Intuition The big pictureA line integral adds up a quantity along a curve (1D). A surface integral adds up a quantity over a surface (2D) living in 3D space.
Scalar surface integral ∬ S f d S \iint_S f\,dS ∬ S f d S f f f density (mass per area) of a curved sheet, what is its total mass?"Vector surface integral / flux ∬ S F ⋅ d S \iint_S \mathbf{F}\cdot d\mathbf{S} ∬ S F ⋅ d S F \mathbf{F} F through the surface per second?"
The whole trick is: a curved surface is hard, but a flat parameter rectangle is easy. We map a flat ( u , v ) (u,v) ( u , v ) stretches area.
Definition Parametrized surfaceA surface is a vector map from a 2D region D D D u v uv uv r ( u , v ) = ( x ( u , v ) , y ( u , v ) , z ( u , v ) ) , ( u , v ) ∈ D . \mathbf{r}(u,v) = \big(x(u,v),\, y(u,v),\, z(u,v)\big), \qquad (u,v)\in D. r ( u , v ) = ( x ( u , v ) , y ( u , v ) , z ( u , v ) ) , ( u , v ) ∈ D . v v v u u u u u u
WHY a stretch factor? When we move the parameters by tiny amounts d u du d u d v dv d v
r u = ∂ r ∂ u , r v = ∂ r ∂ v . \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u}, \qquad \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v}. r u = ∂ u ∂ r , r v = ∂ v ∂ r .
The little parameter square d u d v du\,dv d u d v parallelogram spanned by r u d u \mathbf{r}_u\,du r u d u r v d v \mathbf{r}_v\,dv r v d v
Intuition Area of a parallelogram = magnitude of cross productThe area of the parallelogram built from vectors a \mathbf a a b \mathbf b b ∥ a × b ∥ \|\mathbf a\times\mathbf b\| ∥ a × b ∥ ∥ a × b ∥ = ∥ a ∥ ∥ b ∥ sin θ \|\mathbf a\times\mathbf b\| = \|\mathbf a\|\|\mathbf b\|\sin\theta ∥ a × b ∥ = ∥ a ∥∥ b ∥ sin θ
So the tiny patch of real surface area is
d S = ∥ r u × r v ∥ d u d v . dS = \|\mathbf{r}_u \times \mathbf{r}_v\|\; du\,dv. d S = ∥ r u × r v ∥ d u d v .
This ∥ r u × r v ∥ \|\mathbf{r}_u \times \mathbf{r}_v\| ∥ r u × r v ∥ area scaling factor
If f = 1 f=1 f = 1 surface area ∬ S d S \iint_S dS ∬ S d S
If f = f= f = total mass .
Worked example Graph surfacez = g ( x , y ) z=g(x,y) z = g ( x , y ) u = x u=x u = x v = y v=y v = y r ( x , y ) = ( x , y , g ( x , y ) ) \mathbf{r}(x,y)=(x,y,g(x,y)) r ( x , y ) = ( x , y , g ( x , y )) r x = ( 1 , 0 , g x ) , r y = ( 0 , 1 , g y ) . \mathbf{r}_x=(1,0,g_x),\quad \mathbf{r}_y=(0,1,g_y). r x = ( 1 , 0 , g x ) , r y = ( 0 , 1 , g y ) . Why this step? These are the tangent vectors along the x x x y y y r x × r y = ( − g x , − g y , 1 ) ⇒ ∥ r x × r y ∥ = 1 + g x 2 + g y 2 . \mathbf{r}_x\times\mathbf{r}_y=\big(-g_x,\,-g_y,\,1\big)\;\Rightarrow\;\|\mathbf{r}_x\times\mathbf{r}_y\|=\sqrt{1+g_x^2+g_y^2}. r x × r y = ( − g x , − g y , 1 ) ⇒ ∥ r x × r y ∥ = 1 + g x 2 + g y 2 . Why this step? Cross product gives the normal; its length is the stretch factor.
∬ S f d S = ∬ D f ( x , y , g ) 1 + g x 2 + g y 2 d x d y . \iint_S f\,dS=\iint_D f(x,y,g)\sqrt{1+g_x^2+g_y^2}\,dx\,dy. ∬ S f d S = ∬ D f ( x , y , g ) 1 + g x 2 + g y 2 d x d y .
