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16.6 Parametric Surfaces and Their Areas

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16.6 Parametric Surfaces and Their Areas CHAPTER 16. VECTOR CALCULUS 239 16.6 Parametric Surfaces and Their Areas Comments. On the next page we summarize some ways of looking at graphs in Calc I, Calc II, and Calc III, and pose a question about what comes next. 2 3 2 Definition. Let r : R ! R and let D ⊆ R . A parametric surface S is the set of all points hx; y; zi such that hx; y; zi = r(u; v) for some hu; vi 2 D. In other words, it's the set of points you get using x = some function of u; v y = some function of u; v z = some function of u; v Example 1. (a) Describe the surface z = −x2 − y2 + 2 over the region D : −1 ≤ x ≤ 1; −1 ≤ y ≤ 1 as a parametric surface. Graph it in MATLAB. (b) Describe the surface z = −x2 − y2 + 2 over the region D : x2 + y2 ≤ 1 as a parametric surface. Graph it in MATLAB. Solution: (a) In this case we \cheat": we already have z as a function of x and y, so anything we do to get rid of x and y will work x = u y = v z = −u2 − v2 + 2 −1 ≤ u ≤ 1; −1 ≤ v ≤ 1 [u,v]=meshgrid(linspace(-1,1,35)); x=u; y=v; z=2-u.^2-v.^2; surf(x,y,z) CHAPTER 16. VECTOR CALCULUS 240 How graphs look in different contexts: Pre-Calc, and Calc I graphs: Calc II parametric graphs: y = f(x) x = f(t), y = g(t) • 1 number gets plugged in (usually x) • 1 number gets plugged in (usually t) • 1 number comes out (usually y) • 2 numbers come out (x and y) • Graph is essentially one dimensional (if you zoom in enough it looks like a line, plus you only need • graph is essentially one dimensional (but we one number, x, to specify any location on the don't picture t at all) graph) • We need to picture the graph in two dimensional • We need to picture the graph in a two dimen- world. sional world (to contain 1# in + 1 out) Calc III, chapter 14 and 15 graphs: What goes here? • 2 numbers get plugged in (usually x and y) • 1 number comes out (usually z) • graph is essentially two dimensional (if you zoom in enough it looks like a plane, plus you need two numbers, x and y, to specify a location on the graph). • We need to picture the graph inside a three di- mensional world (to contain 2 #s in + 1 out). (b) In this case, we have to work a little bit more, not because of the equation, but because of the domain. Since the domain is circular, it's convenient to use polar coordinates: x = r cos(θ) y = r sin(θ) z = 2 − r2 0 ≤ θ ≤ 2π; 0 ≤ r ≤ 1 [r,theta]=meshgrid(linspace(-1,1,30),linspace(0,2*pi,35)); x=r.*cos(theta); y=r.*sin(theta); z=2-r.^2; surf(x,y,z) Example 2. Use MATLAB to graph the following paramatric surface: x = (3 + sin(v)) cos(u) y = (3 + sin(v)) sin(u) z = cos(v) 0 ≤ u ≤ 2π; 0 ≤ v ≤ 2π Describe what it looks like. Analyze what u and v represent in the shape. Solution: CHAPTER 16. VECTOR CALCULUS 242 [u,v]= meshgrid(linspace(0,2*pi,50)); x=(3+sin(v)).*cos(u); y =(3+sin(v)).*sin(u); z=cos(v); surf(x,y,z); axis equal It looks like a donut. The parameter u corresponds to how far around the large circle a point is: as u increases you get a path that travel counter clockwise around the shape. The curves that look “flat”, like lines of lattitude, or like the main equator, are generated by holding v equal to a constant and allowing u to increase. In fact, for a constant value of v we have x = ∗ cos(u) and y = ∗ sin(u) where \∗" just represents some number, but the point is that this generates a circle in x and y. The parameter v corresponds to how far around the small circle a point is: as v increases you get a path that travels \up" over the shape, down into the hole, through the hole, and back around back up to where it started. The curves that go up and down, like longitude, are generated by holding u equal to a constant