By Mejlbro L.

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

**Example text**

D 2. The domain is here written (note again the order of x and y): B = {(x, y) ∈ R2 | 0 ≤ y ≤ a, 0 ≤ x ≤ y}. Then we turn to the double integral, a x exp y 3 dS = (4) y 0 B a x exp y 3 dx dy = 0 0 y exp y 3 x dx dy. 0 The inner integral is calculated in the following way: y 1 2 x 2 x dx = 0 y = 0 1 2 y . 2 By insertion in (4) followed by the substitution t = y 3 and dt = 3y 2 dy, where y 2 dy already can be found in the integrand, we get x exp y 3 a dS = B = 3 a 1 1 1 exp y · y 2 dy = exp(t) · · dt 2 2 3 0 0 1 t a3 1 e 0 = exp a3 − 1 .

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 − = .

Examine whether these points are also extrema. D. This is an example where the (r, s, t)-method is of no use at the point (0, 0). ) is not a universal method, which can handle all cases. We shall therefore here use the alternative method of the approximating polynomials. This is in general far better than the (r, s, t)-method, although this is not a universal method either. 4 Aside! None of the two methods above can solve the same problem of extrema for the function 1 for (x, y) = (0, 0), g(x, y) = exp − 2 x + y2 supplied by continuity with g(0, 0) = 0.