Worked example Area of a hemisphere of radiusa a a x 2 + y 2 + z 2 = a 2 x^2+y^2+z^2=a^2 x 2 + y 2 + z 2 = a 2 z = a 2 − x 2 − y 2 z=\sqrt{a^2-x^2-y^2} z = a 2 − x 2 − y 2 D : x 2 + y 2 ≤ a 2 D:\,x^2+y^2\le a^2 D : x 2 + y 2 ≤ a 2 g x = − x a 2 − x 2 − y 2 g_x=\dfrac{-x}{\sqrt{a^2-x^2-y^2}} g x = a 2 − x 2 − y 2 − x g y = − y a 2 − x 2 − y 2 g_y=\dfrac{-y}{\sqrt{a^2-x^2-y^2}} g y = a 2 − x 2 − y 2 − y Why? Differentiate g g g 1 + g x 2 + g y 2 = 1 + x 2 + y 2 a 2 − x 2 − y 2 = a 2 a 2 − x 2 − y 2 . 1+g_x^2+g_y^2=1+\frac{x^2+y^2}{a^2-x^2-y^2}=\frac{a^2}{a^2-x^2-y^2}. 1 + g x 2 + g y 2 = 1 + a 2 − x 2 − y 2 x 2 + y 2 = a 2 − x 2 − y 2 a 2 . d S = a a 2 − x 2 − y 2 d x d y dS=\dfrac{a}{\sqrt{a^2-x^2-y^2}}\,dx\,dy d S = a 2 − x 2 − y 2 a d x d y x 2 + y 2 = ρ 2 x^2+y^2=\rho^2 x 2 + y 2 = ρ 2
=2\pi a\Big[-\sqrt{a^2-\rho^2}\Big]_0^a = 2\pi a^2.$$
*Why polar?* The disk and integrand are radially symmetric → clean integral. Result $2\pi a^2$ = half of $4\pi a^2$. ✔
Definition Unit normal & orientationAn orientable surface has a continuous choice of unit normal
n ^ = r u × r v ∥ r u × r v ∥ . \hat{\mathbf n} = \frac{\mathbf{r}_u\times\mathbf{r}_v}{\|\mathbf{r}_u\times\mathbf{r}_v\|}. n ^ = ∥ r u × r v ∥ r u × r v . n ^ \hat{\mathbf n} n ^ − n ^ -\hat{\mathbf n} − n ^ orients the surface: it picks the "positive" side. (A Möbius strip is non-orientable — no consistent choice exists.)
Intuition What flux measuresImagine fluid with velocity field F \mathbf{F} F d S dS d S n ^ \hat{\mathbf n} n ^ F \mathbf F F along n ^ \hat{\mathbf n} n ^ carries fluid through the surface. Sideways flow slides along, not through.
So the flow rate through the patch is ( F ⋅ n ^ ) d S (\mathbf{F}\cdot\hat{\mathbf n})\,dS ( F ⋅ n ^ ) d S
Intuition Why the magnitude cancellation is beautifulFor scalar integrals you must compute the (often ugly) square root ∥ r u × r v ∥ \|\mathbf r_u\times\mathbf r_v\| ∥ r u × r v ∥ never take that square root — you just dot F \mathbf F F r u × r v \mathbf r_u\times\mathbf r_v r u × r v easier than scalar integrals!
Worked example Flux through a graphz = g ( x , y ) z=g(x,y) z = g ( x , y ) r x × r y = ( − g x , − g y , 1 ) \mathbf{r}_x\times\mathbf{r}_y=(-g_x,-g_y,1) r x × r y = ( − g x , − g y , 1 ) upward since its z z z + 1 > 0 +1>0 + 1 > 0 F = ( P , Q , R ) \mathbf F=(P,Q,R) F = ( P , Q , R ) ∬ S F ⋅ d S = ∬ D ( − P g x − Q g y + R ) d x d y . \iint_S \mathbf F\cdot d\mathbf S=\iint_D\big(-P\,g_x - Q\,g_y + R\big)\,dx\,dy. ∬ S F ⋅ d S = ∬ D ( − P g x − Q g y + R ) d x d y .
