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**Example text**

X2 + y 2 ) dΩ, where the domain of integration A is given by x2 + y 2 ≤ z ≤ h. a 5) The space integral x ≥ 0, y≥0 6) The space integral x2 + y 2 ≤ 2x h a 8) The space integral x2 + y 2 ≤ 3 and A A x2 + y 2 ≤ z ≤ h. a xz dΩ, where the domain of integration A is given by 0≤z≤ x2 + y 2 . (z 2 + y 2 ) dω, where the domain of integration A is given by x2 + y 2 . A and 9) The space integral x2 + y 2 ≤ 3 xy dΩ, where the domain of integration A is given by the conditions and 7) The space integral 0≤z ≤h− A A and (x2 + y 2 ) dΩ, where the domain of integration A is given by 0≤z≤ 1 + x2 + y 2 .

Then a geometrical consideration shows that the distance from origo to the intersection point with the circle of radius a and centrum (0, a) is give by 2a cos θ. This gives us the lower limit for r, thus r ∈ [2a cos θ, 2a]. The domains of the other coordinates are obvious. 2) The integrand is written in spherical coordinates in the following way, f (x, y, z) = z 2 = r2 cos2 θ. 5 2 Figure 30: The meridian cut for a = 1 with a radius vector and a perpendicular line. Then by the reduction theorem for space integrals in spherical coordinates, π 2 z 2 dΩ = 2π A 2a 2a cos θ 0 = 64πa5 5 π 2 0 r2 cos2 θ · r2 sin θ dr dθ = 2π 64πa5 cos − cos θ sin θ dθ = 5 2 7 π 2 0 cos2 θ sin θ · 1 1 cos8 θ − cos3 θ 8 3 5 = 8πa5 64πa 8 − 3 · = .

For x ≥ 0. 8, so we may compare the results. Cf. ] A Space integrals in semi-polar coordinates. D Find the interval [α, β] for ϕ. Describe B(ϕ) in semi-polar coordinates and sketch if necessary B(ϕ) in the meridian half plane. Finally, compute the space integral by using the theorem of reduction in semi-polar coordinates. wanted: ambitious people Please click the advert At NNE Pharmaplan we need ambitious people to help us achieve the challenging goals which have been laid down for the company.