Calculus Lia Vas Parametric Surfaces. Substitution Recall that a curve in space is given by parametric equations as a function of single parameter t x = x(t) y = y(t) z = z(t). A curve is a one-dimensional object in space so its parametrization is a function of one variable. Analogously, a surface is a two-dimensional object in space and, as such can be described using two variables. If a surface is given by an explicit equation z = f(x, y) the dependence on two independent variables is clearlisible. However, a parametric representation of an implicit surface F (x, y, z) = 0 may often be very useful. In particular, it may be useful to parametrize such surface using two parameters possibly different from x and y. In particular, a surface given by the parametric equations x = x(u, v) y = y(u, v) z = z(u, v) is referred to as a parametric surface and the two independent variable u and v as parameters. The equations above are called the parametric equations of the surface. Thus, we have the following. Examples. imension Parameter(s) Equations Curve t r(t) = x(t), y(t), z(t) Surface u, v r(u, v) = x(u, v), y(u, v), z(u, v) A surface of revolution z = f( x + y ) can be parametrized using polar coordinates as follows x = r cos θ, y = r sin θ, z = f(r). In particular, the following surfaces of revolution are parametrized in this way.. The paraboloid z = x + y has parametric representation by x = r cos θ, y = r sin θ, z = r.. The cone z = x + y has parametric representation by x = r cos θ, y = r sin θ, z = r.. The upper hemisphere of the sphere x +y +z = 9 has parametric representation by x = r cos θ, y = r sin θ, z = r. A cylindrical surface obtained from a curve in one of the coordinate planes can be parametrized using the curve parametrization and the remaining variable as the second parameter. For example, consider the cylinders below.
4. The cylinder x +y = 4 is based on the circle x = cos t, and y = sin t. The value of the z-coordinate of a point on the cylinder represents the height h of that point. So, the cylinder can be parametrized as x = cos t, y = sin t, z = h using the angle t and the height h as two parameters. 5. Similarly, the cylinder y + z = 4 is based on the circle y = cos t, and z = sin t in yz-plane. The value of the x-coordinate of a point on the cylinder can also be considered to be the height. Using h for that value, the cylinder can be parametrized as The derivative vectors x u, y u, z u and x v,, z v are in the tangent plane of a parametric surface x = x(u, v), y = y(u, v), z = z(u, v), (also written shortly as r = x(u, v), y(u, v), z(u, v) ). Thus, the tangent plane of the surface has the perpendicular vector ru r v = x u, y u, z u x v,, z v. In case when the surface z = z(x, y) is given by the parameters u = x and v = y, this reduces to the familiar formula ( z x, z y, ). x = h, y = cos t, z = sin t. Tangent Plane and Surface Area The surface area S of this surface over the region in uv-plane can be obtained by integrating surface area elements ds over sub-rectangles of region. The area of each element ds can be approximated with the area of the parallelogram in the tangent plane just as before when we considered surfaces parametrized using x and y. Recall that the area of a parallelogram formed by two vectors is the length of their cross product. Thus, ds = x u, y u, z u x v,, z v dudv or, written shortly, ds = r u r v dudv. The formula for the total surface area S can be obtained by integrating ds over entire region. Thus S = ds = r u r v dudv = x u, y u, z u x v,, z v dudv. In the case when the surface is parametrized by u = x and v = y as z = z(x, y) this formula becomes the one we have encountered before. S = + z x + zy dxdy
General substitution for double integrals. In many cases, a region over which a double integral is being taken may have easier representation in another coordinate system, say in uv-plane, than in xy-plane. In cases like that, one can transform the region in xy-plane to a region in uv-plane by the substitution x = g(u, v) y = h(u, v). Thus, a substitution is just a convenient reparametrization of a surface z = f(x, y) when the parameters x and y are changed to u and v. When evaluating the integral f(x, y)dxdy using substitution, the area element da = dxdy becomes J dudv where the Jacobian determinant J is given by x x J = u v = x u x v. Thus, y u y v f(x, y) dx dy = y u f(x(u, v), y(u, v)) J du dv Note that in one-dimensional case, the Jacobian determinant is simply the derivative of the substitution u = u(x) solved for x so that x = x(u) dx = x (u)du. Jacobian for polar coordinates. The polar coordinates x = r cos θ and y = r sin θ can be considered as a substitution in which u = r and v = θ. Thus, x r = cos θ, x θ = r sin θ and y r = sin θ, y θ = r cos θ. The Jacobian is J = x r y r x θ y θ = cos θ sin θ r sin θ r cos θ = r cos θ + r sin θ = r. This explains the presence of r in the integrals of the section on Polar Coordinates. Practice problems. f(x, y)dxdy = f(r cos θ, r sin θ) r dr dθ.. Find the equation of the tangent plane to given parametric surface at the specified point. a) x = u + v y = u z = u v; (,, 0)
