Math 241: Multivariable calculus, Lecture 32 Parametric Surfaces and Integration on Surfaces, Section 16.6-16.7

go.illinois.edu/math241fa17

Monday, November 27th, 2017

go.illinois.edu/math241fa17. • Point: need parametrization of the curve. • Define the integral of vector field F over C: Z Z b F · dr = F(r(t)) · r0(t) dt. C a • Main theorem: Green says: Integral over the boundary is related to double integral over the interior:

Z ZZ F · dr = Qx − Py dA ∂D D

Review Line Integrals.

• Define the integral of function f over C Z Z b f ds = f (r(t)) kr0(t)k dt. C a

go.illinois.edu/math241fa17. • Define the integral of vector field F over C: Z Z b F · dr = F(r(t)) · r0(t) dt. C a • Main theorem: Green says: Integral over the boundary is related to double integral over the interior:

Z ZZ F · dr = Qx − Py dA ∂D D

Review Line Integrals.

• Define the integral of function f over C Z Z b f ds = f (r(t)) kr0(t)k dt. C a

• Point: need parametrization of the curve.

go.illinois.edu/math241fa17. • Main theorem: Green says: Integral over the boundary is related to double integral over the interior:

Z ZZ F · dr = Qx − Py dA ∂D D

Review Line Integrals.

• Define the integral of function f over C Z Z b f ds = f (r(t)) kr0(t)k dt. C a

• Point: need parametrization of the curve. • Define the integral of vector field F over C: Z Z b F · dr = F(r(t)) · r0(t) dt. C a

go.illinois.edu/math241fa17. Review Line Integrals.

• Define the integral of function f over C Z Z b f ds = f (r(t)) kr0(t)k dt. C a

• Point: need parametrization of the curve. • Define the integral of vector field F over C: Z Z b F · dr = F(r(t)) · r0(t) dt. C a • Main theorem: Green says: Integral over the boundary is related to double integral over the interior:

Z ZZ F · dr = Qx − Py dA ∂D D go.illinois.edu/math241fa17. • Parameterized Curve =⇒ Parameterized surface • Integral over Curve =⇒ Integral over Surface • Green’s Thm =⇒ ????

Goal

Goal: Generalize to next higher dimension: • Curve =⇒ Surface

go.illinois.edu/math241fa17. • Integral over Curve =⇒ Integral over Surface • Green’s Thm =⇒ ????

Goal

Goal: Generalize to next higher dimension: • Curve =⇒ Surface • Parameterized Curve =⇒ Parameterized surface

go.illinois.edu/math241fa17. • Green’s Thm =⇒ ????

Goal

Goal: Generalize to next higher dimension: • Curve =⇒ Surface • Parameterized Curve =⇒ Parameterized surface • Integral over Curve =⇒ Integral over Surface

go.illinois.edu/math241fa17. Goal

Goal: Generalize to next higher dimension: • Curve =⇒ Surface • Parameterized Curve =⇒ Parameterized surface • Integral over Curve =⇒ Integral over Surface • Green’s Thm =⇒ ????

go.illinois.edu/math241fa17. Parametric surfaces

2 A parametric surface S ⊂ R is described by an injective map (the parameterization): 3 r : D → R , (u, v) 7→ (x(u, v), y(u, v), z(u, v)) where x(u, v), y(u, v) and z(u, v) are differentiable functions and 2 D ⊂ R is a domain.

z (x(u, v),y(u, v),z(u, v)) ~r

v D S (u, v)

u x y

go.illinois.edu/math241fa17. • the portion of the plane x + y + z = 1 in the first octant; • the cone z2 + y 2 = x2, x ≥ 0; • the hemisphere x2 + y 2 + z2 = 3, y ≥ 0; 3 • the parallelogram in R with vertices (0, 0, 0), (1, 1, 0), (1, 1, 2) and (2, 2, 2)

Examples of parametric surfaces

Find a parameterization for the following surfaces:

.

go.illinois.edu/math241fa17. • the cone z2 + y 2 = x2, x ≥ 0; • the hemisphere x2 + y 2 + z2 = 3, y ≥ 0; 3 • the parallelogram in R with vertices (0, 0, 0), (1, 1, 0), (1, 1, 2) and (2, 2, 2)

Examples of parametric surfaces

Find a parameterization for the following surfaces: • the portion of the plane x + y + z = 1 in the first octant;

