A surface in 3D is a "rubber sheet" living in space. To describe it cleanly we paint it with two coordinates ( u , v ) (u,v) ( u , v )
Definition Parametric surfaceA parametric surface is a vector function of two parameters:
r ( u , v ) = ⟨ x ( u , v ) , y ( u , v ) , z ( u , v ) ⟩ , ( u , v ) ∈ D \mathbf{r}(u,v) = \langle x(u,v),\; y(u,v),\; z(u,v)\rangle, \qquad (u,v)\in D r ( u , v ) = ⟨ x ( u , v ) , y ( u , v ) , z ( u , v )⟩ , ( u , v ) ∈ D ( u , v ) (u,v) ( u , v ) D D D r \mathbf r r R 3 \mathbb R^3 R 3
WHY two parameters? A surface is intrinsically 2-dimensional — you can wander in two independent directions on it. Curves need 1 parameter (r ( t ) \mathbf r(t) r ( t )
Freeze v = v 0 v=v_0 v = v 0 u u u u u u u = u 0 u=u_0 u = u 0 v v v v v v
Differentiate r \mathbf r r
r u = ∂ r ∂ u = ⟨ ∂ x ∂ u , ∂ y ∂ u , ∂ z ∂ u ⟩ , r v = ∂ r ∂ v = ⟨ ∂ x ∂ v , ∂ y ∂ v , ∂ z ∂ v ⟩ \mathbf r_u = \frac{\partial \mathbf r}{\partial u}=\Big\langle \tfrac{\partial x}{\partial u},\tfrac{\partial y}{\partial u},\tfrac{\partial z}{\partial u}\Big\rangle, \qquad \mathbf r_v = \frac{\partial \mathbf r}{\partial v}=\Big\langle \tfrac{\partial x}{\partial v},\tfrac{\partial y}{\partial v},\tfrac{\partial z}{\partial v}\Big\rangle r u = ∂ u ∂ r = ⟨ ∂ u ∂ x , ∂ u ∂ y , ∂ u ∂ z ⟩ , r v = ∂ v ∂ r = ⟨ ∂ v ∂ x , ∂ v ∂ y , ∂ v ∂ z ⟩
r u \mathbf r_u r u u u u u u u r v \mathbf r_v r v v v v
WHY? r u \mathbf r_u r u u ↦ r ( u , v 0 ) u\mapsto \mathbf r(u,v_0) u ↦ r ( u , v 0 ) in the surface.
Definition Smooth / regular surfaceA surface is smooth at a point if r u × r v ≠ 0 \mathbf r_u \times \mathbf r_v \neq \mathbf 0 r u × r v = 0
WHAT we want: the flat plane that best hugs the surface at P 0 = r ( u 0 , v 0 ) P_0=\mathbf r(u_0,v_0) P 0 = r ( u 0 , v 0 )
HOW: A plane is fixed by a point and a normal vector n \mathbf n n P 0 P_0 P 0 n \mathbf n n both tangent directions r u \mathbf r_u r u r v \mathbf r_v r v
n = r u × r v \boxed{\;\mathbf n = \mathbf r_u \times \mathbf r_v\;} n = r u × r v
WHAT: total area of the surface as ( u , v ) (u,v) ( u , v ) D D D
HOW (build it from tiny patches):
Chop D D D Δ u × Δ v \Delta u \times \Delta v Δ u × Δ v approximately the displacement vectors
