Revisions to Double integral over special region [duplicate]

added 13 characters in body

Source Link

edited Jan 27, 2022 at 15:54

84.1k
6
97
179

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand (6 x^2 + 33 x*y + 42 y^2)^4 is just (u*v)^4 if we set u=3(x+2y) and v=2x+7y.

Then the region became -2<=u<=2 and -2<=v<=2.

To calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)

1/9

Clear[u,v];
Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand (6 x^2 + 33 x*y + 42 y^2)^4 is just (u*v)^4 if we set u=3(x+2y) and v=2x+7y.

Then the region became -2<=u<=2 and -2<=v<=2.

To calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)

1/9

Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand (6 x^2 + 33 x*y + 42 y^2)^4 is just (u*v)^4 if we set u=3(x+2y) and v=2x+7y.

Then the region became -2<=u<=2 and -2<=v<=2.

To calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)

1/9

Clear[u,v];
Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

added 361 characters in body

Source Link

edited Jan 27, 2022 at 15:43

cvgmt

84.1k
6
97
179

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand (6 x^2 + 33 x*y + 42 y^2)^4 is just (u*v)^4 if we set u=3(x+2y) and v=2x+7y. The

Then the region became -2<=u<=2 and -2<=v<=2. To

To calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)

1/9

Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand is just (u*v)^4 if we set u=3(x+2y) and v=2x+7y. The region became -2<=u<=2 and -2<=v<=2. To calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)

1/9

Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand (6 x^2 + 33 x*y + 42 y^2)^4 is just (u*v)^4 if we set u=3(x+2y) and v=2x+7y.

Then the region became -2<=u<=2 and -2<=v<=2.

To calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)

1/9

Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

added 361 characters in body

Source Link

edited Jan 27, 2022 at 15:37

cvgmt

84.1k
6
97
179

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand is just (u*v)^4 if we set u=3(x+2y) and v=2x=7yv=2x+7y. and theThe region became -2<=u<=2 and -2<=v<=2. and we toTo calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{u3(x+2y), v2x+7y}, {x, y}] // Det // Abs)

1/9

Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand is (u*v)^4 if we set u=3(x+2y) and v=2x=7y. and the region became -2<=u<=2 and -2<=v<=2. and we to calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{u, v}, {x, y}] // Det // Abs)

1/9

Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

Method 1

Integrate[
 Boole[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}](6 x^2 + 
    33 x*y + 42 y^2)^4, {x, -∞, ∞}, {y, -∞, ∞}]

Method 2

Integrate[(6 x^2 + 33 x*y + 42 y^2)^4, {x, y} ∈ 
  ImplicitRegion[{ -2 <= 2 x + 7 y <= 2 && -2 <= 3 x + 6 y <= 2}, {x, 
    y}]]

Method 3

(* change of variables *)
sol = Solve[{u == 2 x + 7 y, v == 3 x + 6 y}, {x, y}][[1]];
(*  {x -> 1/9 (-6 u + 7 v), y -> 1/9 (3 u - 2 v)}  *)
expr = (6 x^2 + 33 x*y + 42 y^2)^4 /. sol // Simplify;
(* u^4 v^4 *)
jacobian = Grad[{x, y} /. sol, {u, v}] // Det // Abs
(* 1/9 *)
Integrate[expr*jacobian, {u, -2, 2}, {v, -2, 2}]

Method 4

Factor[6 x^2 + 33 x*y + 42 y^2]
(* 3 (x + 2 y) (2 x + 7 y) *)

It means that the integrand is just (u*v)^4 if we set u=3(x+2y) and v=2x+7y. The region became -2<=u<=2 and -2<=v<=2. To calculate Grad[{x,y},{u,v}]//Det, we just need to calculate 1/Grad[{u,v},{x,y}]//Det.

1/(Grad[{3(x+2y),2x+7y}, {x, y}] // Det // Abs)

1/9

Integrate[(u*v)^4*1/9,{u,-2,2},{v,-2,2}]

added 361 characters in body

Source Link

edited Jan 27, 2022 at 15:31

cvgmt

84.1k
6
97
179

Loading

deleted 38 characters in body

Source Link

edited Jan 27, 2022 at 15:08

cvgmt

84.1k
6
97
179

Loading

added 45 characters in body

Source Link

edited Jan 27, 2022 at 15:02

cvgmt

84.1k
6
97
179

Loading

added 9 characters in body

Source Link

edited Jan 27, 2022 at 14:55

cvgmt

84.1k
6
97
179

Loading

added 261 characters in body

Source Link

edited Jan 27, 2022 at 14:48

cvgmt

84.1k
6
97
179

Loading

added 2 characters in body

Source Link

edited Jan 27, 2022 at 14:27

cvgmt

84.1k
6
97
179

Loading

Source Link

created Jan 27, 2022 at 14:14

cvgmt

84.1k
6
97
179

Loading

Stack Exchange Network

Return to Answer