solution_exam2

solution_exam2 - Math 251:H1 Exam #2 Name : Problem Number...

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

View Full Document Right Arrow Icon
Math 251:H1 Exam #2 Name : Problem Number Points Possible Points 1 15 2 10 3 15 4 10 5 10 6 10 7 10 8 10 9 10 10 (Bonus) 10 Total
Background image of page 1

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

View Full DocumentRight Arrow Icon
1. [15 pts] Consider the square R with vertices (0 , 0), (1 , 1), (2 , 0) and (1 , - 1). Using the change of coordinates T : ± x = ( u + v ) / 2 y = ( - u + v ) / 2 , evaluate ZZ R e x - y dA. Hint : The inverse change of coordinates is given by T - 1 : ± u = x - y v = x + y Solution: The region R can be written R = { ( x,y ) | 0 x - y 2 , 0 x + y 2 } . Then, we have that ZZ R 2 e x - y dA = ZZ S e u ² ² ² ² ( x,y ) ( u,v ) ² ² ² ² du dv, where S = { ( u,v ) | 0 u 2 , 0 v 2 } and ( x,y ) ( u,v ) = ∂x ∂u ∂y ∂v - ∂x ∂v ∂y ∂u = 1 2 · 1 2 - 1 2 · ³ - 1 2 ´ = 1 2 . Hence, ZZ R e x - y dA = Z 2 0 Z 2 0 e u 1 2 du dv = e 2 - 1 .
Background image of page 2
2. [10 pts] Consider the function f ( x,y ) = ye x 2 x 3 and D = { ( x,y ) | y x 1 , 0 y 1 } . Note that f is always positive on D . Find the volume of E = { ( x,y,z ) | ( x,y ) D, 0 z f ( x,y ) } . Solution: We have that the volume is given by V = ZZ D f ( x,y ) dA = Z 1 0 Z 1 y ye x 2 x 3 dx dy. We cannot integrate first with respect to x . Hence, we reverse the order of integration. Notice that we can also express D as D = { ( x,y ) | 0 y x 2 , 0 x 1 } . Then, we have that V = ZZ D f ( x,y ) dA = Z 1 0 Z x 2 0 ye x 2 x 3 dy dx = Z 1 0 x 2 e x 2 dx = 1 4 ( e - 1) .
Background image of page 3

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

View Full DocumentRight Arrow Icon
3. [15 pts] Consider the function f ( x,y ) = 4 x 2 + 3 y 2 - 4 x - 5. 1) [5 pts] Find and classify the critical point(s) of f . 2) [10 pts] Find the absolute minimum and maximum of f inside the region R = { ( x,y ) | x 2 + y 2 1 } . Solution: 1) Solving f ( x,y ) = h 8 x - 4 , 6 y i = h 0 , 0 i , we get that the only critical point of f is ( x,y ) = ( 1 2 , 0). Now, since f xx ( 1 2 , 0) f yy ( 1 2 , 0) - ± f xy ( 1 2 , 0) ² 2 = 8 · 6 - 0 · 0 = 48 > 0 and f xx ( 1 2 , 0) = 8 > 0, then we have that the critical point ( x,y ) = (2 , 0) is a local minimum. 2) We already showed in 1) that
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.

This note was uploaded on 08/10/2008 for the course MATH 251 taught by Professor Beck during the Spring '08 term at Rutgers.

Page1 / 11

solution_exam2 - Math 251:H1 Exam #2 Name : Problem Number...

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