Normal Vector
A normal vector to a surface at a point is a non-zero vector perpendicular to the surface at that point.
A surface can be defined in 3 different forms.
A surface σ \sigma σ
r ( u , v ) = x ( u , v ) i ^ + y ( u , v ) j ^ + z ( u , v ) k ^ , ( u , v ) ∈ R \boldsymbol{r}(u,v) = x(u,v)\hat{i} + y(u,v)\hat{j} + z(u,v)\hat{k}, \quad (u,v) \in R r ( u , v ) = x ( u , v ) i ^ + y ( u , v ) j ^ + z ( u , v ) k ^ , ( u , v ) ∈ R Here:
R R R region in the u v uv uv x , y , z x, y, z x , y , z component functions of u u u v v v
A normal vector to σ \sigma σ
N = ∂ r ∂ u × ∂ r ∂ v \boldsymbol{N} = \frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v} N = ∂ u ∂ r × ∂ v ∂ r Examples:
Sphere of radius a a a r ( ϕ , θ ) = a sin ϕ cos θ i ^ + a sin ϕ sin θ j ^ + a cos ϕ k ^ \boldsymbol{r}(\phi,\theta) = a\sin\phi\cos\theta\,\hat{i} + a\sin\phi\sin\theta\,\hat{j} + a\cos\phi\,\hat{k} r ( ϕ , θ ) = a sin ϕ cos θ i ^ + a sin ϕ sin θ j ^ + a cos ϕ k ^
Plane through r 0 \boldsymbol{r}_0 r 0 r ( u , v ) = r 0 + u a + v b \boldsymbol{r}(u,v) = \boldsymbol{r}_0 + u\boldsymbol{a} + v\boldsymbol{b} r ( u , v ) = r 0 + u a + v b
A surface σ \sigma σ ( x , y , z ) (x,y,z) ( x , y , z )
G ( x , y , z ) = 0 G(x,y,z) = 0 G ( x , y , z ) = 0 A normal vector at ( x , y , z ) (x,y,z) ( x , y , z ) σ \sigma σ
N = ∇ G ( x , y , z ) \boldsymbol{N} = \nabla G(x,y,z) N = ∇ G ( x , y , z ) Examples:
Sphere x 2 + y 2 + z 2 − a 2 = 0 x^2 + y^2 + z^2 - a^2 = 0 x 2 + y 2 + z 2 − a 2 = 0
Plane x + y + z − 1 = 0 x + y + z - 1 = 0 x + y + z − 1 = 0
A surface σ \sigma σ
z = g ( x , y ) , ( x , y ) ∈ R z = g(x,y), \quad (x,y) \in R z = g ( x , y ) , ( x , y ) ∈ R Here R R R σ \sigma σ x y xy x y
An upward-pointing normal vector is
N = ⟨ − g x ( x , y ) , − g y ( x , y ) , 1 ⟩ \boldsymbol{N} = \langle -g_x(x,y), -g_y(x,y), 1 \rangle N = ⟨ − g x ( x , y ) , − g y ( x , y ) , 1 ⟩ A graph is a parametric surface with r ( x , y ) = x i ^ + y j ^ + g ( x , y ) k ^ \boldsymbol{r}(x,y) = x\hat{i} + y\hat{j} + g(x,y)\hat{k} r ( x , y ) = x i ^ + y j ^ + g ( x , y ) k ^ G ( x , y , z ) = 0 G(x,y,z) = 0 G ( x , y , z ) = 0 G z ≠ 0 G_z \neq 0 G z = 0 z z z
Examples:
Paraboloid z = x 2 + y 2 z = x^2 + y^2 z = x 2 + y 2
Plane z = 1 − x − y z = 1 - x - y z = 1 − x − y
Smoothness
A surface is smooth iff it has a non-zero normal vector at each point. A surface is piecewise smooth iff it is composed of finitely many smooth pieces.