r u Δ u and r v Δ v \mathbf r_u\,\Delta u \quad\text{and}\quad \mathbf r_v\,\Delta v r u Δ u and r v Δ v
Why? Moving Δ u \Delta u Δ u u u u ≈ r u Δ u \approx \mathbf r_u\,\Delta u ≈ r u Δ u
The little patch is approximately a parallelogram spanned by those two edge vectors. The area of a parallelogram spanned by vectors a , b \mathbf a,\mathbf b a , b ∣ a × b ∣ |\mathbf a\times\mathbf b| ∣ a × b ∣
Δ S ≈ ∣ r u Δ u × r v Δ v ∣ = ∣ r u × r v ∣ Δ u Δ v \Delta S \approx |\mathbf r_u\,\Delta u \times \mathbf r_v\,\Delta v| = |\mathbf r_u\times\mathbf r_v|\,\Delta u\,\Delta v Δ S ≈ ∣ r u Δ u × r v Δ v ∣ = ∣ r u × r v ∣ Δ u Δ v
Summing and taking the limit gives an integral:
Intuition Why a cross product (not a dot)?Area is about how much the two edges spread apart . The cross-product magnitude ∣ r u ∣ ∣ r v ∣ sin θ |\mathbf r_u||\mathbf r_v|\sin\theta ∣ r u ∣∣ r v ∣ sin θ sin θ = 1 \sin\theta=1 sin θ = 1 sin θ = 0 \sin\theta=0 sin θ = 0
Parametrize with r ( x , y ) = ⟨ x , y , f ( x , y ) ⟩ \mathbf r(x,y)=\langle x,y,f(x,y)\rangle r ( x , y ) = ⟨ x , y , f ( x , y )⟩ r x = ⟨ 1 , 0 , f x ⟩ \mathbf r_x=\langle1,0,f_x\rangle r x = ⟨ 1 , 0 , f x ⟩ r y = ⟨ 0 , 1 , f y ⟩ \mathbf r_y=\langle0,1,f_y\rangle r y = ⟨ 0 , 1 , f y ⟩ r x × r y = ⟨ − f x , − f y , 1 ⟩ ⇒ ∣ r x × r y ∣ = f x 2 + f y 2 + 1 \mathbf r_x\times\mathbf r_y=\langle -f_x,-f_y,1\rangle \Rightarrow |\mathbf r_x\times\mathbf r_y|=\sqrt{f_x^2+f_y^2+1} r x × r y = ⟨ − f x , − f y , 1 ⟩ ⇒ ∣ r x × r y ∣ = f x 2 + f y 2 + 1 A = ∬ D f x 2 + f y 2 + 1 d A A=\iint_D \sqrt{f_x^2+f_y^2+1}\;dA A = ∬ D f x 2 + f y 2 + 1 d A
Worked example (a) Tangent plane to a paraboloidSurface r ( u , v ) = ⟨ u , v , u 2 + v 2 ⟩ \mathbf r(u,v)=\langle u,\,v,\,u^2+v^2\rangle r ( u , v ) = ⟨ u , v , u 2 + v 2 ⟩ ( u , v ) = ( 1 , 1 ) (u,v)=(1,1) ( u , v ) = ( 1 , 1 ) ( 1 , 1 , 2 ) (1,1,2) ( 1 , 1 , 2 )
r u = ⟨ 1 , 0 , 2 u ⟩ = ⟨ 1 , 0 , 2 ⟩ \mathbf r_u=\langle1,0,2u\rangle=\langle1,0,2\rangle r u = ⟨ 1 , 0 , 2 u ⟩ = ⟨ 1 , 0 , 2 ⟩ Why: differentiate each component in u u u r v = ⟨ 0 , 1 , 2 v ⟩ = ⟨ 0 , 1 , 2 ⟩ \mathbf r_v=\langle0,1,2v\rangle=\langle0,1,2\rangle r v = ⟨ 0 , 1 , 2 v ⟩ = ⟨ 0 , 1 , 2 ⟩ n = r u × r v = ⟨ ( 0 ) ( 2 ) − ( 2 ) ( 1 ) , ( 