Calculus 2c, Examples of Maximum and Minimum Integration and by Mejlbro L.

By Mejlbro L.

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At SimCorp, you will be part of a large network of competent and skilled colleagues who all aspire to reach common goals with dedication and team spirit. We invest in our employees to ensure that you can meet your ambitions on a personal as well as on a professional level. SimCorp employs the best qualified people within economics, finance and IT, and the majority of our colleagues have a university or business degree within these fields. Ambitious? 3 A. Calculate the space integral x2 yz dΩ, I= A where A = {(x, y, z) ∈ R3 | x2 + y 2 ≤ a2 , y ≥ 0, 0 ≤ z ≤ h}.

I 1. Here the domain is written (note the order of x and y): B = {(x, y) ∈ R2 | 0 ≤ x ≤ a, x ≤ y ≤ a}. Then we can write down the double integral: a x exp y 3 dS = 0 B a a x exp y 3 dy dx = x a x 0 exp y 3 dy dx. com 41 Calculus 2c The plane integral does not just look impossible to calculate; it is impossible to calculate with our arsenal of functions! Therefore we give up this variant. Instead we examine, if we shall be more successful by interchanging the order of integration. 2 Figure 24: The domain B for a = 1 with a horizontal line of integration from x = 0 to x = y where y is kept fixed.

0 0 We first calculate the inner integral, where θ is considered as a constant, √ a cos 2θ 0 1 4 r r dr = 4 3 √ a cos 2θ = 0 a4 cos2 2θ. com 62 Calculus 2c The space integral Then by insertion, I π 4 = 2π 0 cos θ sin θ · a4 cos2 2θ dθ 4 π π 4 4 πa4 1 πa4 cos2 2θ sin 2θ dθ = cos2 2θ − = 4 0 4 θ=0 2 πa4 πa4 π πa4 1 cos3 2θ = {1 − 0} = . = − 8 3 24 24 0 4 d cos 2θ I 2. Let us now turn to the semi-polar variant. √ The problem with this is to find = P (z) as a function of z for the boundary curve r = a cos 2θ for the domain in the meridian half-plane.

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