a a 2 + b 2 \lVert \mathbf{r}' \times \mathbf{r}'' \rVert = \sqrt{a^2 b^2 + a^4} = a\sqrt{a^2 + b^2} ∥ r ′ × r ′′ ∥ = a 2 b 2 + a 4 = a a 2 + b 2
κ = a a 2 + b 2 ( a 2 + b 2 ) 3 / 2 = a a 2 + b 2 \kappa = \frac{a\sqrt{a^2 + b^2}}{(a^2 + b^2)^{3/2}} = \frac{a}{a^2 + b^2} κ = ( a 2 + b 2 ) 3/2 a a 2 + b 2 = a 2 + b 2 a
For the torsion:
( r ′ × r ′ ′ ) ⋅ r ′ ′ ′ = a b sin t ⋅ a sin t + ( − a b cos t ) ( − a cos t ) + a 2 ⋅ 0 = a 2 b (\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}^{\prime\prime\prime} = ab\sin t \cdot a\sin t + (-ab\cos t)(-a\cos t) + a^2 \cdot 0 = a^2 b ( r ′ × r ′′ ) ⋅ r ′′′ = ab sin t ⋅ a sin t + ( − ab cos t ) ( − a cos t ) + a 2 ⋅ 0 = a 2 b
τ = a 2 b a 2 ( a 2 + b 2 ) = b a 2 + b 2 \tau = \frac{a^2 b}{a^2(a^2 + b^2)} = \frac{b}{a^2 + b^2} τ = a 2 ( a 2 + b 2 ) a 2 b = a 2 + b 2 b
■ \blacksquare ■
Remark. The helix has constant curvature and constant torsion, reflecting its uniform geometry.
A parametric surface is a C 1 C^1 C 1 r : D ⊆ R 2 → R 3 \mathbf{r} : D \subseteq \mathbb{R}^2 \to \mathbb{R}^3 r : D ⊆ R 2 → R 3 r ( u , v ) = ( x ( u , v ) , y ( u , v ) , z ( u , v ) ) \mathbf{r}(u, v) = (x(u,v),\, y(u,v),\, z(u,v)) r ( u , v ) = ( x ( u , v ) , y ( u , v ) , z ( u , v ))
The tangent plane at r ( u 0 , v 0 ) \mathbf{r}(u_0, v_0) r ( u 0 , v 0 )
r u = ∂ r ∂ u , r v = ∂ r ∂ v \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u}, \quad \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} r u = ∂ u ∂ r , r v = ∂ v ∂ r
The unit normal to the surface is
n = r u × r v ∥ r u × r v ∥ \mathbf{n} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{\lVert \mathbf{r}_u \times \mathbf{r}_v \rVert} n = ∥ r u × r v ∥ r u × r v
Examples of parametric surfaces:
Sphere (spherical coordinates): r ( θ , ϕ ) = ( ρ sin ϕ cos θ , ρ sin ϕ sin θ , ρ cos ϕ ) \mathbf{r}(\theta, \phi) = (\rho\sin\phi\cos\theta,\, \rho\sin\phi\sin\theta,\, \rho\cos\phi) r ( θ , ϕ ) = ( ρ sin ϕ cos θ , ρ sin ϕ sin θ , ρ cos ϕ ) Cylinder: r ( θ , z ) = ( r cos θ , r sin θ , z ) \mathbf{r}(\theta, z) = (r\cos\theta,\, r\sin\theta,\, z) r ( θ , z ) = ( r cos θ , r sin θ , z ) Graph of z = f ( x , y ) z = f(x,y) z = f ( x , y ) r ( x , y ) = ( x , y , f ( x , y ) ) \mathbf{r}(x, y) = (x,\, y,\, f(x,y)) r ( x , y ) = ( x , y , f ( x , y )) For the graph z = f ( x , y ) z = f(x,y) z = f ( x , y ) n = ( − f x , − f y , 1 ) 1 + f x 2 + f y 2 \mathbf{n} = \frac{(-f_x,\, -f_y,\, 1)}{\sqrt{1 + f_x^2 + f_y^2}} n = 1 + f x 2 + f y 2 ( − f x , − f y , 1 )
