Final Exam Solution

# To determine d project the triangle down to the xy

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Unformatted text preview: and (0, 1, 1). Evaluate the surface integral S f (x, y, z ) dS . Solution: This problem was done in class. For full details, refer to your lecture notes. The triangle lies on a plane. First ﬁnd a normal vector to the plane. This is done by taking the cross product of two non-parallel vectors that lie on the plane. Let α be the vector from (1, 0, 0) to (0, 2, 0) and β be the vector from (1, 0, 0) to (0, 1, 1). α = (−1, 2, 0) , β = (−1, 1, 1) . Then α × β = (2, 1, 1) is a normal vector to the plane. Using this and the point (1, 0, 0), he point-normal form equation of the plane is: (2, 1, 1) · (x − 1, y, z ) = 0 ⇒ z = 2 − 2x − y. This equation for z should be the same regardless the method you use to ﬁnd it. The triangle can be given by Φ (x, y ) = (x, y, 2 − 2x − y ) for (x, y ) ∈ D . To determine D , project the triangle down to the XY plane. So D is a triangle on R2 with vertices (1, 0), (0, 2) and (0, 1). D = {(x, y ) |x ∈ [0, 1] , y ∈ [1 − x, 2(1 − x)]} . Also, Tx = (1, 0, −2) and Ty = (0, 1, −1). Then Tx × Ty = (2, 1, 1) = √ f (x, y, z ) 6 dy dx f (x, y, z ) dS = S D = √ 1 2(1−x) 1 1−x 2(1−x) 6 0 = √ = √ 6 0 1−x 1 y3 x (1 − x)y − 3 0 = = 2√ x 2(1 − x)y − y 2 dy dx 2(1−x) 2 6 √ xy (2 − 2x − y ) dy dx 1 6 0 x 3− 1 6 3 2√ 6 = 3 2√ = 6 3 √ 6 = 30 0 7 3 dx 1−x (1 − x)3 dx x (1 − x)3 dx 1 0 x − 3x2 + 3x3 − x4 dx 31 1 dx −1+ − 2 45 √ 6. Math 3010 Final Exam - Page 8 of 14 Dec. 8, 2012 5. Let S ⊂ R3 be the oriented surface...
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