2011_Solution_8 - MA 1505 Mathematics I Tutorial 8...

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Unformatted text preview: MA 1505 Mathematics I Tutorial 8 Solutions ￿x=a ￿ ￿ b￿ a ￿ b￿ ￿ b￿ 13 13 2 2 2 2 1. (a) (x + y ) dxdy = x + xy dy = a + ay dy 3 3 0 0 0 ￿0 ￿b x=0 13 1 1 1 = a y + ay 3 = a3 b + ab3 . 3 3 3 3 0 (b) ￿ 1 2￿ 1 0 ￿ 2￿ ￿￿x=1 xy 1￿ 2 1/2 √ dxdy = − y 2(4 − x ) dy 2 4 − x2 x=0 ￿1 2 = −y (31/2 − 41/2 ) dy 1 ￿ ￿ √ 1 2 y=2 3√ = (2 − 3) y = 3− 3. 2 2 y =1 2. (a) The region can be regarded as a Type A region D: ￿ 0 1￿ x x2 e 0 dydx = ￿ 1￿ x2 ￿y=x 0 ≤ y ≤ x, ￿ ye dx = 0￿ ￿1 y=0 1 x2 1 = e = (e − 1). 2 2 0 1 0 ≤ x ≤ 1. 2 xex dx 0 (b) The region can be regarded as a type A region with bottom boundary y = x2 and top √ boundary y = x. Since the two curves intersect at x = 0 and x = 1, the left and right are bounded by x = 0 and x = 1 respectively. So D: ￿ 0 1￿ √ x x2 x2 ≤ y ≤ √ x, 0 ≤ x ≤ 1. ￿√ ￿1 1 2 y= x 1 1 (x + y ) dydx = xy + y dx = (x3/2 + x − x3 − x4 ) dx 2 2 2 0 y =x2 ￿0 ￿1 1 5/2 1 2 1 4 1 3 = 2x + x − x − x5 = . 5 4 4 10 10 0 ￿ 1￿ 3. The line joining (1, 0) and (4, 2) has equation y−0 2−0 2 2 2 3 = = ⇐⇒ y = x − ⇐⇒ x = y + 1. x−1 4−1 3 3 3 2 The line joining (1, 0) and (9, −3) has equation y−0 (−3) − 0 3 3 3 8 = = − ⇐⇒ y = − x + ⇐⇒ x = − y + 1. x−1 9−1 8 8 8 3 The region D is is the union of D1 and D2 , where MA1505 Tutorial 8 Solutions 3 y 2 ≤ x ≤ y + 1, 2 8 y 2 ≤ x ≤ − y + 1, 3 D1 : D2 : Hence the required answer is ￿￿ ￿￿ x dA = D = ￿ 0 = since ￿ 0 ￿ 0 −3 2 ￿ (3y/2)+1 x dA + D1 2 ￿ (3y/2)+1 −(8y/3)+1 − 3 ≤ y ≤ 0. x dA D2 x dxdy + y2 x dxdy = ￿ 2 0 x dxdy = y2 ￿ 0 −3 ￿ 0 −3 19 106 + = 25, 5 5 y2 ￿ ￿￿ 0 ≤ y ≤ 2, ￿ −(8y/3)+1 x dxdy y2 1 19 (9y 2 + 12y + 4 − 4y 4 ) dy = , 8 5 1 106 (64y 2 − 48y + 9 − 9y 4 ) dy = . 18 5 4. The region in Cartesian coordinates is given by ￿ D : 0 ≤ y ≤ 1 − x2 , 0≤x≤1 This is a type A region with x-axis as the bottom boundary and upper half of the unit circle as the upper boundary. Since the range of x is from 0 to 1, the region D is the first quadrant of the unit disk. y . . . . .. . ... ... . . . . . ....... ........... . ...... . ..... . . .... .... . . ... . ... . ... . ... . ... . ... . ... . ... . . .. . .. . .. . . .. . .. . . .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .............................................. . .. .............................................. ......................................... ... .............................................. .. . 1 y= √ 1 − x2 R 0 1 x In polar coordinates, this is given by D: ￿ 0 1￿ 0 √ 1−x2 ex 2 +y 2 dydx = ￿ 0 ≤ r ≤ 1, π /2 ￿ 1 er r drdθ = 2 0 ≤ θ ≤ π /2. ￿ 0￿ 0￿ 0 π 1 r2 1 1 = e = 4 π (e − 1). 22 0 2 π /2 dθ ￿ 0 1 2 er r dr MA1505 Tutorial 8 Solutions 5. (a) The type B region R is given by √ 3 y ≤ x ≤ 2, 0 ≤ y ≤ 8. √ It is bounded on the left by the cubic curve 3 y = x and on the right by the vertical line x = 2. Below it is bounded by the x-axis, and on top the left and right boundaries intersect at y = 8. y . . . .. . . .. ... . . . . . .............. . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . . . . . . . . . . . . . . .. . .. . . . .. . .. . .. . .. .. . .. . . .. . .. .. . . . . .. . 3 . .. . . . . .. . . .. . . . . . . . .. .. . . . . ... . . ... . . . . . ... ... . . . .. . . . . ... . . ... . . . . . . .... . . .... . . . .. ................................................ . . .. ....... . .................................................. .............................................. .............................................. . . .. . . 