Answers to additional exercises for Colley, Chapter 5Section 5.1S1. 3S2. 8Section 5.2S1. 32S2. TRUE. Everything about the integral on the left hand side is symmetric with respectto both the x−axis and the y−axis. The first of these implies that the integral over the squaredefined by −1 ≤ x ≤ 1 and −1 ≤ y ≤ 1 is twice the integral over the rectangle define by 0 ≤ x ≤ 1and −1 ≤ y ≤ 1, and the second implies that the integral over the latter rectangle is equal to twicethe integral over the square defined by 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Combining these, we concludethat the integral over the larger square is four times the integral over the smaller one.S3. loge3 −32loge2S4.13S5.23S6. The region A is given by the inequalities 0 ≤ y ≤√3 and −y ≤ x ≤ 2y − y2. In orderto interchange the order of integration, it is necesary to split this up into two regions, one of whichis defined by 0 ≤ x ≤ 1 and 1 −√1 − x ≤ y ≤ 1 +√1 − x, and the other of which is defined by−p(3) ≤ x ≤ 0 and −x ≤ y ≤ 1 +√1 − x. The area is equal to92−√3 .S7. 1Section 5.3S1.Z10Z1yp1 − x2dx dyS2.Z1−1Z√1−x20loge(1 + x2y2) dy dxS3.Z10Z3√xx2tan(xy) dy dx1S4.12(cos 1 − 1), where the θ in cos θ is given by radian measure.Section 5.4S1. 12S2.38S3.(e − 1)22S4. The region is defined by the inequalities −1 ≤ z ≤ 1, −p(1 − z2)/2 ≤ x ≤p(1 − z2)/2, and −√1 − x2− z2≤ y ≤√1 − x2− z2.S5. The region is defined by the inequalities 0 ≤ x ≤ 4, 0 ≤ y ≤ 4−x, and 0 ≤ z ≤ x+y+1.S6.760S7.Z√3/2−√3/2Z√34−x2−√34−x2Z√1−x2−y212f(x, y, z) dz dy dxSection 5.5S1. The corresponding region in the uv−plane is the triangle whose vertices are (0, 0),(0, 1) and (1, 0).S2. The corresponding region in the uv−plane is the square whose vertices are (0, 0),(0, 1), (1, 0) and (1, 1).S3.a33S4.23S5. 0S6.π8S7.83S8. 0S9. uv(u + v − uv)S10. 17S11.π2is the value of the original integral. The Jacobian of (x, y) with respect to (u, v)turns out to be 4(u2+ v2), so the correspondint uv-integrals will be given as follows:Z ZRx dx dy =Z ZS(u2− v2) · 4(u2+ v2) du dv2Z ZRy dx dy =Z ZS(2uv) · 4(u2+ v2) du dvFor computational purposes it is convenient to rewrite the right hand sides in polar coordinates. Ifmake such a change of variables, these are the iterated integrals we obtain (in order):Zπ/20Z104r5(cos2θ − sin2θ) dr dθ ,Zπ/20Z108r5(cos θ sin θ) dr dθSection 5.6S1.x = y = 0, z = h/4S2. x = y = 0, z = 3r/8S3. x = y = 0,z =3(R4− r4)8(R3− r3)S4.12(1 + e +