b) x = uv y = ue v z = ve u ; (0, 0, 0). Find the area of the surface. a) The part of the plane z = x + y that lies above the triangle with vertices (0,0), (,) and (0,). b) The part of the surface z = y + x that lies between the cylinders x + y = and x + y = 4. Write down the parametric equations of the paraboloid and use them to find the surface area. c) The part of the cone z = x + y that lies between the cylinders x + y = 4 and x + y = 9. Write down the parametric equations of the cone first. Then find the surface area using the parametric equations.. Consider the cylinder x + z = 4. a) Write down the parametric equations of this cylinder. b) Using the parametric equations, find the tangent plane to the cylinder at the point (0,, ). c) Using the parametric equations and formula for the surface area for parametric curves, show that the surface area of the cylinder x + z = 4 for 0 y 5 is 0π. 4. Use the given substitution to evaluate the integral. a) (x+4y) dx dy where is the region bounded by the lines y = x, y = x, y = x, and y = x. The substitution x = (u + v), y = (v u) transforms the region to a rectangle 0 u and 0 v. b) xy dx dy where is the region in the first quadrant bounded by the curves y = x, y = x, y =, and y =. The substitution x = u, y = v transforms the region into a x x v region with bounds u and v. c) xy dx dy where is the region in the first quadrant bounded by the curves y = x, y = x,y =, and y =. The substitution x =, y = v transforms the region into x x v a square u and v. Solutions.. a) x u, y u, z u =, 6u, and x v,, z v =, 0,. The cross product of these vectors is 6u,, 6u. Now we need to find u-value (v-value is not needed since v does not appear in this product) that corresponds to the point (,, 0). When (x, y, z) = (,, 0), x = u + v = and z = u v = 0. From this we have that u = v =. So, the perpendicular vector is 6,, 6. The equation of the plane is 6(x ) + (y ) 6(z 0) = 0 x y + z = b) x u, y u, z u = v, e v, ve u and x v,, z v = u, ue v, e u. The cross product of these vectors is e uv ( uv), ve u (u ), ue v (v ). Now we need to find the u and v values that correspond to the point (0, 0, 0). When (x, y, z) = (0, 0, 0), ue v = 0 u = 0 and ve u = 0 v = 0. From this we have that u = v = 0. So, the perpendicular vector is, 0, 0. The equation of the plane is (x 0) + 0(y 0) + 0(z 0) = 0 x = 0 (yz-plane).
. a) You can use the standard xy-parametrization here and the formula S = T + z x + zy dxdy where T is the given triangle. z = x + y z x = and z y =. The bounds for x are 0 and and y is bounded by the line connecting (0,0) and (,) from below and by the line that connects (0,) and (,) from above. S = 0 x + + 4 dxdy = 6 0 ( x)dx = 6( ) = 6 =.. Thus x y. The surface area is b) The paraboloid can be parametrized by x = r cos t, y = r sin t, z = x + y = r. x r, y r, z r = cos t, sin t, r and x t, y t, z t = r sin t, r cos t, 0. The cross product of these vectors is r cos t, r sin t, r. The length of this product is 4r 4 cos t + 4r 4 sin t + r = 4r 4 + r = r (4r + ) = r 4r +. The bounds for the integration are determined by the projection in the xy-plane which is the region between the circles x + y = and x + y = 4. Thus 0 t π and r. So, the surface area is S = π 0 dt r 4r + dr = π4.9 = 0.85. c) The cone can be parametrized by x = r cos t, y = r sin t, z = x + y = r. x r, y r, z r = cos t, sin t, and x t, y t, z t = r sin t, r cos t, 0. The cross product of these vectors is r cos t, r sin t, r. The length of this product is r cos t + r sin t + r = r + r = r = r. The bounds for the integration are determined by the projection in the xy-plane which is the region between the circles x + y = 4 and x + y = 9. Thus 0 t π and r. So, the surface area is S = π 0 dt rdr = π ( 9 4) = 5π.. a) The cylinder can be parametrized as x = cos t, y = y, z = sin t. b) x t, y t, z t = sin t, 0, cos t and x y, y y, z y = 0,, 0. The cross product is cos t, 0, sin t. The t-value that corresponds to (0,, ) can be obtained from x = cos t = 0 and z = sin t =. Thus t = π. The perpendicular vector is 0, 0, and the equation of the plane is 0(x 0) + 0(y ) + (z ) = 0 z =. c) We already computed the cross product cos t, 0, sin t. Its length is 4 cos t + 4 sin t = 4 =. The t-bounds are 0 and π and the y-bounds are given to be 0 y 5. So, the surface area is S = π 5 0 0 dydt = π(5)() = 0π. x 4. a) Calculate the Jacobian J = u x v y u = 0 0 (u + v + 4(v u)) du dv = (6 + 9 + 6) = b) The Jacobian is J = x u y u x v udu ln v = udu ln = ln u c) The Jacobian is J = v x u y u v du dv = v u v x v v 0 (uv + + 4v 8uv 6 ) 0 du = = + =. 9 9 (x + 4y) dx dy = 0 (u + 9 + 6 8u) du = u = v v 0 =. v xy dx dy = = 4 ln = ln =.97 = uv v v v du dv = u 4 ln v = ln =.97 u v du dv = v v = + =. 4v 4v v xy dx dy =