s8 - Solution to Set 8 1 1 0 2 1. (a) 0 ex+y dxdy = 4−y 2...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution to Set 8 1 1 0 2 1. (a) 0 ex+y dxdy = 4−y 2 1 0 2 ex dx x2 2 √ 0 1 0 4−y 2 ey dy 1 2 = (e − 1)2 2 0 √ 0 (b) 0 xdxdy = 0 dy = (4 − y 2 )dy = 8 3 2. The reversed interated integral is 1 x 0 (a) 0 ex dydx = √ x 0 2 1 0 ex xdx = 2 1 x2 e 2 1 = 0 1 (e − 1) 2 π 2/3 2 4 − cos x3/2 = 3 3 0 y /2 0 4 π 2/3 (b) 0 sin x 3/2 π 2/3 dydx = y /2 0 0 (sin x3/2 )x1/2 dx = 4 4 3. (a) R (x + y )dA = 0 0 √ 0 (x + y )dxdy = 0 0 x2 + yx 2 dy = 0 52 40 y dy = 8 3 16−x2 (b) R xdA = −4 xdydx = −4 x 16 − x2 dx = − 64 3 5 2π 5 0 4. (a) R xydA = 0 π 0 r2 cos θ sin θ rdrdθ = 2π 0 π 0 cos θ sin θdθ 0 r3 dr =0 (b) R x2 dA = 4 sin θ 0 r2 cos2 θ rdrdθ = 43 cos2 θ sin4 θdθ = 4π 5. (a) The projection R of the portion of the plane on the xy-plane is the triangle bounded by the two axes and the line x = 8 − 2y . 1 A unit normal to the surface is n = √ (i + 2j + 3k). 14 3 n·k = √ 14 √ √ 4 8−2y 16 14 dA 14 dxdy = . Surface area = = 3 3 |n · k | 00 R (b) The intersection of the paraboloid with the xy-plane is the circle x2 + y 2 = 9. The projection R of the portion of paraboloid on the xy-plane is the disk with radius 3. 1 A unit normal vector to the paraboloid is n = (−2xi − 2y j − k ). 2 + 4y 2 + 1 4x dA 4x2 + 4y 2 + 1 dA. Surface area = = |n · k | R In terms of polar coordinates, the integral becomes 2π 3 π 4r2 + 1 rdrdθ = (373/2 − 1) ≈ 37.3 π . 6 0 0 1 6. (a) R is the annular region between x2 + y 2 = 1 and x2 + y 2 = 4; z 2 dS = σ R (x2 + y 2 ) √ x2 x2 y2 +2 + 1 dA 2 +y x + y2 2π 2 1 = √ 2 R (x2 + y 2 )dA = 2 0 r3 drdθ = 15 √ π 2. 2 √ (b) The projection of the surface σ (parameterized by z = + 1 − x2 ) on the xy -plane is the region R = [−1, 1] × [0, 1]. The suface integral becomes x2 ydS = σ 1 0 1 −1 x2 y −x √ 1 − x2 2 + 1 dxdy = 1 2 1 −1 x2 √ dx. 1 − x2 π /2 With the substitution u = sin x, the integral can be evaluated as Therefore, σ −π/2 sin2 u du = π/2. x y dS = π/4. 2 2 ...
View Full Document

This note was uploaded on 02/27/2010 for the course MATH MATH101 taught by Professor Chan during the Fall '09 term at HKUST.

Page1 / 2

s8 - Solution to Set 8 1 1 0 2 1. (a) 0 ex+y dxdy = 4−y 2...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online