Explanation Given: A map T ( u , v ) = ( u 2 − v 2 , 2 u v ) , a triangle D o = { ( u , v ) : 0 ≤ v ≤ u ≤ 1 } , bounded domain x = 0 , y = 0 , y 2 = 4 − 4 x . Double integral: ∬ D x 2 + y 2 d x d y Formula used: Jacobian = [ ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ] Calculation: From the given map, we have: ⇒ x = u 2 − v 2 ⇒ y = 2 u v Let us solve this system in terms of u and v . ⇒ x = u 2 − v 2 ⇒ y = 2 u v ⇒ ( u 2 + v 2 ) 2 = ( u 2 − v 2 ) 2 + 4 u 2 v 2 ⇒ ( u 2 + v 2 ) 2 = x 2 + y 2 ⇒ u 2 + v 2 = x 2 + y 2 Let us now solve the system: ⇒ u 2 + v 2 = x 2 + y 2 ⇒ u 2 − v 2 = x On adding the two equations: ⇒ u 2 = x 2 + y 2 + x ⇒ u = x 2 + y 2 + x On subtracting the two equations: ⇒ v 2 = x 2 + y 2 − x ⇒ v = x 2 + y 2 − x Upon plugging these expressions in the given triangular region, we can set up: ⇒ 0 ≤ v ≤ u ≤ 1 ⇒ 0 ≤ x 2 + y 2 − x ≤ x 2 + y 2 + x ≤ 1 ⇒ 0 ≤ x 2 + y 2 − x ≤ x 2 + y 2 + x ≤ 1 ⇒ 0 ≤ x 2 + y 2 − x ⇒ y = 0 , x = 0 ⇒ x 2 + y 2 + x ≤ 1 ⇒ y 2 = 4 − 4 x Thus, T ( u , v ) = ( u 2 − v 2 , 2 u v ) maps the triangle D o = { ( u , v ) : 0 ≤ v ≤ u ≤ 1 } to the domain D bounded by x = 0 , y = 0 , y 2 = 4 − 4 x ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 16.6, Problem 33E

To determine

To prove:

That $T (u, v) = (u^{2} - v^{2}, 2 u v)$ maps the triangle $D_{o} = {(u, v) : 0 \leq v \leq u \leq 1}$ to the domain D bounded by $x = 0, y = 0, y^{2} = 4 - 4 x$ . Further, we have to evaluate the given double integral using the map.

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Chapter 16.6, Problem 32E

Chapter 16.6, Problem 34E

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Chapter 16 Solutions

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To prove: That T ( u , v ) = ( u 2 − v 2 , 2 u v ) maps the triangle D o = { ( u , v ) : 0 ≤ v ≤ u ≤ 1 } to the domain D bounded by x = 0 , y = 0 , y 2 = 4 − 4 x . Further, we have to evaluate the given double integral using the map.

Solution Summary: The author proves that T maps the triangle D o = ( u, v ) to a bounded domain D, and evaluates the given double integral using the map.

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To prove:

That $T (u, v) = (u^{2} - v^{2}, 2 u v)$ maps the triangle $D_{o} = {(u, v) : 0 \leq v \leq u \leq 1}$ to the domain D bounded by $x = 0, y = 0, y^{2} = 4 - 4 x$ . Further, we have to evaluate the given double integral using the map.

Want to see the full answer?

Chapter 16 Solutions

To prove: That T ( u , v ) = ( u 2 − v 2 , 2 u v ) maps the triangle D o = { ( u , v ) : 0 ≤ v ≤ u ≤ 1 } to the domain D bounded by x = 0 , y = 0 , y 2 = 4 − 4 x . Further, we have to evaluate the given double integral using the map.

Solution Summary: The author proves that T maps the triangle D o = ( u, v ) to a bounded domain D, and evaluates the given double integral using the map.

Videos

To prove: That T(u,v)=(u2−v2,2uv) maps the triangle Do={(u,v):0≤v≤u≤1} to the domain D bounded by x=0,y=0,y2=4−4x. Further, we have to evaluate the given double integral using the map.

Want to see the full answer?

Chapter 16 Solutions

To prove:

That $T (u, v) = (u^{2} - v^{2}, 2 u v)$ maps the triangle $D_{o} = {(u, v) : 0 \leq v \leq u \leq 1}$ to the domain D bounded by $x = 0, y = 0, y^{2} = 4 - 4 x$ . Further, we have to evaluate the given double integral using the map.