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25−y 2 4y 3 1−y y 2 −4 4 f (x, y) dx dy. 1−y − f (x, y) dx dy. √ 1−y 2 f (x, y) dx dy. √ 25−y 2 0 f (x, y) dx dy. A Interchange of the order of integrations in double integrals. D Sketch the set B and set up the double integral in the reverse order. Notice that a nice description in one case does not imply a nice description in the reverse order. 1. I 1) The domain is given by B = {(x, y) | 0 ≤ x ≤ 1, x2 ≤ y ≤ x} = {(x, y) | 0 ≤ y ≤ 1, y ≤ x ≤ √ y}. In fact, it follows from the inner integral that x2 ≤ y ≤ x, from which it is easy to derive √ y ≤ x ≤ y.

6. 4. We can therefore choose the truncation Bt = {(x, y) | t ≤ x ≤ 1, x ≤ y ≤ 1 + x2 }, 0 < t < 1, where we remove a strip along the Y -axis. Then by integrating over Bt , 1 dS = x+y Bt = x ln(x + 1 t √ 1+x2 x 1 + x2 ) 1 t − t2 ) 1 dy x+y 1 t √ 1 dx = t {ln(x + 1 + x2 ) − ln 2 − ln x}dx x dx − [ln 2 · x + x ln x − x]1t 1 + x2 1 = ln 2 − t ln(t + 1 + − 1 + t2 + (1 − ln 2) · (1 − t) − t ln t t √ = 1 − 2 + 1 + t2 − t ln(t + 1 + t2 ) − t ln t − t(1 − ln 2). Due to the rules of magnitudes, t · ln t → 0− for t → 0+, so we conclude that the improper plane integral exists and it has the value B 1 dS = lim t→0+ x+y Bt √ 1 dS = 2 − 2.

Then compute the plane integrals and D 1 dS. com 44 Calculus 2c-5 Plane integrals, polar coordinates A and D Sketch of a domain; computation of a plane integral. 5 2 x Figure 34: The domain B. 5 2 x Figure 35: The domain D. I The domains are sketched on the two figures. We see that B ∪ D = [1, 2] × [1, 2], which may be exploited in one of the variants, because B and D have just one boundary curve in common and are otherwise disjoint; cf. the alternative below. From B= 1 ≤ x ≤ 2, (x, y) 2 ≤y≤2 x , follows that B 1 dS = xy 2 1 2 2 x 1 dy xy 2 dx = 1 1 [ln y]22 dx = x x 2 1 1 1 ln x dx = (ln 2)2 .

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