Closed Surface A surface σ \sigma σ iff it is the boundary of a solid region in R 3 \mathbb{R}^3 R 3
Surface Integral
Let f f f σ \sigma σ σ \sigma σ n n n k k k Δ S k \Delta S_k Δ S k ( x k ∗ , y k ∗ , z k ∗ ) (x_k^*, y_k^*, z_k^*) ( x k ∗ , y k ∗ , z k ∗ ) k k k f f f σ \sigma σ
∬ σ f ( x , y , z ) d S = lim n → ∞ ∑ k = 1 n f ( x k ∗ , y k ∗ , z k ∗ ) Δ S k \iint_\sigma f(x,y,z)\,\text{d}S = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*, y_k^*, z_k^*)\,\Delta S_k ∬ σ f ( x , y , z ) d S = n → ∞ lim k = 1 ∑ n f ( x k ∗ , y k ∗ , z k ∗ ) Δ S k provided the limit exists and does not depend on the partition or the choice of sample points.
The limit is evaluated through the parametric or graph form of σ \sigma σ
For a Parametric Surface
Let σ \sigma σ r ( u , v ) = x ( u , v ) i ^ + y ( u , v ) j ^ + z ( u , v ) k ^ \boldsymbol{r}(u,v) = x(u,v)\hat{i} + y(u,v)\hat{j} + z(u,v)\hat{k} r ( u , v ) = x ( u , v ) i ^ + y ( u , v ) j ^ + z ( u , v ) k ^ ( u , v ) (u,v) ( u , v ) R R R u v uv uv f f f σ \sigma σ
∬ σ f ( x , y , z ) d S = ∬ R f ( x ( u , v ) , y ( u , v ) , z ( u , v ) ) ∥ ∂ r ∂ u × ∂ r ∂ v ∥ d A u v \iint_\sigma f(x,y,z)\,\text{d}S = \iint_R f(x(u,v),y(u,v),z(u,v)) \left\lVert \frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v} \right\rVert \text{d}A_{uv} ∬ σ f ( x , y , z ) d S = ∬ R f ( x ( u , v ) , y ( u , v ) , z ( u , v )) ∂ u ∂ r × ∂ v ∂ r d A uv For a Graph
Let σ \sigma σ z = g ( x , y ) z = g(x,y) z = g ( x , y ) R R R x y xy x y g g g g x g_x g x g y g_y g y R R R f f f σ \sigma σ
∬ σ f ( x , y , z ) d S = ∬ R f ( x , y , g ( x , y ) ) 1 + ( g x ( x , y ) ) 2 + ( g y ( x , y ) ) 2 d A \iint_\sigma f(x,y,z)\,\text{d}S = \iint_R f(x,y,g(x,y)) \sqrt{1 + (g_x(x,y))^2 + (g_y(x,y))^2}\,\text{d}A ∬ σ f ( x , y , z ) d S = ∬ R f ( x , y , g ( x , y )) 1 + ( g x ( x , y ) ) 2 + ( g y ( x , y ) ) 2 d A Worked Examples
Part of a Plane
Evaluate ∬ σ z d S \iint_\sigma z\,\text{d}S ∬ σ z d S σ \sigma σ x + y + z = 1 x + y + z = 1 x + y + z = 1
Solution Reveal Hide Write σ \sigma σ
z = g ( x , y ) = 1 − x − y z = g(x,y) = 1 - x - y z = g ( x , y ) = 1 − x − y Compute the area factor.