.

go.illinois.edu/math241fa17. • the hemisphere x2 + y 2 + z2 = 3, y ≥ 0; 3 • the parallelogram in R with vertices (0, 0, 0), (1, 1, 0), (1, 1, 2) and (2, 2, 2)

Examples of parametric surfaces

Find a parameterization for the following surfaces: • the portion of the plane x + y + z = 1 in the first octant; • the cone z2 + y 2 = x2, x ≥ 0;

.

go.illinois.edu/math241fa17. 3 • the parallelogram in R with vertices (0, 0, 0), (1, 1, 0), (1, 1, 2) and (2, 2, 2)

Examples of parametric surfaces

Find a parameterization for the following surfaces: • the portion of the plane x + y + z = 1 in the first octant; • the cone z2 + y 2 = x2, x ≥ 0; • the hemisphere x2 + y 2 + z2 = 3, y ≥ 0;

.

go.illinois.edu/math241fa17. Examples of parametric surfaces

Find a parameterization for the following surfaces: • the portion of the plane x + y + z = 1 in the first octant; • the cone z2 + y 2 = x2, x ≥ 0; • the hemisphere x2 + y 2 + z2 = 3, y ≥ 0; 3 • the parallelogram in R with vertices (0, 0, 0), (1, 1, 0), (1, 1, 2) and (2, 2, 2).

go.illinois.edu/math241fa17. • x is positive, so x = pz2 + y 2. Use polar coordinates in the z, y-plane: r(ρ, θ) = (ρ, ρ cos(θ), ρ sin(θ)). No restriction of the parameters. √ • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x, z plane: p √ r(ρ, θ) = (ρ cos(θ), 3 − ρ2, ρ sin(θ)). 0 ≤ ρ ≤ 3, 0 ≤ θ ≤ 2π • Use s, t as coordinates in the direction of the sides: r(s, t) = s(1, 1, 0) + t(1, 1, 2) = (s + t, s + t, 2t), with 0 ≤ s, t ≤ 1

Solutions

• z = 1 − x − y, so use x, y as parameters. So r(u, v) = (u, v, 1 − u − v), for 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u.

go.illinois.edu/math241fa17. √ • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x, z plane: p √ r(ρ, θ) = (ρ cos(θ), 3 − ρ2, ρ sin(θ)). 0 ≤ ρ ≤ 3, 0 ≤ θ ≤ 2π • Use s, t as coordinates in the direction of the sides: r(s, t) = s(1, 1, 0) + t(1, 1, 2) = (s + t, s + t, 2t), with 0 ≤ s, t ≤ 1

Solutions

• z = 1 − x − y, so use x, y as parameters. So r(u, v) = (u, v, 1 − u − v), for 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u. • x is positive, so x = pz2 + y 2. Use polar coordinates in the z, y-plane: r(ρ, θ) = (ρ, ρ cos(θ), ρ sin(θ)). No restriction of the parameters.

go.illinois.edu/math241fa17. • Use s, t as coordinates in the direction of the sides: r(s, t) = s(1, 1, 0) + t(1, 1, 2) = (s + t, s + t, 2t), with 0 ≤ s, t ≤ 1

Solutions

• z = 1 − x − y, so use x, y as parameters. So r(u, v) = (u, v, 1 − u − v), for 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u. • x is positive, so x = pz2 + y 2. Use polar coordinates in the z, y-plane: r(ρ, θ) = (ρ, ρ cos(θ), ρ sin(θ)). No restriction of the parameters. √ • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x, z plane: p √ r(ρ, θ) = (ρ cos(θ), 3 − ρ2, ρ sin(θ)). 0 ≤ ρ ≤ 3, 0 ≤ θ ≤ 2π

go.illinois.edu/math241fa17. Solutions

• z = 1 − x − y, so use x, y as parameters. So r(u, v) = (u, v, 1 − u − v), for 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u. • x is positive, so x = pz2 + y 2. Use polar coordinates in the z, y-plane: r(ρ, θ) = (ρ, ρ cos(θ), ρ sin(θ)). No restriction of the parameters. √ • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x, z plane: p √ r(ρ, θ) = (ρ cos(θ), 3 − ρ2, ρ sin(θ)). 0 ≤ ρ ≤ 3, 0 ≤ θ ≤ 2π • Use s, t as coordinates in the direction of the sides: r(s, t) = s(1, 1, 0) + t(1, 1, 2) = (s + t, s + t, 2t), with 0 ≤ s, t ≤ 1

go.illinois.edu/math241fa17. Tangent plane and normal to a surface

3 For a parameterized surface r : D → R , the tangent plane at r(u, v) is spanned by the tangent vectors:

ru = hxu, yu, zui, rv = hxv , yv , zv i, and a normal vector at r(u, v) is given by:

n = ru × rv . z ~n ~r ~rv ~ru v D (u, v) S

u x y

go.illinois.edu/math241fa17. In particular, the area of S is defined by:

