{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

practiceexam3-sol

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

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

Math 2163, Practice Exam III, Solution 1. 2. 1

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

View Full Document
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

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

View Full Document
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.
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
Ask a homework question - tutors are online