g x = − 1 , g y = − 1 , 1 + g x 2 + g y 2 = 3 g_x = -1, \quad g_y = -1, \quad \sqrt{1 + g_x^2 + g_y^2} = \sqrt{3} g x = − 1 , g y = − 1 , 1 + g x 2 + g y 2 = 3 The projection R R R x y xy x y 0 ≤ x ≤ 1 0 \le x \le 1 0 ≤ x ≤ 1 0 ≤ y ≤ 1 − x 0 \le y \le 1 - x 0 ≤ y ≤ 1 − x
∬ σ z d S = ∬ R ( 1 − x − y ) 3 d A = 3 ∫ 0 1 ∫ 0 1 − x ( 1 − x − y ) d y d x \iint_\sigma z\,\text{d}S = \iint_R (1 - x - y)\sqrt{3}\,\text{d}A = \sqrt{3} \int_0^1 \int_0^{1-x} (1 - x - y)\,\text{d}y\,\text{d}x ∬ σ z d S = ∬ R ( 1 − x − y ) 3 d A = 3 ∫ 0 1 ∫ 0 1 − x ( 1 − x − y ) d y d x Integrate with respect to y y y
∫ 0 1 − x ( 1 − x − y ) d y = ( 1 − x ) 2 2 \int_0^{1-x} (1 - x - y)\,\text{d}y = \frac{(1-x)^2}{2} ∫ 0 1 − x ( 1 − x − y ) d y = 2 ( 1 − x ) 2 Integrate with respect to x x x
∬ σ z d S = 3 ∫ 0 1 ( 1 − x ) 2 2 d x = 3 6 \iint_\sigma z\,\text{d}S = \sqrt{3} \int_0^1 \frac{(1-x)^2}{2}\,\text{d}x = \frac{\sqrt{3}}{6} ∬ σ z d S = 3 ∫ 0 1 2 ( 1 − x ) 2 d x = 6 3
Sphere
Evaluate ∬ σ z 2 d S \iint_\sigma z^2\,\text{d}S ∬ σ z 2 d S σ \sigma σ x 2 + y 2 + z 2 = a 2 x^2 + y^2 + z^2 = a^2 x 2 + y 2 + z 2 = a 2 a > 0 a > 0 a > 0
Solution Reveal Hide Parametrize σ \sigma σ
r ( ϕ , θ ) = a sin ϕ cos θ i ^ + a sin ϕ sin θ j ^ + a cos ϕ k ^ , 0 ≤ ϕ ≤ π , 0 ≤ θ ≤ 2 π \boldsymbol{r}(\phi,\theta) = a\sin\phi\cos\theta\,\hat{i} + a\sin\phi\sin\theta\,\hat{j} + a\cos\phi\,\hat{k}, \quad 0 \le \phi \le \pi, \ 0 \le \theta \le 2\pi r ( ϕ , θ ) = a sin ϕ cos θ i ^ + a sin ϕ sin θ j ^ + a cos ϕ k ^ , 0 ≤ ϕ ≤ π , 0 ≤ θ ≤ 2 π Differentiate.
∂ r ∂ ϕ = a cos ϕ cos θ i ^ + a cos ϕ sin θ j ^ − a sin ϕ k ^ \frac{\partial \boldsymbol{r}}{\partial \phi} = a\cos\phi\cos\theta\,\hat{i} + a\cos\phi\sin\theta\,\hat{j} - a\sin\phi\,\hat{k} ∂ ϕ ∂ r = a cos ϕ cos θ i ^ + a cos ϕ sin θ j ^ − a sin ϕ k ^ ∂ r ∂ θ = − a sin ϕ sin θ i ^ + a sin ϕ cos θ j ^ \frac{\partial \boldsymbol{r}}{\partial \theta} = -a\sin\phi\sin\theta\,\hat{i} + a\sin\phi\cos\theta\,\hat{j} ∂ θ ∂ r = − a sin ϕ sin θ i ^ + a sin ϕ cos θ j ^ Compute the cross product.