and allowing v to increase. In fact, for a constant value of u, and say when y = 0, then we have x = 3 + sin(v) and z = cos(v). This generates a circle in the (x; z) plane where the center is z = 0 and x = 3. This is where we ended on Wednesday, April 24 Example 3. Based on the previous example, what will happen to the graph if we change the parametric equations so that z = u + cos(v) and 0 ≤ u ≤ 6π? Solution: Everything is the same, except now z will increase because of the \u+" term. Thus, we will have a donut, but it won't close, it will go around and around. If we had enough batter we could make an endless donut. CHAPTER 16. VECTOR CALCULUS 243 [u,v]= meshgrid(linspace(0,6*pi,75),linspace(0,2*pi,15)); x=(3+sin(v)).*cos(u); y =(3+sin(v)).*sin(u); z=u+cos(v); surf(x,y,z); axis equal 16.6.1 Tangent planes Definition. Let r(u; v) be a parametric surface, and write the component functions as x(u; v), y(u; v) and z(u; v). The tangent plane at the point (u0; v0) is the plane that contains the point r(u0; v0) and which is normal to rv(u0; v0) × ru(u0; v0) where @x @y @z r (u ; v ) = (u ; v ); (u ; v ); (u ; v ) v 0 0 @v 0 0 @v 0 0 @v 0 0 @x @y @z r (u ; v ) = (u ; v ); (u ; v ); (u ; v ) u 0 0 @u 0 0 @u 0 0 @u 0 0 Recall that the equation of a plane is easily found by specifying what a point that it should contain, and a normal vector (see Section 12.5 Definition 12.5.2.) Example 4. Find the equation of the tangent plane of the donut from Example2 at point u = π, v = π=4. Graph the result. Solution: r(π; π=4) = h(3 + sin(π=4)) cos(π); (3 + sin(π=4)) sin(π); cos(π=4)i D p p E = −(3 + 1= 2); 0; 1= 2 ru = h−(3 + sin(v)) sin(u); (3 + sin(v)) cos(u); 0i π,π=4 = h−(3 + sin(π=4)) sin(π); (3 + sin(π=4)) cos(π); 0i CHAPTER 16. VECTOR CALCULUS 244 D p E = 0; −(3 + 1= 2); 0 rv = h(cos(v)) cos(u); (cos(v)) sin(u); − sin(v)i π,π=4 = hcos(π=4) cos(π); cos(π=4) sin(π); − sin(π=4)i D p p E = −1= 2; 0; −1= 2 i j k p ru × rv = 0 −(3 + 1= 2) 0 p p −1= 2 0 −1= 2 p p p p = i(3 + 1= 2)(1= 2) − j0) − k(3 + 1= 2)(1= 2) 3 1 3 1 = p + ; 0; −p − 2 2 2 2 3 1 p 3 1 p Plane: p + x + 3 + 1= 2 + 0 − p + z − 1= 2 = 0 2 2 2 2 p p x + 3 + 1= 2 − z − 1= 2 = 0 p z = x + 3 + 2 [u,v]= meshgrid(linspace(0,2*pi,50)); x=(3+sin(v)).*cos(u); y =(3+sin(v)).*sin(u); z=cos(v); surf(x,y,z, 'FaceAlpha ' ,0.5 , 'EdgeAlpha ' ,0); axis equal hold on u= pi; v=pi/4; x=(3+sin(v)).*cos(u); y =(3+sin(v)).*sin(u); z=cos(v); CHAPTER 16. VECTOR CALCULUS 245 plot3(x,y,z, 'or ',' MarkerSize ' ,10, ' MarkerFaceColor ','r'); [x,y]=meshgrid(linspace(-4.5,-3,10),linspace(-0.5,0.5,10)); z=x+3+sqrt(2); surf(x,y,z, 'FaceColor ','green ','FaceAlpha ' ,0.5); hold off 16.6.2 Surface Area Definition. If S is the surface parameterized by r(u; v) for (u; v) 2 D, then the surface area is ZZ A(S) = jru × rvj dA D where @x @y @z r = ; ; v @v @v @v @x @y @z r = ; ; u @u @u @u jru × rvj = length of the vector ru × rv Example 5. Find the surface area of the donut from Example2 Solution: ru = h−(3 + sin(v)) sin(u); (3 + sin(v)) cos(u); 0i rv = hcos(v) cos(u); cos(v) sin(u); − sin(v)i D ru × rv = − (3 + sin(v)) cos(u) sin(v); −(3 + sin(v)) sin(u) sin(v); E −(3 + sin(v)) sin2(u) cos(v) − cos(v) cos2(u)(3 + sin(v)) For the last component, we factor out −(3+sin(v)) cos(v) and this leaves behind sin2(u)+ cos2(u) = 1. ru × rv = h−(3 + sin(v)) cos(u) sin(v); −(3 + sin(v)) sin(u) sin(v); −(3 + sin(v)) cos(v)i q 2 2 2 2 2 2 2 2 jru × rvj = (3 + sin(v)) cos (u) sin (v) + (3 + sin(v)) sin (u) sin (v) + (3 + sin(v)) cos (v) From the first two terms we factor out (3 + sin(v))2 sin2(v) and this leaves behind cos2(u) + sin2(u) = 1: q 2 2 2 2 jru × rvj = (3 + sin(v)) sin (v) + (3 + sin(v)) cos (v) r = (3 + sin(v))2 sin2(v) + cos2(v) = p(sin(v) + 3)2 = sin(v) + 3 ZZ A(S) = sin(v) + 3 dA D CHAPTER 16.
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