MATH 251 Fall 2023 Section 16 6 Parametric Surfaces and Their Areas In the same way that we describe a space curve by a vector function r t we can describe a surface by a vector function r u v of two parameters u and v In particular is a vector valued function de ned on a region D in the uv plane r u v x u v i y u v j z u v k The set x u v y u v z u v u v 2 D is called a parametric surface S and the equations x x u v y y u v z z u v are called parametric equations of S Example 1 Parameterize the following surfaces 1 2x 5y 3z 60 2 x2 y2 z2 36 3 x2 z2 16 4 The quarter cylinder x2 z2 16 with z 0 and x 0 1 2 5 Part of the sphere x2 y2 z2 1 that is above the cone z p3x2 3y2 6 Part of the plane y z 3 that lies inside the cylinder x2 z2 4 Surface Area Given the parametric surface S r u v x u v i y u v j z u v k we wish to compute its area over a region D in its domain in the uv plane We partition D into subrectangles Rij s The corner point u i v i of the rectangle Rij gives the point Pij r u i v i on the surface S In fact r maps the subrectangle Rij to the surface Sij by mapping the edges of Rij to the edges of Sij We shall estimate the area Sij One edge of Sij is along the curve r u i v This edge can be approximated by the vector rv u i v i v Another edge of Sij is along the curve r u v i This edge can be approximated by the vector ru u i v i u Hence the area of Sij is approximately rv u i v i v ru u i v i u rv ru u v rv ru dA Therefore the area for the whole surface S over the region D in the uv plane is 3 Theorem Let S be given by the parametric equation and S is covered just once as u v ranges through D then the surface area of S is r u v x u v i y u v j z u v k u v 2 D A S x D ru rv dA A S x D ru rv dA where ru i j k and rv i j k x u y u z u x v y v z v Example 2 Find the area of the part of the plane x 2y z 10 that lies inside the cylinder x2 y2 4 4 Example 3 Find the area of the part of the plane 2x 3y 5z 30 that lies in the rst octant Example 4 Find the area of the part of the surface z x2 5y 10 that lies above the triangle in the xy plane with vertices 0 0 2 0 and 2 4 5 Recall that we can parameterize the sphere of radius as r h sin cos sin sin cos i r h cos cos cos sin sin i and r h sin sin sin cos 0i r r h 2 sin2 cos 2 sin2 sin 2 sin cos i r r 2 sin Example 5 Find the area of the part of the surface the sphere x2 y2 z2 16 that lies above the 6 Hence You can compute that cone z p3x2 3y2
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