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Extra resources for Calculus 2b, Real Functions in Several Variables
Example text
8 1 –1 Figure 35: The domain A for a = 1. 2 Figure 36: The cut in the meridian half-plane for a = 1. I. The cut B(ϕ), which is revolved around the z-axis, must be independent of ϕ, so have in the meridian half plane B(ϕ) = {( , z) | 0 ≤ z ≤ a, 0 ≤ ≤ z}. Then we get by the reduction theorem that the ϕ-integral can be factored out, I = = = = π 2 −π 2 π 2 −π 2 sin2 ϕ · cos ϕ dϕ · 1 sin3 ϕ 3 2 15 cos ϕ · B(ϕ) a 0 π 2 −π 2 · z 6 dz = 2 sin2 ϕ · z · d dz 4 z d dz B(ϕ) a z z 0 dϕ 4 d dz = 0 2 · 3 a 0 z· 1 5 z dz 5 2 1 7 2 7 · a = a .
It is therefore sufficient to examine g(u, v) in the neighbourhood of (4, 4). The approximating polynomial for g(u, v) expanded from (4, 4) of at most degree 2 is found by using: g(u, v) = 1 − 4u − 4v + uv, g(4, 4) = −15, ∂g = −4 + v, ∂u gu (4, 4) = 0, ∂g = −4 + u, ∂v gv (4, 4) = 0, ∂2g ∂v = 1, ∂u ∂2g ∂2g = = 0, ∂u2 ∂v 2 , Brain power Please click the advert By 2020, wind could provide one-tenth of our planet’s electricity needs. Already today, SKF’s innovative knowhow is crucial to running a large proportion of the world’s wind turbines.
D 2. Here we first integrate horizontally. I 2. The domain is written B = {(x, y) ∈ R2 | 0 ≤ y ≤ 1, 2 − 2y ≤ x ≤ 2}, because y ∈ [0, 1] is now the outer variable of integration (lying between two constants), and where 2 − 2y ≤ x ≤ 2 for the inner variable of integration x for any fixed y. The double integral becomes here 1 2 xy dS = (3) 0 B 1 2 xy dx dy = 2−2y x dx dy. 0 2−2y Let us first calculate the inner integral, 2 1 2 x 2 x dx = 2(1−y) 2 = 2(1−y) 2 1 2 2 · 2 1 − (1 − y)2 = 2(2y − y 2 ) 2 = 4y − 2y .