Area(S) = x 1 dS = x |ru × rv | dA. S D

Motivation: Recall the change of variables formula!

Surface integrals

3 3 Given a surface S ⊂ R with parameterization r : D → R and a function f : S → R, the surface integral of f over S is:

x f (x, y, z) dS = x f (r(u, v))|ru × rv | dA. S D

go.illinois.edu/math241fa17. Motivation: Recall the change of variables formula!

Surface integrals

3 3 Given a surface S ⊂ R with parameterization r : D → R and a function f : S → R, the surface integral of f over S is:

x f (x, y, z) dS = x f (r(u, v))|ru × rv | dA. S D

In particular, the area of S is defined by:

Area(S) = x 1 dS = x |ru × rv | dA. S D

go.illinois.edu/math241fa17. Surface integrals

3 3 Given a surface S ⊂ R with parameterization r : D → R and a function f : S → R, the surface integral of f over S is:

x f (x, y, z) dS = x f (r(u, v))|ru × rv | dA. S D

In particular, the area of S is defined by:

Area(S) = x 1 dS = x |ru × rv | dA. S D

Motivation: Recall the change of variables formula!

go.illinois.edu/math241fa17. Problem: Calculate the integral ZZ xz dS S where S is the portion of the sphere of radius 2, centered at (0, 0, 0) that lies in the first octant (x, y, z ≥ 0).

go.illinois.edu/math241fa17. • Calculate the tangent vectors rφ and rθ. • Calculate the length of the normal vector 2 krφ × rθk = ρ sin(φ), see Example 16.6.10 in Book. •

x xz dA = S Z π/2 Z π/2 = ((ρ sin(φ) cos(θ)) · ρ cos(φ)) · ρ2 sin(φ) dφ dθ = 0 0 = ...

Solution

• Parametrize the surface using spherical coordinates: ρ = 2 and 0 ≤ φ, θ ≤ π/2, so r(φ, θ) = (ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ)).

go.illinois.edu/math241fa17. • Calculate the length of the normal vector 2 krφ × rθk = ρ sin(φ), see Example 16.6.10 in Book. •

x xz dA = S Z π/2 Z π/2 = ((ρ sin(φ) cos(θ)) · ρ cos(φ)) · ρ2 sin(φ) dφ dθ = 0 0 = ...

Solution

• Parametrize the surface using spherical coordinates: ρ = 2 and 0 ≤ φ, θ ≤ π/2, so r(φ, θ) = (ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ)).

• Calculate the tangent vectors rφ and rθ.

go.illinois.edu/math241fa17. •

x xz dA = S Z π/2 Z π/2 = ((ρ sin(φ) cos(θ)) · ρ cos(φ)) · ρ2 sin(φ) dφ dθ = 0 0 = ...

Solution

• Parametrize the surface using spherical coordinates: ρ = 2 and 0 ≤ φ, θ ≤ π/2, so r(φ, θ) = (ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ)).

• Calculate the tangent vectors rφ and rθ. • Calculate the length of the normal vector 2 krφ × rθk = ρ sin(φ), see Example 16.6.10 in Book.

go.illinois.edu/math241fa17. Solution

• Parametrize the surface using spherical coordinates: ρ = 2 and 0 ≤ φ, θ ≤ π/2, so r(φ, θ) = (ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ)).

• Calculate the tangent vectors rφ and rθ. • Calculate the length of the normal vector 2 krφ × rθk = ρ sin(φ), see Example 16.6.10 in Book. •

x xz dA = S Z π/2 Z π/2 = ((ρ sin(φ) cos(θ)) · ρ cos(φ)) · ρ2 sin(φ) dφ dθ = 0 0 = ...

go.illinois.edu/math241fa17.