E24.1 The surface area of S is given by s 1 dS = u vdudv, where...

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E24.1 The surface area of S is given by s 1 dS = \u v\dudv, where D is the domain of integration on the u-v plane. (We will talk about why this is true on Friday. ) Let S be the part of the paraboloid z = x2 y2 that lies below z = 4. Use the following two parameterizations to find the area of S (you should get the same answer). (a) r(u,v) = (u,v,u + v), and D is the region in the u-v plane with u + v 4. (b) r(u,v) = (v cos u, sin u, v), and D is the region in the u-v plane with 0 v 4 and 0 u 2. E24.1 The surface area of S is given by s 1 dS = \u v\dudv, where D is the domain of integration on the u-v plane. (We will talk about why this is true on Friday. ) Let S be the part of the paraboloid z = x2 y2 that lies below z = 4. Use the following two parameterizations to find the area of S (you should get the same answer). (a) r(u,v) = (u,v,u + v), and D is the region in the u-v plane with u + v 4. (b) r(u,v) = (v cos u, sin u, v), and D is the region in the u-v plane with 0 v 4 and 0 u 2.