Definition. The area of a parametric surface r : D → R 3 \mathbf{r} : D \to \mathbb{R}^3 r : D → R 3
A ( S ) = ∬ D ∥ r u × r v ∥ d u d v A(S) = \iint_D \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert\, du\, dv A ( S ) = ∬ D ∥ r u × r v ∥ d u d v
Derivation. Partition D D D D i j D_{ij} D ij Δ u Δ v \Delta u\, \Delta v Δ u Δ v r ( D i j ) \mathbf{r}(D_{ij}) r ( D ij ) r u Δ u \mathbf{r}_u\, \Delta u r u Δ u r v Δ v \mathbf{r}_v\, \Delta v r v Δ v ∥ r u × r v ∥ Δ u Δ v \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert\, \Delta u\, \Delta v ∥ r u × r v ∥ Δ u Δ v ■ \blacksquare ■
Problem. Find the surface area of the part of the paraboloid z = x 2 + y 2 z = x^2 + y^2 z = x 2 + y 2 z = 4 z = 4 z = 4
Solution Parametrise by r ( x , y ) = ( x , y , x 2 + y 2 ) \mathbf{r}(x, y) = (x,\, y,\, x^2 + y^2) r ( x , y ) = ( x , y , x 2 + y 2 ) x 2 + y 2 ≤ 4 x^2 + y^2 \leq 4 x 2 + y 2 ≤ 4
r x = ( 1 , 0 , 2 x ) \mathbf{r}_x = (1,\, 0,\, 2x) r x = ( 1 , 0 , 2 x ) r y = ( 0 , 1 , 2 y ) \mathbf{r}_y = (0,\, 1,\, 2y) r y = ( 0 , 1 , 2 y )
r x × r y = ∣ i j k 1 0 2 x 0 1 2 y ∣ = ( − 2 x , − 2 y , 1 ) \mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2x \\ 0 & 1 & 2y \end{vmatrix} = (-2x,\, -2y,\, 1) r x × r y = i 1 0 j 0 1 k 2 x 2 y = ( − 2 x , − 2 y , 1 )
∥ r x × r y ∥ = 4 x 2 + 4 y 2 + 1 \lVert \mathbf{r}_x \times \mathbf{r}_y \rVert = \sqrt{4x^2 + 4y^2 + 1} ∥ r x × r y ∥ = 4 x 2 + 4 y 2 + 1
A = ∬ x 2 + y 2 ≤ 4 4 x 2 + 4 y 2 + 1 d x d y A = \iint_{x^2+y^2 \leq 4} \sqrt{4x^2 + 4y^2 + 1}\, dx\, dy A = ∬ x 2 + y 2 ≤ 4 4 x 2 + 4 y 2 + 1 d x d y
Use polar coordinates: x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ 0 ≤ r ≤ 2 0 \leq r \leq 2 0 ≤ r ≤ 2 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π
A = ∫ 0 2 π ∫ 0 2 4 r 2 + 1 r d r d θ A = \int_0^{2\pi} \int_0^2 \sqrt{4r^2 + 1}\, r\, dr\, d\theta A = ∫ 0 2 π ∫ 0 2 4 r 2 + 1 r d r d θ
Let u = 4 r 2 + 1 u = 4r^2 + 1 u = 4 r 2 + 1 d u = 8 r d r du = 8r\, dr d u = 8 r d r
= 2 π ⋅ 1 8 ∫ 1 17 u d u = π 4 [ 2 u 3 / 2 3 ] 1 17 = π 6 ( 17 3 / 2 − 1 ) = 2\pi \cdot \frac{1}{8} \int_1^{17} \sqrt{u}\, du = \frac{\pi}{4}\left[\frac{2u^{3/2}}{3}\right]_1^{17} = \frac{\pi}{6}(17^{3/2} - 1) = 2 π ⋅ 8 1 ∫ 1 17 u d u = 4 π [ 3 2 u 3/2 ] 1 17 = 6 π ( 1 7 3/2 − 1 )
