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Examples7.6 - 1 Example 1 Evaluate the surface integral of...

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1 Example 1 Evaluate the surface integral of the vector field F = 3 x 2 i - 2 yx j + 8 k over the surface S that is the graph of z = 2 x - y over the rectangle [0 , 2] × [0 , 2] . Solution. Use the formula for a surface integral over a graph z = g ( x, y ) : ZZ S F · d S = ZZ D F · - ∂g ∂x i - ∂g ∂y j + k dx dy. In our case we get Z 2 0 Z 2 0 (3 x 2 , - 2 yx, 8) · ( - 2 , 1 , 1) dx dy = Z 2 0 Z 2 0 ( - 6 x 2 - 2 yx + 8) dx dy = Z 2 0 - 2 x 3 - yx 2 + 8 x 2 x =0 dy = Z 2 0 - 4 y dy = - 2 y 2 | 2 0 = - 8 .
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2 Example 2 Let S be the triangle with vertices (1 , 0 , 0) , (0 , 2 , 0) , and (0 , 1 , 1) , and let F = xyz ( i + j ) . Calculate the surface integral ZZ S F · d S , if the triangle is oriented by the “downward” normal. Solution. Since S lies in a plane (see the right hand part of the Figure), it is part of the graph of a linear function z = ax + by + c. v = 2 - 2 u v = 1 - u (0, 2) (0, 1) (1, 0) z y x (0,1,1) (0,2,0) D S v u Substituting the vertices of the triangle for ( x, y, z ) , we get the equation 0 = a + c, 0 = 2 b + c, 1 = b + c, which we can solve to find b = - 1 , c = 2 , a = - 2 , i . e ., z = - 2 x - y + 2 . We may take x and y as parameters; i.e., x = u, y = v, z = - 2 u - v + 2 , or Φ ( u, v ) = ( u, v, - 2 u - v +2) .
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