2 ) ( 0 ) − ( 1 ) ( 2 ) , ( 1 ) ( 1 ) − ( 0 ) ( 0 ) ⟩ = ⟨ − 2 , − 2 , 1 ⟩ \mathbf n=\mathbf r_u\times\mathbf r_v=\langle (0)(2)-(2)(1),\,(2)(0)-(1)(2),\,(1)(1)-(0)(0)\rangle=\langle-2,-2,1\rangle n = r u × r v = ⟨( 0 ) ( 2 ) − ( 2 ) ( 1 ) , ( 2 ) ( 0 ) − ( 1 ) ( 2 ) , ( 1 ) ( 1 ) − ( 0 ) ( 0 )⟩ = ⟨ − 2 , − 2 , 1 ⟩ Why: cross product gives the normal.Plane: − 2 ( x − 1 ) − 2 ( y − 1 ) + 1 ( z − 2 ) = 0 ⇒ − 2 x − 2 y + z + 2 = 0 -2(x-1)-2(y-1)+1(z-2)=0 \Rightarrow -2x-2y+z+2=0 − 2 ( x − 1 ) − 2 ( y − 1 ) + 1 ( z − 2 ) = 0 ⇒ − 2 x − 2 y + z + 2 = 0 z = 2 x + 2 y − 2 z=2x+2y-2 z = 2 x + 2 y − 2
Worked example (b) Surface area of a sphere of radiusR R R r ( ϕ , θ ) = ⟨ R sin ϕ cos θ , R sin ϕ sin θ , R cos ϕ ⟩ \mathbf r(\phi,\theta)=\langle R\sin\phi\cos\theta,\,R\sin\phi\sin\theta,\,R\cos\phi\rangle r ( ϕ , θ ) = ⟨ R sin ϕ cos θ , R sin ϕ sin θ , R cos ϕ ⟩ ϕ ∈ [ 0 , π ] , θ ∈ [ 0 , 2 π ] \phi\in[0,\pi],\theta\in[0,2\pi] ϕ ∈ [ 0 , π ] , θ ∈ [ 0 , 2 π ]
r ϕ = ⟨ R cos ϕ cos θ , R cos ϕ sin θ , − R sin ϕ ⟩ \mathbf r_\phi=\langle R\cos\phi\cos\theta,\,R\cos\phi\sin\theta,\,-R\sin\phi\rangle r ϕ = ⟨ R cos ϕ cos θ , R cos ϕ sin θ , − R sin ϕ ⟩ r θ = ⟨ − R sin ϕ sin θ , R sin ϕ cos θ , 0 ⟩ \mathbf r_\theta=\langle -R\sin\phi\sin\theta,\,R\sin\phi\cos\theta,\,0\rangle r θ = ⟨ − R sin ϕ sin θ , R sin ϕ cos θ , 0 ⟩ Compute ∣ r ϕ × r θ ∣ = R 2 sin ϕ |\mathbf r_\phi\times\mathbf r_\theta|=R^2\sin\phi ∣ r ϕ × r θ ∣ = R 2 sin ϕ Why: the cross product magnitude simplifies using sin 2 + cos 2 = 1 \sin^2+\cos^2=1 sin 2 + cos 2 = 1
A = ∫ 0 2 π ∫ 0 π R 2 sin ϕ d ϕ d θ = R 2 ( 2 π ) [ − cos ϕ ] 0 π = R 2 ( 2 π ) ( 2 ) = 4 π R 2 . A=\int_0^{2\pi}\!\!\int_0^{\pi} R^2\sin\phi\,d\phi\,d\theta = R^2\,(2\pi)\,[-\cos\phi]_0^\pi = R^2(2\pi)(2)=4\pi R^2. A = ∫ 0 2 π ∫ 0 π R 2 sin ϕ d ϕ d θ = R 2 ( 2 π ) [ − cos ϕ ] 0 π = R 2 ( 2 π ) ( 2 ) = 4 π R 2 .
Worked example (c) Area under a graphPart of z = x 2 + y 2 z=x^2+y^2 z = x 2 + y 2 z = 2 x + 2 y z=2x+2y z = 2 x + 2 y D D D f x = 2 , f y = 2 f_x=2,f_y=2 f x = 2 , f y = 2 = 4 + 4 + 1 = 3 =\sqrt{4+4+1}=3 = 4 + 4 + 1 = 3 = 3 ⋅ area ( D ) = 3 π =3\cdot\text{area}(D)=3\pi = 3 ⋅ area ( D ) = 3 π Why: constant stretch factor pulls out of the integral.