■ \blacksquare ■
Definition (Scalar surface integral). The integral of a scalar function f f f S S S
∬ S f d S = ∬ D f ( r ( u , v ) ) ∥ r u × r v ∥ d u d v \iint_S f\, dS = \iint_D f(\mathbf{r}(u,v))\, \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert\, du\, dv ∬ S f d S = ∬ D f ( r ( u , v )) ∥ r u × r v ∥ d u d v
Definition (Vector surface integral / flux). The flux of a vector field F \mathbf{F} F S S S
∬ S F ⋅ d S = ∬ D F ( r ( u , v ) ) ⋅ ( r u × r v ) d u d v \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\, du\, dv ∬ S F ⋅ d S = ∬ D F ( r ( u , v )) ⋅ ( r u × r v ) d u d v
Where the orientation is determined by the choice of normal r u × r v \mathbf{r}_u \times \mathbf{r}_v r u × r v r v × r u \mathbf{r}_v \times \mathbf{r}_u r v × r u
Problem. Evaluate ∬ S F ⋅ d S \iint_S \mathbf{F} \cdot d\mathbf{S} ∬ S F ⋅ d S F = ( x , y , z 2 ) \mathbf{F} = (x,\, y,\, z^2) F = ( x , y , z 2 ) S S S x 2 + y 2 + z 2 = 4 x^2 + y^2 + z^2 = 4 x 2 + y 2 + z 2 = 4 z ≥ 0 z \geq 0 z ≥ 0
Solution Use the divergence theorem on the closed hemisphere plus the disk at z = 0 z = 0 z = 0 E E E
∬ c l o s e d S F ⋅ d S = ∭ E ∇ ⋅ F d V = ∭ E ( 1 + 1 + 2 z ) d V \iint_{\mathrm{closed\ S} \mathbf{F} \cdot d\mathbf{S} = \iiint_E \nabla \cdot \mathbf{F}\, dV = \iiint_E (1 + 1 + 2z)\, dV} ∬ closed S F ⋅ d S = ∭ E ∇ ⋅ F d V = ∭ E ( 1 + 1 + 2 z ) d V
= 2 V + 2 ∭ E z d V = 2V + 2\iiint_E z\, dV = 2 V + 2 ∭ E z d V
Where V = 1 2 ⋅ 4 3 π ( 2 3 ) = 16 π 3 V = \frac{1}{2} \cdot \frac{4}{3}\pi(2^3) = \frac{16\pi}{3} V = 2 1 ⋅ 3 4 π ( 2 3 ) = 3 16 π
By symmetry, the centroid of a hemisphere of radius R = 2 R = 2 R = 2 z = 3 R / 8 = 3 / 4 z = 3R/8 = 3/4 z = 3 R /8 = 3/4
∭ E z d V = z ˉ ⋅ V = 3 4 ⋅ 16 π 3 = 4 π \iiint_E z\, dV = \bar{z} \cdot V = \frac{3}{4} \cdot \frac{16\pi}{3} = 4\pi ∭ E z d V = z ˉ ⋅ V = 4 3 ⋅ 3 16 π = 4 π
= 2 ⋅ 16 π 3 + 2 ⋅ 4 π = 32 π 3 + 8 π = 56 π 3 = 2 \cdot \frac{16\pi}{3} + 2 \cdot 4\pi = \frac{32\pi}{3} + 8\pi = \frac{56\pi}{3} = 2 ⋅ 3 16 π + 2 ⋅ 4 π = 3 32 π + 8 π = 3 56 π
The flux through the disk z = 0 z = 0 z = 0 x 2 + y 2 ≤ 4 x^2 + y^2 \leq 4 x 2 + y 2 ≤ 4 − k -\mathbf{k} − k ∬ d i s k F ⋅ ( − k ) d S = ∬ d i s k 0 d S = 0 \iint_{\mathrm{disk} \mathbf{F} \cdot (-\mathbf{k})\, dS = \iint_{\mathrm{disk} 0\, dS = 0}} ∬ disk F ⋅ ( − k ) d S = ∬ disk 0 d S = 0
