Dr. Z’s Math251 Handout #16.4 (2nd ed.) [Parametrized Surfaces and Surface Integrals]By Doron ZeilbergerProblem Type 16.4a: Find an equation of the tangent plane to the given parametric surface atthe specified point.x = x(u, v) , y = y(u, v) , z = z(u, v) ; u = 1, v = 1 .Example Problem 16.4a: Find an equation of the tangent plane to the given parametric surfaceat the specified point.x = u2, y = v2, z = uv ; u = 1, v = 1 .Steps Example1. Set-upr = x i + y j + z k ,and compute the partial derivatives w.r.t.u and w.r.t. v :ru= xui + yuj + zuk ,rv= xvi + yvj + zvk .Then plug-in the given values of u andv.1. In this problemr = u2i + v2j + uv k ,We haveru= 2u i + 0 j + v k ,rv= 0 i + 2v j + u k .Now plug-in u = 1, v = 1 to get numer-ical vectors.ru(1, 1) = 2 i + 0 j + 1 k ,rv(1, 1) = 0 i + 2 j + 1 k .So at this point, ru= h2, 0, 1 i, rv=h0, 2, 1 i.12. Find the cross-product ru× rv. Thisis a vector normal to the tangent plane.2.ru× rv=i j k2 0 10 2 1= −2 i − 2 j + 4 k .Or in hi notationN = h−2, −2, 4i .3. Find the point (x0, y0, z0) by plugginginto x, y, z the specific values of u and vgiven in the problem The desired equationof the tangent plane isa(x − x0) + b(y − y0) + c(z − z0) = 0 .where N = ha, b, ci and the point is (x0, y0, z0).3. The point is (12, 12, 1 · 1) = (1, 1, 1).The desired equation of the tangent planeis(−2)(x − 1) − 2(y − 1) + 4(z − 1) = 0 .Or, in expanded form−2x − 2y + 4z = 0 .Dividing by −2 (to make it nicer), we get:Ans.: x + y − 2z = 0.A Problem from a previous Final: Find an equation for the tangent plane to the parametricsurfacex = u2, y = u + v , z = v2,at the point (1, 2, 1). Simplify as much as you can!Ans.: x − 2y + z = −2.Another Problem from a Previous Final: Evaluate the surface integralZ ZS√3 x dS ,where S is the triangular region with vertices (1, 0, 0), (0, 1, 0), (0, 0,
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