Explanation G is a coordinate transformation from x y -plane to u v -plane then G ( u , v ) = ( x , y ) ---------------- (i) Given: G ( u , v ) = ( u − u v , u v ) ---------- (ii) Here, x = u − u v ; y = u v Calculation: Substitute y = u v in x x = u − y x + y = u and y = u v ⇒ v = y u ⇒ v = y x + y u = x + y ; v = y x + y ---------------- (iii) a) To find the image of v = c Consider x = u − u v x = u ( 1 − v ) x 1 − v = u , substitute in y y = u v y = x 1 − v v Substitute v = c in y y = x c 1 − c ⇒ y = c 1 − c x Thus image of v = c is y = c 1 − c x , if c ≠ 1 To find image of v = c , if c = 1 Substitute v = c is x = u − u v x = u ( 1 − v ) = u ( 1 − c ) = u ( 1 − 1 ) = u ( 0 ) = 0 = u ( 1 − c ) , when c = 1 Thus, x = 0 is the equation of y − a x i s Thus the image of v = c , if c = 1 is y − a x i s b) Given region From the figure the equation of straight lines are x + y = a , x + y = b By the transformation from (iv) we have x + y = a ⇒ u = a x + y = b ⇒ u = b x = 0 ⇒ u − u v = 0 u ( 1 − v ) = 0 u = 0 or 1 − v = 0 ⇒ v = 1 y = 0 u v = 0 u = 0 or v = 0 Region in u v plane in A ' B ' C ' D ' i.e., the region enclosed between u = a , u = b ; v = 0 , v = 1 Image of vertical lines are u = a is x + y = a u = b is x + y = b c) To find Jacobian Jacobian of the transformation is J ( u , v ) = ∂ ( x , y ) ∂ ( u , v ) = | ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v | Where ∂ x ∂ u , ∂ x ∂ v , ∂ y ∂ u , ∂ y ∂ v are first order partial derivatives ∂ x ∂ u = ∂ ∂ u ( u − u v ) = ∂ ∂ u ( u ) − ∂ ∂ u ( u v ) = 1 − v ∂ ∂ u ( u ) = 1 − v ( 1 ) [While differentiating partially with respect to u treat v as constant] ∂ x ∂ u = 1 − v ∂ x ∂ v = ∂ ∂ v ( u − u v ) = ∂ ∂ v ( u ) − u ∂ ∂ v ( v ) = 0 − u ( 1 ) = − u [While differentiating partially with respect to v treat u as constant] ∂ x ∂ v = u ∂ y ∂ u = ∂ ∂ u ( u v ) = v ∂ ∂ u ( u ) = v ( 1 ) = v ∂ y ∂ u = v ∂ y ∂ v = ∂ ∂ v ( u v ) = u ∂ ∂ v ( v ) = u ( 1 ) = u ∂ y ∂ v = u Substitute these in (iii) J ( u , v ) = | 1 − v − u v u | 2 × 2 determinant | a b c d | = a d − b c J = ( 1 − v ) u − v ( − u ) = u − v u + v u = u Thus, J ( u , v ) = u d) Area of the triangle is 1 2 ( b a s e ) ( h e i g h t ) Given figure:

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 16.6, Problem 22E

To determine

a) The image of $v = c$

b) The image of vertical lines in the $u v$ plane

c) Jacobian

d) Area

e) $\iint_{D} x y d x d y$

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Chapter 16.6, Problem 21E

Chapter 16.6, Problem 23E

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Chapter 16 Solutions

CALCULUS (CLOTH)

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a) The image of v = c b) The image of vertical lines in the u v plane c) Jacobian d) Area e) ∬ D x y d x d y

Solution Summary: The author explains that G is a coordinate transformation from x y to u v -plane.

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a) The image of $v = c$

b) The image of vertical lines in the $u v$ plane

c) Jacobian

d) Area

e) $\iint_{D} x y d x d y$

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Chapter 16 Solutions

a) The image of v = c b) The image of vertical lines in the u v plane c) Jacobian d) Area e) ∬ D x y d x d y

Solution Summary: The author explains that G is a coordinate transformation from x y to u v -plane.

Videos

a) The image of v=c b) The image of vertical lines in the uvplane c) Jacobian d) Area e) ∬Dxydxdy

Want to see the full answer?

Chapter 16 Solutions

a) The image of $v = c$

b) The image of vertical lines in the $u v$ plane

c) Jacobian

d) Area

e) $\iint_{D} x y d x d y$