So the flux through the hemisphere alone is 56 π 3 \frac{56\pi}{3} 3 56 π ■ \blacksquare ■
Problem. Evaluate ∬ S z d S \iint_S z\, dS ∬ S z d S S S S 2 x + 2 y + z = 4 2x + 2y + z = 4 2 x + 2 y + z = 4
Solution Parametrise the surface. Solve for z = 4 − 2 x − 2 y z = 4 - 2x - 2y z = 4 − 2 x − 2 y x ≥ 0 x \geq 0 x ≥ 0 y ≥ 0 y \geq 0 y ≥ 0 z ≥ 0 z \geq 0 z ≥ 0 2 x + 2 y ≤ 4 2x + 2y \leq 4 2 x + 2 y ≤ 4 x + y ≤ 2 x + y \leq 2 x + y ≤ 2
r ( x , y ) = ( x , y , 4 − 2 x − 2 y ) \mathbf{r}(x,y) = (x,\, y,\, 4 - 2x - 2y) r ( x , y ) = ( x , y , 4 − 2 x − 2 y ) D = ( x , y ) : x ≥ 0 , y ≥ 0 , x + y ≤ 2 D = \\{(x,y) : x \geq 0,\, y \geq 0,\, x + y \leq 2\\} D = ( x , y ) : x ≥ 0 , y ≥ 0 , x + y ≤ 2
r x = ( 1 , 0 , − 2 ) \mathbf{r}_x = (1,\, 0,\, -2) r x = ( 1 , 0 , − 2 ) r y = ( 0 , 1 , − 2 ) \mathbf{r}_y = (0,\, 1,\, -2) r y = ( 0 , 1 , − 2 )
r x × r y = ( 2 , 2 , 1 ) \mathbf{r}_x \times \mathbf{r}_y = (2,\, 2,\, 1) r x × r y = ( 2 , 2 , 1 )
∥ r x × r y ∥ = 4 + 4 + 1 = 3 \lVert \mathbf{r}_x \times \mathbf{r}_y \rVert = \sqrt{4 + 4 + 1} = 3 ∥ r x × r y ∥ = 4 + 4 + 1 = 3
∬ S z d S = ∬ D ( 4 − 2 x − 2 y ) ⋅ 3 d x d y = 3 ∫ 0 2 ∫ 0 2 − x ( 4 − 2 x − 2 y ) d y d x \iint_S z\, dS = \iint_D (4 - 2x - 2y) \cdot 3\, dx\, dy = 3 \int_0^2 \int_0^{2-x} (4 - 2x - 2y)\, dy\, dx ∬ S z d S = ∬ D ( 4 − 2 x − 2 y ) ⋅ 3 d x d y = 3 ∫ 0 2 ∫ 0 2 − x ( 4 − 2 x − 2 y ) d y d x
= 3 ∫ 0 2 [ ( 4 − 2 x ) y − y 2 ] 0 2 − x d x = 3 ∫ 0 2 ( 2 − x ) ( 2 − x ) d x = 3 \int_0^2 \left[(4-2x)y - y^2\right]_0^{2-x}\, dx = 3 \int_0^2 (2-x)(2-x)\, dx = 3 ∫ 0 2 [ ( 4 − 2 x ) y − y 2 ] 0 2 − x d x = 3 ∫ 0 2 ( 2 − x ) ( 2 − x ) d x
= 3 ∫ 0 2 ( 2 − x ) 2 d x = 3 [ − ( 2 − x ) 3 3 ] 0 2 = 3 ⋅ 8 3 = 8 = 3 \int_0^2 (2-x)^2\, dx = 3\left[-\frac{(2-x)^3}{3}\right]_0^2 = 3 \cdot \frac{8}{3} = 8 = 3 ∫ 0 2 ( 2 − x ) 2 d x = 3 [ − 3 ( 2 − x ) 3 ] 0 2 = 3 ⋅ 3 8 = 8
■ \blacksquare ■
:::caution Common Pitfalls
Parameterisation domain. Always verify that the parameterisation covers the entire surface and that the map is one-to-one (except possibly on the boundary).Normal orientation. The cross product r u × r v \mathbf{r}_u \times \mathbf{r}_v r u × r v Surface area vs. Flux. Surface area uses ∥ r u × r v ∥ \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert ∥ r u × r v ∥ r u × r v \mathbf{r}_u \times \mathbf{r}_v r u × r v :::