F = ( x , y , z ) \mathbf F=(x,y,z) F = ( x , y , z ) r ( ϕ , θ ) = ( sin ϕ cos θ , sin ϕ sin θ , cos ϕ ) \mathbf r(\phi,\theta)=(\sin\phi\cos\theta,\ \sin\phi\sin\theta,\ \cos\phi) r ( ϕ , θ ) = ( sin ϕ cos θ , sin ϕ sin θ , cos ϕ ) ϕ ∈ [ 0 , π ] \phi\in[0,\pi] ϕ ∈ [ 0 , π ] θ ∈ [ 0 , 2 π ] \theta\in[0,2\pi] θ ∈ [ 0 , 2 π ] Why this step? On the unit sphere r \mathbf r r F = r \mathbf F=\mathbf r F = r r ϕ × r θ = sin ϕ r \mathbf r_\phi\times\mathbf r_\theta = \sin\phi\,\mathbf r r ϕ × r θ = sin ϕ r sin ϕ \sin\phi sin ϕ F ⋅ ( r ϕ × r θ ) = r ⋅ ( sin ϕ r ) = sin ϕ ∥ r ∥ 2 = sin ϕ \mathbf F\cdot(\mathbf r_\phi\times\mathbf r_\theta)=\mathbf r\cdot(\sin\phi\,\mathbf r)=\sin\phi\,\|\mathbf r\|^2=\sin\phi F ⋅ ( r ϕ × r θ ) = r ⋅ ( sin ϕ r ) = sin ϕ ∥ r ∥ 2 = sin ϕ ∥ r ∥ = 1 \|\mathbf r\|=1 ∥ r ∥ = 1 ∬ S F ⋅ d S = ∫ 0 2 π ∫ 0 π sin ϕ d ϕ d θ = 2 π ⋅ 2 = 4 π . \iint_S\mathbf F\cdot d\mathbf S=\int_0^{2\pi}\!\!\int_0^{\pi}\sin\phi\,d\phi\,d\theta = 2\pi\cdot 2 = 4\pi. ∬ S F ⋅ d S = ∫ 0 2 π ∫ 0 π sin ϕ d ϕ d θ = 2 π ⋅ 2 = 4 π . Check via Divergence Theorem: ∇ ⋅ F = 3 \nabla\cdot\mathbf F=3 ∇ ⋅ F = 3 = 4 3 π =\tfrac{4}{3}\pi = 3 4 π = 3 ⋅ 4 3 π = 4 π =3\cdot\tfrac{4}{3}\pi=4\pi = 3 ⋅ 3 4 π = 4 π
Common mistake "Flux needs the square root∥ r u × r v ∥ \|\mathbf r_u\times\mathbf r_v\| ∥ r u × r v ∥ Why it feels right: d S dS d S definition ∬ ( F ⋅ n ^ ) d S \iint(\mathbf F\cdot\hat{\mathbf n})dS ∬ ( F ⋅ n ^ ) d S d S dS d S The fix: n ^ d S = r u × r v ∥ ⋯ ∥ ⋅ ∥ ⋯ ∥ d u d v = ( r u × r v ) d u d v \hat{\mathbf n}\,dS = \frac{\mathbf r_u\times\mathbf r_v}{\|\cdots\|}\cdot\|\cdots\|\,du\,dv = (\mathbf r_u\times\mathbf r_v)\,du\,dv n ^ d S = ∥ ⋯ ∥ r u × r v ⋅ ∥ ⋯ ∥ d u d v = ( r u × r v ) d u d v F ⋅ ( r u × r v ) \mathbf F\cdot(\mathbf r_u\times\mathbf r_v) F ⋅ ( r u × r v ) no root .
Common mistake Forgetting that flux has a SIGN (orientation).Why it feels right: Scalar integrals ∬ f d S ≥ 0 \iint f\,dS\ge0 ∬ f d S ≥ 0 f ≥ 0 f\ge0 f ≥ 0 The fix: Flux can be negative. Swapping 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
F ⋅ r \mathbf F\cdot\mathbf r F ⋅ r F ⋅ ( r u × r v ) \mathbf F\cdot(\mathbf r_u\times\mathbf r_v) F ⋅ ( r u × r v ) Why it feels right: On the sphere r \mathbf r r is the normal direction, so it accidentally works there.
The fix: The vector area element is the cross product of the tangents , not the position vector. Only by luck do they align on the sphere.
f ( x , y , z ) f(x,y,z) f ( x , y , z ) Why it feels right: You "see" x , y , z x,y,z x , y , z f f f The fix: Everything must become a function of u , v u,v u , v x → x ( u , v ) x\to x(u,v) x → x ( u , v )
Recall Active recall — cover the answers
What is d S dS d S ∥ 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
Why is flux often easier than a scalar integral? → The norm cancels; no square root.
What does flux physically measure? → Net flow of a field through the surface.
What changes if you reverse orientation? → Flux flips sign; scalar integral unchanged.
Recall Feynman: explain to a 12-year-oldImagine a bumpy net held under a waterfall. Scalar integral: if I want to paint the whole net, how much paint do I need? I add up the area of every little square — but slanted squares are bigger than they look from above, so I stretch them out first (that stretch is ∥ r u × r v ∥ \|\mathbf r_u\times\mathbf r_v\| ∥ r u × r v ∥ Flux: if I want to know how much water gushes through the net, I only count water going straight through the holes, not water sliding along the net. The arrow sticking out of the net (n ^ \hat{\mathbf n} n ^
Mnemonic Remember the two formulas"Scalar Stretches, Flux Flows."