∂ r ∂ ϕ × ∂ r ∂ θ = a 2 sin ϕ ( sin ϕ cos θ i ^ + sin ϕ sin θ j ^ + cos ϕ k ^ ) \frac{\partial \boldsymbol{r}}{\partial \phi} \times \frac{\partial \boldsymbol{r}}{\partial \theta} = a^2\sin\phi\left(\sin\phi\cos\theta\,\hat{i} + \sin\phi\sin\theta\,\hat{j} + \cos\phi\,\hat{k}\right) ∂ ϕ ∂ r × ∂ θ ∂ r = a 2 sin ϕ ( sin ϕ cos θ i ^ + sin ϕ sin θ j ^ + cos ϕ k ^ ) The vector in brackets is a unit vector and sin ϕ ≥ 0 \sin\phi \ge 0 sin ϕ ≥ 0 a 2 sin ϕ a^2\sin\phi a 2 sin ϕ
Swapping the order of the cross product reverses the vector, but the norm stays a 2 sin ϕ a^2\sin\phi a 2 sin ϕ
On σ \sigma σ z = a cos ϕ z = a\cos\phi z = a cos ϕ
∬ σ z 2 d S = ∫ 0 2 π ∫ 0 π a 2 cos 2 ϕ ⋅ a 2 sin ϕ d ϕ d θ \iint_\sigma z^2\,\text{d}S = \int_0^{2\pi} \int_0^\pi a^2\cos^2\phi \cdot a^2\sin\phi\,\text{d}\phi\,\text{d}\theta ∬ σ z 2 d S = ∫ 0 2 π ∫ 0 π a 2 cos 2 ϕ ⋅ a 2 sin ϕ d ϕ d θ Integrate with respect to ϕ \phi ϕ
∫ 0 π cos 2 ϕ sin ϕ d ϕ = [ − cos 3 ϕ 3 ] 0 π = 2 3 \int_0^\pi \cos^2\phi\sin\phi\,\text{d}\phi = \left[ -\frac{\cos^3\phi}{3} \right]_0^\pi = \frac{2}{3} ∫ 0 π cos 2 ϕ sin ϕ d ϕ = [ − 3 cos 3 ϕ ] 0 π = 3 2 Integrate with respect to θ \theta θ
∬ σ z 2 d S = a 4 ⋅ 2 π ⋅ 2 3 = 4 π a 4 3 \iint_\sigma z^2\,\text{d}S = a^4 \cdot 2\pi \cdot \frac{2}{3} = \frac{4\pi a^4}{3} ∬ σ z 2 d S = a 4 ⋅ 2 π ⋅ 3 2 = 3 4 π a 4
Area of a Paraboloid
Find the area of the part of the paraboloid z = x 2 + y 2 z = x^2 + y^2 z = x 2 + y 2 z = 1 z = 1 z = 1
Solution Reveal Hide The area is ∬ σ 1 d S \iint_\sigma 1\,\text{d}S ∬ σ 1 d S
Graph Form
Write σ \sigma σ
z = g ( x , y ) = x 2 + y 2 , g x = 2 x , g y = 2 y z = g(x,y) = x^2 + y^2, \quad g_x = 2x, \quad g_y = 2y z = g ( x , y ) = x 2 + y 2 , g x = 2 x , g y = 2 y The paraboloid meets z = 1 z = 1 z = 1 x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1 R R R
S = ∬ R 1 + 4 x 2 + 4 y 2 d A S = \iint_R \sqrt{1 + 4x^2 + 4y^2}\,\text{d}A S = ∬ R 1 + 4 x 2 + 4 y 2 d A The integrand depends only on x 2 + y 2 = r 2 x^2 + y^2 = r^2 x 2 + y 2 = r 2 d A = r d r d θ \text{d}A = r\,\text{d}r\,\text{d}\theta d A = r d r d θ
S = ∫ 0 2 π ∫ 0 1 1 + 4 r 2 r d r d θ S = \int_0^{2\pi} \int_0^1 \sqrt{1 + 4r^2}\,r\,\text{d}r\,\text{d}\theta S = ∫ 0 2 π ∫ 0 1 1 + 4 r 2 r d r d θ Substitute u = 1 + 4 r 2 u = 1 + 4r^2 u = 1 + 4 r 2 r d r = 1 8 d u r\,\text{d}r = \frac{1}{8}\text{d}u r d r = 8 1 d u u u u 1 1 1 5 5 5
