practiceexam3-sol - Math 2163 Practice Exam III Solution 1...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 2163, Practice Exam III, Solution 1. 2. 1
Background image of page 1

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

View Full Document Right Arrow Icon
2
Background image of page 2
3. (a) i 1 0 i 1 0 ye xy dx dy = i 1 0 e xy | x =1 x =0 dy = i 1 0 ( e y 1) dy = e 2 (b) i 1 0 i y 0 i 1 x 6 xyz dz dx dy = i 1 0 i y 0 (3 xy 3 x 3 y ) dx dy = i 1 0 ( 3 2 y 3 3 4 y 5 ) dy = 1 4 (c) i π/ 2 0 i sin 2 θ 0 r dr dθ = i π/ 2 0 1 2 sin 2 (2 θ ) = i π/ 2 0 1 4 (1 cos 4 θ ) = π/ 8 (d) i π/ 2 0 i π/ 2 0 i 2 1 ρ 2 sin φ dρ dφ dθ = i π/ 2 0 i π/ 2 0 7 3 sin φ dφ dθ = i π/ 2 0 7 3 = 7 π 6 4. D = { ( r, θ ) | 0 θ 2 π, 0 r 4 + 3 cos θ } , so A ( D ) = ii D dA = i 2 π 0 i 4+3 cos θ 0 r dr dθ = i 2 π 0 1 2 (4 + 3 cos θ ) 2 = 1 2 i 2 π 0 (16 + 24 cos θ + 9 cos 2 θ ) = 1 2 i 2 π 0 (16 + 24 cos θ + 9(1 + cos 2 θ ) / 2) = 41 2 π 5. Use polar coordinates, D = { ( r, θ ) | 0 θ 2 π, 0 r 3 } , and f x = 2 x, f y = 2 y. So A = D r (2 x ) 2 + (2 y ) 2 + 1 dA = i 2 π 0 i 3 0 ( 4 r 2 + 1) r dr dθ = π 6 (37 37 1) 3
Background image of page 3

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

View Full Document Right Arrow Icon
6. Solution 1: Use a double integral, D is the intersection of b z = 3 x 2 + 3 y 2 z = 4 x 2 y 2 x 2 + y 2 = 1 . So D = { ( r, θ ) | 0 θ 2 π, 0 r 1 } , and V = ii D ( top surface bottom surface ) dA = i 2 π 0 i 1 0 ((4 r 2 ) 3 r 2 ) r dr dθ = 2 π.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 6

practiceexam3-sol - Math 2163 Practice Exam III Solution 1...

This preview shows document pages 1 - 5. Sign up to view the full document.

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