S calar → keep the S tretch factor ∥ r u × r v ∥ \|\mathbf r_u\times\mathbf r_v\| ∥ r u × r v ∥ F lux → F ield dotted into the cross product, norm F alls away.
What is the area element d S dS d S r ( u , v ) \mathbf r(u,v) r ( u , v ) d S = ∥ r u × r v ∥ d u d v dS=\|\mathbf r_u\times\mathbf r_v\|\,du\,dv d S = ∥ r u × r v ∥ d u d v For a graph z = g ( x , y ) z=g(x,y) z = g ( x , y ) ∥ r x × r y ∥ \|\mathbf r_x\times\mathbf r_y\| ∥ r x × r y ∥ 1 + g x 2 + g y 2 \sqrt{1+g_x^2+g_y^2} 1 + g x 2 + g y 2 State the scalar surface integral formula. ∬ S f d S = ∬ D f ( r ) ∥ r u × r v ∥ d u d v \iint_S f\,dS=\iint_D f(\mathbf r)\,\|\mathbf r_u\times\mathbf r_v\|\,du\,dv ∬ S f d S = ∬ D f ( r ) ∥ r u × r v ∥ d u d v State the flux integral formula in terms of parameters. ∬ S F ⋅ d S = ∬ D F ( r ) ⋅ ( r u × r v ) d u d v \iint_S\mathbf F\cdot d\mathbf S=\iint_D \mathbf F(\mathbf r)\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv ∬ S F ⋅ d S = ∬ D F ( r ) ⋅ ( r u × r v ) d u d v Why does flux avoid the square root that scalar integrals need? n ^ d S = r u × r v ∥ ⋯ ∥ ∥ ⋯ ∥ d u d v = ( r u × r v ) d u d v \hat{\mathbf n}\,dS=\frac{\mathbf r_u\times\mathbf r_v}{\|\cdots\|}\|\cdots\|du\,dv=(\mathbf r_u\times\mathbf r_v)du\,dv n ^ d S = ∥ ⋯ ∥ r u × r v ∥ ⋯ ∥ d u d v = ( r u × r v ) d u d v — the norm cancels
What physical quantity does the flux integral compute? The net flow of the field
F \mathbf F F through the surface (e.g. fluid volume per unit time)
What happens to flux if you reverse the surface's orientation? Its sign flips (the normal reverses)
What is the vector area element d S d\mathbf S d S d S = n ^ d S = ( r u × r v ) d u d v d\mathbf S=\hat{\mathbf n}\,dS=(\mathbf r_u\times\mathbf r_v)\,du\,dv d S = n ^ d S = ( r u × r v ) d u d v Flux of F = ( x , y , z ) \mathbf F=(x,y,z) F = ( x , y , z ) 4 π 4\pi 4 π (also from div thm:
3 ⋅ 4 3 π 3\cdot\frac43\pi 3 ⋅ 3 4 π )
Surface area of a sphere of radius a a a 4 π a 2 4\pi a^2 4 π a 2 (each hemisphere
2 π a 2 2\pi a^2 2 π a 2 )
What makes a surface non-orientable (give an example)? No continuous choice of unit normal; e.g. the Möbius strip
For a graph with upward normal, flux of ( P , Q , R ) (P,Q,R) ( P , Q , R ) ∬ D ( − P g x − Q g y + R ) d x d y \iint_D(-Pg_x-Qg_y+R)\,dx\,dy ∬ D ( − P g x − Q g y + R ) d x d y magnitude = parallelogram area
Parametrized surface r of u,v
Tangent vectors r_u and r_v
Graph formula sqrt 1+gx^2+gy^2
Line integral over curve 1D
Intuition Hinglish mein samjhoDekho, surface integral ka matlab simple hai: jaise line integral ek curve (1D) ke upar cheezein add karta hai, waise hi surface integral ek curved chaadar (2D) ke upar add karta hai. Problem ye hai ki curved surface pe directly integrate karna mushkil hai — isliye hum ek flat ( u , v ) (u,v) ( u , v ) d u du d u d v dv 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 stretch factor hai — bilkul Jacobian jaisa.
Do type hote hain. Scalar surface integral ∬ S f d S \iint_S f\,dS ∬ S f d S f f f ∥ r u × r v ∥ \|\mathbf r_u\times\mathbf r_v\| ∥ r u × r v ∥ zaroori hai (square root b