Common mistake Using a dot product for areaTempting because dot products show up everywhere and feel "natural". But r u ⋅ r v \mathbf r_u\cdot\mathbf r_v r u ⋅ r v opposite of zero area. Fix: area needs the cross product magnitude ∣ r u × r v ∣ |\mathbf r_u\times\mathbf r_v| ∣ r u × r v ∣
Common mistake Forgetting the Jacobian factorWriting A = ∬ D d A A=\iint_D dA A = ∬ D d A parameter region. Feels right because d A dA d A But that's area in the flat ( u , v ) (u,v) ( u , v ) ∣ r u × r v ∣ |\mathbf r_u\times\mathbf r_v| ∣ r u × r v ∣
Common mistake Cross product order / signr v × r u = − ( r u × r v ) \mathbf r_v\times\mathbf r_u = -(\mathbf r_u\times\mathbf r_v) r v × r u = − ( r u × r v ) tangent plane the sign doesn't matter (any normal works). For orientation/flux it does. For area we take magnitude, so sign is irrelevant.
∣ r u ∣ ∣ r v ∣ |\mathbf r_u||\mathbf r_v| ∣ r u ∣∣ r v ∣ ∣ r u × r v ∣ |\mathbf r_u\times\mathbf r_v| ∣ r u × r v ∣ θ = 90 ° \theta=90° θ = 90° ∣ r u × r v ∣ = ∣ r u ∣ ∣ r v ∣ sin θ ≤ ∣ r u ∣ ∣ r v ∣ |\mathbf r_u\times\mathbf r_v|=|\mathbf r_u||\mathbf r_v|\sin\theta \le |\mathbf r_u||\mathbf r_v| ∣ r u × r v ∣ = ∣ r u ∣∣ r v ∣ sin θ ≤ ∣ r u ∣∣ r v ∣
Recall Feynman: explain to a 12-year-oldImagine wrapping a bumpy hill with a stretchy grid blanket made of tiny squares. On the flat ground each square is the same size, but when you press it onto the hill the squares get pulled and tilted, so each one covers more (or less) hill. To find how much hill there is, you measure how stretched each tiny square got and add them all up. The "stretch number" for each square is found by taking two little arrows along its sides and seeing how much area they fence off — that's the cross product. The flat plane that just touches the hill at a point, like a sheet of glass resting on it, is the tangent plane, and its perpendicular direction is found the same way.
CR oss for AR ea, N ormal is the same CR oss"
C ross → A rea factor ∣ r u × r v ∣ |\mathbf r_u\times\mathbf r_v| ∣ r u × r v ∣ N ormal r u × r v \mathbf r_u\times\mathbf r_v r u × r v same cross product. One vector (normal) for planes, its length for area.
How many parameters describe a surface, and why? Two — a surface is intrinsically 2-dimensional; you can move in two independent directions on it.
What are r u \mathbf r_u r u r v \mathbf r_v r v Tangent vectors to the
u u u -curve and
v v v -curve; they lie in the surface and span the tangent plane.
What is the normal vector to a parametric surface? n = r u × r v \mathbf n=\mathbf r_u\times\mathbf r_v n = r u × r v (perpendicular to both tangent vectors).
State the tangent plane equation. a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 a(x-x_0)+b(y-y_0)+c(z-z_0)=0 a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 with
⟨ a , b , c ⟩ = r u × r v \langle a,b,c\rangle=\mathbf r_u\times\mathbf r_v ⟨ a , b , c ⟩ = r u × r v at the point.
Give the surface area formula and the meaning of its integrand. A = ∬ D ∣ r u × r v ∣ d A A=\iint_D|\mathbf r_u\times\mathbf r_v|\,dA A = ∬ D ∣ r u × r v ∣ d A ; the integrand is the local area-stretch factor.
