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. p • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x; z plane: p 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. p • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x; z plane: p 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. p • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x; z plane: p 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. p • y is positive so we can write y = 3 − x2 − z2. Use polar coordinates in x; z plane: p 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 jru × rv j 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))jru × rv j 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))jru × rv j dA: S D In particular, the area of S is defined by: Area(S) = x 1 dS = x jru × rv j dA: S D go.illinois.edu/math241fa17.