By Mejlbro L.
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Extra info for Calculus 2b, Real Functions in Several Variables
Example text
1 A. Calculate the space integral I= A (3 + y − z) x dΩ, where A = {(x, y, z) ∈ R3 | (x, y) ∈ B, 0 ≤ z ≤ 2y}, and B is the upper triangle shown on the figure. e. the projection of the domain A onto the (x, y)-plane. D Apply the first rectangular reduction theorem in 3 dimensions. We have according to the first rectangular reduction theorem that (7) I = A (3 + y − z)x dΩ = 2y B 0 (3 + y − z) dz dS. Considering x and y as constants, we calculate the inner and concrete integral, 2y 0 2y (3 + y − z)x dz = x 0 (3 + y − z = dz = x (3 + y)z − 1 2 z 2 2y z=0 = x (3 + y) · 2y − 2y 2 = x · 2y{3 + y − y} = 6xy.
In this case it is possible to calculate the integral by using either rectangular or polar coordinates. D 1. In rectangular coordinates the domain B is described by ax − x2 ≤ y ≤ B = {(x, y) | 0 ≤ 0 ≤ a, − ax − x2 }. I 1. The rectangular double integral is given by √ a x dS = I= x 0 B ax−x2 √ − ax−x2 dy dx = 2a a ax − x2 dx. The trick in problems of this type is to call the “ugly” part something different. We put t = ax − x2 , dt = (a − 2x) dx. Then by adding the right term and subtract it again we get a I 2x = 0 = − = − = − a 0 a a ax − x2 dx = − 0 (a − 2x − a) ax − x2 · (a − 2x) dx + a √ a t dt + a x=0 2 ax − x2 3 3 2 0 a 0 ax − x2 dx ax − x2 dx a +a 0 a ax − x2 dx 0 ax − x2 dx = 0 + a a 0 ax − x2 dx.
4) z = f4 (x, y) = x2 − y 2 = (x + y)(x − y). The graph is a hyperbolic paraboloid. There is no extremum at (0, 0). An analysis of the sign shows that f4 (x, y) is 0 on the lines x + y = 0 and x − y = 0, and that f4 (x, y) attains both positive and negative values in any neighbourhood of (0, 0). It is finally also possible to consider the restrictions x − axis: y − axis: f4 (x, 0) = x2 > 0 f5 (0, y) = −y 2 < 0 for x = 0, for y = 0, Please click the advert from which we arrive to the same conclusion.