�� �� � ������������30. Surface integralsSuppose we are given a smooth 2-manifold M ⊂ R3 . Let�g : U −→ M ∩ W, be a diffeomorphism, where U ⊂ R2, with coordinates s and t. We can define two tangent vectors, which span the tangent plane to M at P = �g(s0,t0): �∂�gTs(s0,t0)= (s0,t0)∂s T�t(s0,t0)= ∂�g (s0,t0). ∂t We get an element of area on M, dS = �T�s × T�t� ds dt. Using this we can define the area of M ∩ W to be area(M ∩ W )= dS = �T�s × T�t� ds dt. M∩WU Example 30.1. We can parametrise the torus, M = { (x,y,z) | (a − x2 + y2)2 + z 2 = b2 }, as follows. Let U = (0, 2π) × (0, 2π), and W = R3 \{ (x,y,z) | x ≥ 0 and y =0, or x 2 + y 2 ≥ a 2 and z =0 }. Let �g : U −→ M ∩ W, be the function �g(s,t) = ((a + b cos t)cos s, (a + b cos t)sin s,b sin t). Let’s calculate the tangent vectors, T�s = ∂�g =(−(a + b cos t)sin s, (a + b cos t)cos s, 0),∂s T�t = ∂�g =(−b sin t cos s, −b sin t sin s,b cos t). ∂t SoT�s × T�t =ˆıjˆkˆ−(a + b cos t)sin s (a + b cos t)cos s 0 −b sin t cos s −b sin t sin sb cos t=(a + b cos t)b cos s cos tˆı +(a + b cos t)b sin s cos tˆk. j +(a + b cos t)b sin tˆ1 �� Therefore, �T�s × T�t� =(a + b cos t)b(cos2 s cos 2 t + sin2 s cos 2 t + sin2 t)1/2 =(a + b cos t)b. As a ≥ b, note that (a + b cos t)b> 0. Hence area(M) = area(M ∩ W ) =dS �� M∩W = �T�s × T�t� ds dt U� 2π � 2π =(a + b cos t)b ds dt 00� 2π =2πb (a + b cos t)dt 0 =4π2ab. Notice that this is the surface area of a cylinder of radius b and height 2πa, as expected. Example 30.2. We can parametrise the sphere, M = { (x,y,z) | x 2 + y 2 + z 2 = a 2 }, as follows. Let U = (0,π) × (0, 2π), and W = R3 \{ (x,y,z) | x ≥ 0 and y =0 }. Let �g : U −→ M ∩ W, be the function �g(φ,θ)=(a sin φ cos θ,a sin φ sin θ,a cos φ). Let’s calculate the tangent vectors, ∂�gT�φ = ∂φ =(a cos φ cos θ,a cos φ sin θ, −a sin φ), T�θ = ∂�g =(−a sin φ sin θ,a sin φ cos θ, 0). ∂θ 2 �������������� SoT�φ × T�θ =ˆıjˆkˆ−a sin φ sin θa sin φ cos θa cos φ cos θa cos φ sin θ −a sin φ 0 = a 2 sin2 φ cos θˆı + a 2 sin2 φ sin θjˆ+ a 2 cos φ sin φˆk. Therefore, �T�φ × T�θ� = a 2 sin φ(sin2 φ cos 2 θ + sin2 φ sin2 θ + cos2 φ)1/2 = a 2 sin φ. As 0 <φ<π, note that a2 sin φ> 0. Hence area(M) = area(M ∩ W ) =dS �� M∩W = �T�φ × T�θ� dφ dθ U� 2π � π = a 2 sin φ dφ dθ 00� 2π =2a 2 dt 0 =4πa2 . Notice that this is the surface area of a sphere of radius a. Let’s now suppose that there are two different ways to parametrise the same piece M ∩ W of the manifold M: �g : U −→ M ∩ W and �h: V −→ M ∩ W. Let use (u,v) coordinates for U ⊂ R2 and (s,t) coordinates for V ⊂ R2 . Then f�=(�h)−1 ◦ �g : U −→ V, is a diffeomorphism. Note that �g = �h f�. We then have ◦∂�g (u,v)= ∂(�h ◦ f�)(u,v)∂u∂u ∂�h ∂s ∂�h ∂t =(s,t)(u,v)+ (s,t)(u,v). ∂s ∂u∂t ∂u3 � � � � � � �� �� �� Similarly ∂�g (u,v)= ∂(�h ◦ f�)(u,v)∂v ∂v ∂�h ∂s ∂�h ∂t =(s,t)(u,v)+ (s,t)(u,v). ∂s ∂v ∂t ∂v ∂�g ∂�g ∂�h ∂s ∂�h ∂t ∂�h ∂s ∂�h ∂t ∂u × ∂v = ∂s ∂u + ∂t ∂u × ∂s ∂v + ∂t ∂v ∂�h ∂�h ∂s ∂t ∂s ∂t = ∂s × ∂t ∂u ∂v − ∂v ∂u ∂�h ∂�h ∂(s,t) = . ∂s × ∂t ∂(u,v)It follows that∂�g ∂�g ∂�h ∂�h ∂(s,t).�∂u × ∂v � = � ∂s × ∂t �|∂(u,v)|Hence∂�g ∂�g ∂�h ∂�h ∂(s,t) U �∂u × ∂v � du dv = U � ∂s × ∂t �|∂(u,v)| du dv ∂�h ∂�h = � ∂s × ∂t � ds dt. V Notice that the first term is precisely the integral we use to define the area of M ∩W . This formula then says that the area is independent of the choice of parametrisation. 4 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.