Why a cross product (not dot) for area? Area = parallelogram spread =
∣ r u ∣ ∣ r v ∣ sin θ |\mathbf r_u||\mathbf r_v|\sin\theta ∣ r u ∣∣ r v ∣ sin θ , which the cross-product magnitude gives; dot gives alignment.
Surface area formula for a graph z = f ( x , y ) z=f(x,y) z = f ( x , y ) A = ∬ D f x 2 + f y 2 + 1 d A A=\iint_D\sqrt{f_x^2+f_y^2+1}\,dA A = ∬ D f x 2 + f y 2 + 1 d A .
Condition for a surface to be smooth at a point? r u × r v ≠ 0 \mathbf r_u\times\mathbf r_v\neq\mathbf 0 r u × r v = 0 there.
Surface area of a sphere of radius R R R ∫ 0 2 π ∫ 0 π R 2 sin ϕ d ϕ d θ = 4 π R 2 \int_0^{2\pi}\!\int_0^\pi R^2\sin\phi\,d\phi\,d\theta=4\pi R^2 ∫ 0 2 π ∫ 0 π R 2 sin ϕ d ϕ d θ = 4 π R 2 .
Cross Product — source of both the normal vector and area factor.Tangent Lines and Velocity Vectors — 1-parameter analogue (r ′ ( t ) \mathbf r'(t) r ′ ( t ) Double Integrals — the machinery that sums tiny patch areas.Change of Variables and the Jacobian — ∣ r u × r v ∣ |\mathbf r_u\times\mathbf r_v| ∣ r u × r v ∣ Surface Integrals and Flux — uses r u × r v \mathbf r_u\times\mathbf r_v r u × r v Arc Length — 1D cousin: ∫ ∣ r ′ ( t ) ∣ d t \int|\mathbf r'(t)|\,dt ∫ ∣ r ′ ( t ) ∣ d t ∬ ∣ r u × r v ∣ d A \iint|\mathbf r_u\times\mathbf r_v|\,dA ∬ ∣ r u × r v ∣ d A
magnitude gives patch area
Parametric surface r of u,v
dS = magnitude of r_u x r_v
Intuition Hinglish mein samjhoDekho, ek surface ko describe karne ke liye humein do parameters chahiye — u u u v v v r ( u , v ) = ⟨ x , y , z ⟩ \mathbf r(u,v)=\langle x,y,z\rangle r ( u , v ) = ⟨ x , y , z ⟩ v v v u u u r u \mathbf r_u r u r v \mathbf r_v r v
Ab normal kaise nikaalein? Humein ek vector chahiye jo dono r u \mathbf r_u r u r v \mathbf r_v r v n = r u × r v \mathbf n=\mathbf r_u\times\mathbf r_v n = r u × r v a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 a(x-x_0)+b(y-y_0)+c(z-z_0)=0 a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0
Surface area ka idea simple hai: parameter plane ko chhoti rectangles me kaato. Har rectangle surface pe ek tilted parallelogram ban jaata hai jiske do edges r u d u \mathbf r_u\,du r u d u r v d v \mathbf r_v\,dv r v d v = ∣ r u × r v ∣ d u d v =|\mathbf r_u\times\mathbf r_v|\,du\,dv = ∣ r u × r v ∣ d u d v A = ∬ D ∣ r u × r v ∣ d A A=\iint_D|\mathbf r_u\times\mathbf r_v|\,dA A = ∬ D ∣ r u × r v ∣ d A
Yaad rakho: area ke liye cross product, dot nahi — dot to alignment batata hai, area to spread chahiye. Aur parameter plane ka simple d A dA d A ∣ r u × r v ∣ |\mathbf r_u\times\mathbf r_v| ∣ r u × r v ∣ R 2 sin ϕ R^2\sin\phi R 2 sin ϕ = 4 π R 2 =4\pi R^2 = 4 π R 2
