 By Mejlbro L.

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Extra info for Calculus 2c, Examples of Maximum and Minimum Integration and Vector Analysis

Example text

Here we can expect quite a few diﬃculties, no matter which version we are choosing. In fact, the integrand is suited for the polar coordinates, while the domain B in the (x, y)-plane is best described in rectangular coordinates. By experience, such a mixture of polar and rectangular coordinates will always give computational problems. Then we notice that if we start by ﬁrst integrating after x, then we shall immediately run into troubles with this ﬁrst integral dx . (x2 + y 2 + z 2 )2 It is possible to go through with the calculations, but they are far from elementary.

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 . 15 7 105 C. It is seen as a weak control that the result is of the form constant·a7 as mentioned in A. 5 A. Let A be an upper half sphere of radius 2a, from which we have removed a cylinder of radius a and then halved the resulting domain by the plane x + y = 0.

In this case we write the domain in the form B = {(x, y) ∈ R2 | 0 ≤ x ≤ 2, 1 − 1 x ≤ y ≤ 1}. 2 Notice that the outer variable x must always lie between two constants, 0 ≤ x ≤ 2. e. in this particular case 1 − x ≤ y ≤ 1. 2 Then write down the double integral: 2 (2) 1 xy dS = 1− 12 x 0 B 2 xy dy dx = 1 x 1− 12 x 0 y dy dx. Calculate the inner integral, 1 1− 12 y dy = x 1 2 y 2 1 = 1− 12 x 1 2 1− 1− 2 1 x 2 = 1 2 1− 1 2 x 4 = 1 1 x − x2 . 2 8 By insertion in (2), we get 2 xy dS = x 0 B = 1 1 x − x2 dx = 2 8 1 3 1 4 x − x 6 32 2 = 0 2 1 2 1 3 x − x dx 2 8 0 5 8 1 − = .