8 √ x= y R 0 2 x Converting to type A region, the lower boundary is y = 0, the top boundary is the cubic curve y = x3 . On the left, these two boundaries intersect at x = 0 and on the right, it is bounded by x = 2. So the region is given by 0 ≤ y ≤ x3 , 0 ≤ x ≤ 2. ￿ 0 8￿ 2 √ 3y 4 ex dxdy = ￿ 2 ￿ x3 4 ex dydx = ￿ 2 0 ￿0 0 ￿2 1 x4 1 16 = e = (e − 1). 4 4 0 3 4 ex [y ]y=x dx = y =0 ￿ 2 4 x3 ex dx 0 (b) The type B region R is given by y/2 ≤ x ≤ √ ln 3, √ 0 ≤ y ≤ 2 ln 3. It is bounded on the left by the straight line x = y/2 and on the right by the vertical line √ x = ln 3. Below it is bounded by the x-axis, and on top the left and right boundaries intersect at √ y = 2 ln 3. y . . . .. . .. ... . . . . √ . . . .. .. . . . 2 ln 3 ... ... ... ... ... ... ... ... ... ... ... ... ................ . . .. . . . . . . . . . ... . ... . . .. . . . . ... . . ... . . . . y ...... . . ... . . . . ... . . .. . . . 2.... . . . . ... . . . . ... ... . . . . . . . ... . . ... . . . . . ... . . ... . . . . . ... . . ... . . . . . ... . ..... . . . . .. . . . ..... . . . .... . ..... .. . ........................................................................................... ....................................................................................... ... .. . . x= R √ ln 3 0 x Converting to type A region, the lower boundary is y = 0, the top boundary is the line y = 2x. On the left, these two boundaries intersect at x = 0 and on the right, it is bounded by √ x = ln 3. So the region is given by √ 0 ≤ y ≤ 2x, 0 ≤ x ≤ ln 3. 3 MA1505 Tutorial 8 Solutions ￿ 0 √ √ 2 ln 3 ￿ ln 3 x2 e dxdy = y/2 ￿ ￿ √ 0 ln 3 ￿ 2x x2 e dydx = 0 x2 =e ￿√ln 3 0 ￿ √ ln 3 e 0 =e ln 3 x2 [y ]y=2x dx y =0 = ￿ √ ln 3 2 2xex dx 0 − 1 = 2. 6. The volume is given by the double integral ￿￿ V= f (x, y )dA D where D is the region bounded by the parabola y = 4 − x2 and straight line y = 3x and f (x, y ) is the function whose graph is the plane x − z + 4 = 0. Writing the equation of the plane as z = x + 4, we get the function f (x, y ) = x + 4. A rough sketch of the region D is shown below: D can be regarded as type A region D: 3x ≤ y ≤ 4 − x2 , − 4 ≤ x ≤ 1. (The two limits −4 and 1 of x are obtained by solving the two equation y = 3x and y = 4 − x2 .) Hence ￿ V= 1 −4 ￿ 4−x2 3x ￿ ￿ 73 141 625 2 (x +4)dydx = (x +4)(4 − x − 3x)dx = 16x − 4x − x − x = 3 4 12 −4 −4 ￿ 1 2 7. By symmetry, the desired volume V is 8 times the volume V1 in the first octant. Now, V1 = ￿ r ￿ √r2 −y2 ￿ ￿ r ￿￿ ￿x=√r2 −y2 x r2 − y 2 dy r2 − y 2 dxdy = ￿ ￿0 ￿0 r 0 13r 2 2 2 2 = (r − y ) dy = r y − y = r3 . 3 3 0 0 Therefore, V = x=0 16 3 3r . x 4 z . . . .. .. ... . . . . . . . ............. ............... .. . .... .... ... . ... . .. ... ... ... . ... . ... . ... . . ... ... ... . . .. . . .. ... ... . .. . . . .. ... ... . .. . . . ... . ... . . . . . . . . ....... . .......... . ... .... . . .. . ....... . . ...... . . . . . ..... . . ..... . . .. . . . . . . .. . .. . .......... ........................... . . ....................................... . . . .. . . .. . ... .. .................... .. ... ................ . . . . . ... . ... . . . . . .. ... . .. ... . . . . ... ... . ... ... . .. .. . ... ... . ... ... . ..... . .. .... ..... . ..... .. .. ..... ...... ...... . . .................... ........ ....... . ................... ... ... .... .... . .... .... y ...
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