- August 4, 2020

1. [i] (Concept Checks): Determine whether the following statements are true or false. Circle your answer. • (True/False) Z 2 0 Z 1 0 xy2dydx = Z 2 0 xdx Z 1 0 y2dy. • (True/False) Z 4 1 x2 p ydy is a number. • (True/False) Z 1 0 Z x 0 f(x, y)dydx = Z 1 0 Z y 0 f(x, y)dxdy • (True/False) In polar coordinates, ZZ R drd✓ is the area of the closed and bounded region R. • (True/False) In cylindrical coordinates, ZZZ D rdrd✓dz is the volume of the solid D. • (True/False) Without calculating the integral, we can tell that Z R (x x2)dA where R = {(x, y) | 1 x 1 1 y 1} is negative. • (True/False) Z 1 0 Z 1 x dydx represents the area of a triangular region in the xy-plane. • (True/False) Given f(x, y) is a positive function and assuming the following integrals exist. We have Z 1 0 Z 1 0 f(x, y)dydx < Z 1 0 Z 1 x2 f(x, y)dydx < Z 1 0 Z 1 x3 f(x, y)dydx • (True/False) Z 3 0 Z 4x 0 f(x, y)dydx = Z 12 0 Z 3 y/4 f(x, y)dxdy. • (True/False) The region of the integration for Z r r Z pr2 x2 pr2 x2 Z 100 0 f(x, y, z)dzdydx is a cylin- der. • (True/False) Z 1 0 Z 1 0 Z 1 0 dzdydx represents the volume of a cube with side-length 1. • (True/False) Using polar coordinates, we haveZ 3 0 Z 2⇡ ⇡ rd✓dr = Z 3 0 Z p9 y2 p 9 y2 dydx . • (True/False) Using cylindrical coordinates, we haveZ 5 5 Z 3 0 Z p25 x2 p25 x2 xdydzdx = Z 5 0 Z 3 0 Z 2⇡ 0 r cos ✓d✓dzdr 3 [ii] (Skill Builders): • Let ZZ R f(x, y)dA where R is the region bounded by the curves x = p y and x = y and the line y = 1. Rewrite the integral as an iterated integral. • Rewrite Z 2 0 Z p4 x2 0 (x2 + y2)dydx using polar coordinates and evaluate it. • Give a geometric interpretation for the integral Z 1 1 Z p1 x2 p1 x2 Z 2p1 x2 y2 0 dzdydx and eval- uate it using cylindrical coordinates. 4 • Calculate ZZZ D ydV where D is the solid enclosed by 2x+ y = 0, 2x+ y = 3, x z = 0,x z = 2,z = 0, and z = 4. • We can calculate the area of the region R by evaluating Z 3⇡/4 ⇡/4 Z 3 2 rdrd✓. What does this area look like? Sketch or describe it. • We can calculate the volume of a solid S by evaluating Z 1 0 Z 1 0 (4 x 2y)dxdy. What does this solid look like? Sketch or describe it. • Without calculating the integral, explain why ZZ R xy 1 + x4 dA = 0 where R = {(x, y) | 1 x 1 , 0 y 1}. 5 2. With so many tools available to calculate the integrals, sometimes all you need to do is to set up the integral that calculates what you need. This may not be the case at some other times. In this question, just set up the integrals that calculate the followings, do not evaluate the integrals! • Find the area of the region in the first quadrant enclosed by y = 6x x2 and y = 4x 8. • Find the volume under z = f(x, y) = 1 x2 y2 and above R = {(x, y) | 1 x 1 , 0 y 1}. • Find the volume of the region bounded by both of x2 + y2 = 1 and x2 + z2 = 1. • Consider the solid that is enclosed by z = 16 4x2 y2 and the xy-plane. Now only take the part of this solid that is in the first octant. Find its volume. • Find the volume of the solid enclosed by the upper half (z 0) of the sphere x2+y2+z2 = 4 and the cylinder (x 1)2 + y2 = 1. 6 3. Is it possible to invent an invertible transformation that maps the ellipsoid x2 + 8y2 + 6z2 + 4xy 2xz + 4yz = 9 onto the unit sphere? 4. Let V denote the volume of region enclosed by the ellipsoid (x2/a2) + (y2/b2) + (z2/c2) = 1. Set up the integral that calculates V and use a transformation. Then interpret the integral geometrically and find the answer to this integral without calculating the integral. 7 5. a) Show that ZZ D(R) e (x 2+y2) dx dy = ⇡(1 e R2) where D(R) is the disc of radius R, centred at (0, 0). b) Let D be the square {(x, y) | |x| b, |y| b}. Show that ZZ D e (x 2+y2) dx dy = 4 [email protected] bZ 0 e x 2 dx 1A2 . c) Hence, prove that 1Z 0 e x 2 dx = p ⇡ 2 a result which is important in probability theory and in other applications. This integral cannot be evaluated directly. 8