∫ 0 1 1 + 4 r 2 r d r = 1 8 ∫ 1 5 u 1 / 2 d u = 1 8 ⋅ 2 3 ( 5 5 − 1 ) = 5 5 − 1 12 \int_0^1 \sqrt{1 + 4r^2}\,r\,\text{d}r = \frac{1}{8} \int_1^5 u^{1/2}\,\text{d}u = \frac{1}{8} \cdot \frac{2}{3}\left(5\sqrt{5} - 1\right) = \frac{5\sqrt{5} - 1}{12} ∫ 0 1 1 + 4 r 2 r d r = 8 1 ∫ 1 5 u 1/2 d u = 8 1 ⋅ 3 2 ( 5 5 − 1 ) = 12 5 5 − 1 Integrate with respect to θ \theta θ
S = 2 π ⋅ 5 5 − 1 12 = π ( 5 5 − 1 ) 6 S = 2\pi \cdot \frac{5\sqrt{5} - 1}{12} = \frac{\pi\left(5\sqrt{5} - 1\right)}{6} S = 2 π ⋅ 12 5 5 − 1 = 6 π ( 5 5 − 1 ) Parametric Form
Parametrize σ \sigma σ r r r θ \theta θ
r ( r , θ ) = r cos θ i ^ + r sin θ j ^ + r 2 k ^ , 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π \boldsymbol{r}(r,\theta) = r\cos\theta\,\hat{i} + r\sin\theta\,\hat{j} + r^2\,\hat{k}, \quad 0 \le r \le 1, \ 0 \le \theta \le 2\pi r ( r , θ ) = r cos θ i ^ + r sin θ j ^ + r 2 k ^ , 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π Differentiate.
∂ r ∂ r = cos θ i ^ + sin θ j ^ + 2 r k ^ , ∂ r ∂ θ = − r sin θ i ^ + r cos θ j ^ \frac{\partial \boldsymbol{r}}{\partial r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j} + 2r\,\hat{k}, \quad \frac{\partial \boldsymbol{r}}{\partial \theta} = -r\sin\theta\,\hat{i} + r\cos\theta\,\hat{j} ∂ r ∂ r = cos θ i ^ + sin θ j ^ + 2 r k ^ , ∂ θ ∂ r = − r sin θ i ^ + r cos θ j ^ Compute the cross product.
∂ r ∂ r × ∂ r ∂ θ = − 2 r 2 cos θ i ^ − 2 r 2 sin θ j ^ + r k ^ \frac{\partial \boldsymbol{r}}{\partial r} \times \frac{\partial \boldsymbol{r}}{\partial \theta} = -2r^2\cos\theta\,\hat{i} - 2r^2\sin\theta\,\hat{j} + r\,\hat{k} ∂ r ∂ r × ∂ θ ∂ r = − 2 r 2 cos θ i ^ − 2 r 2 sin θ j ^ + r k ^ Take the norm. Since r ≥ 0 r \ge 0 r ≥ 0
∥ ∂ r ∂ r × ∂ r ∂ θ ∥ = 4 r 4 + r 2 = r 1 + 4 r 2 \left\lVert \frac{\partial \boldsymbol{r}}{\partial r} \times \frac{\partial \boldsymbol{r}}{\partial \theta} \right\rVert = \sqrt{4r^4 + r^2} = r\sqrt{1 + 4r^2} ∂ r ∂ r × ∂ θ ∂ r = 4 r 4 + r 2 = r 1 + 4 r 2 Integrate over the unit disk.
S = ∫ 0 2 π ∫ 0 1 r 1 + 4 r 2 d r d θ = π ( 5 5 − 1 ) 6 S = \int_0^{2\pi} \int_0^1 r\sqrt{1 + 4r^2}\,\text{d}r\,\text{d}\theta = \frac{\pi\left(5\sqrt{5} - 1\right)}{6} S = ∫ 0 2 π ∫ 0 1 r 1 + 4 r 2 d r d θ = 6 π ( 5 5 − 1 ) 