Vector Calculus. Dr. D. Sukumar

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Transcription:

Vector Calculus Dr. D. Sukumar

Space co-ordinates Change of variable

Cartesian co-ordinates < x <

Cartesian co-ordinates < x < < y <

Cartesian co-ordinates < x < < y < < z <

Cylindrical

Cylindrical

Cylindrical

Cylindrical 0 r <

Cylindrical 0 r <

Cylindrical 0 r <

Cylindrical 0 r <

Cylindrical 0 r < 0 θ 2π

Cylindrical 0 r < 0 θ 2π

Cylindrical 0 r < 0 θ 2π

Cylindrical 0 r < 0 θ 2π

Cylindrical 0 r < 0 θ 2π < z <

Spherical

Spherical

Spherical

Spherical 0 ρ<

Spherical 0 ρ<

Spherical 0 ρ<

Spherical 0 ρ<

Spherical 0 ρ< 0 θ 2π

Spherical 0 ρ< 0 θ 2π

Spherical 0 ρ< 0 θ 2π

Spherical 0 ρ< 0 θ 2π

Spherical 0 ρ< 0 θ 2π 0 φ π

Multivariable function A function f is given on a domain D f : D R 3 R

Multivariable function A function f is given on a domain D D:= Set form f : D R 3 R

Multivariable function A function f is given on a domain D f : D R 3 R D:= Set form {(x, y, z) a x b, c y d, e z f }

Multivariable function A function f is given on a domain D f : D R 3 R D:= Set form {(x, y, z) a x b, c y d, e z f } D:= Description by words

Multivariable function A function f is given on a domain D f : D R 3 R D:= Set form {(x, y, z) a x b, c y d, e z f } D:= Description by words Tetrahedron cut from the first octant by the plane 6x + 3y + 2z = 6

Multivariable function A function f is given on a domain D f : D R 3 R D:= Set form {(x, y, z) a x b, c y d, e z f } D:= Description by words Tetrahedron cut from the first octant by the plane 6x + 3y + 2z = 6 D:= Picture

Triple Integral(Rectangular co-ordinates) Describe the regions and setup the integral 1. Write six different iterated triple integrals for the volume of the rectangular solids in the first octant bounded by the co-ordinates planes and the planes x = 1,y = 2,z = 3 2. Region in the first octant enclosed by the cylinder x 2 + z 2 = 4 and the plane y = 3. 3. Region bounded by paraboloids z = 8 x 2 y 2 and z = x 2 + y 2

Triple integral (Cylindrical co-ordinates) 1. Solid enclosed by the cylinder x 2 + y 2 = 4 bounded above by the paraboloid z = x 2 + y 2 2. Cylinder whose base is the circle r = 3cosθ and whose top lies in the plane z = 5 x 3. Find the average value of the function f (r, θ, z) = r over the region bounded by the cylinder r = 1 between the planes z = 1 and z = 1.

Triple integral (Spherical co-ordinates) 1. Region bounded below by z = 0, and above by the sphere x 2 + y 2 + z 2 = 4 side by x 2 + y 2 = 1. 2. Find the average value of the function f (ρ, φ, θ) over the solid ball ρ 1.

Cylindrical to Rectangular

Cylindrical to Rectangular P(r, θ, z)

Cylindrical to Rectangular P(r, θ, z)

Cylindrical to Rectangular P(r, θ, z) A(r, θ, 0)

Cylindrical to Rectangular B r cos θ O θ r P(r, θ, z) r sin θ A(r, θ, 0)

Cylindrical V = dzr drdθ

Cylindrical V = dzr drdθ x = r cos θ y = r sin θ z = z

Cylindrical V = dzr drdθ x = r cos θ y = r sin θ z = z

Spherical to Rectangular O

Spherical to Rectangular P(ρ, φ, θ) O

Spherical to Rectangular B ρ cos φ φ P(ρ, φ, θ) O

Spherical to Rectangular B ρ cos φ φ P(ρ, φ, θ) O ρ sin φ

Spherical to Rectangular B ρ cos φ φ P(ρ, φ, θ) O θ ρ sin φ A

Spherical to Rectangular B ρ cos φ φ P(ρ, φ, θ) O θ C ρ sin φ ρ sin φ cos θ D ρ sin φ sin θ A

Spherical V = ρ 2 sin φdρdφdθ x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ

Spherical V = ρ 2 sin φdρdφdθ x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ

Change of variable In evaluating the integration b f (x)dx we do a change of a variable t = 2x ˆ b a f (x) dx

Change of variable In evaluating the integration b f (x)dx we do a change of a variable t = 2x ˆ b ( t f dx 2) a t = 2x x = t 2

Change of variable In evaluating the integration b f (x)dx we do a change of a variable t = 2x ˆ 2b ( t f dx 2) 2a t = 2x x = t 2 x = a t = 2a, x = b t = 2b

Change of variable In evaluating the integration b f (x)dx we do a change of a variable t = 2x ˆ 2b ( t ) dt f 2 2 2a t = 2x x = t 2 x = a t = 2a, x = b t = 2b x = t 2 dx = dt 2

Change of variable Let G be the region in u, v plane. It is transformed into another region R in the xy-plane as a one-to-one map by the equation. x = g(u, v), y = h(u, v). We call, R is the image of G, G is the pre-image of R. Any function f defined on R can be thought of as a function f (g(u, v), h(u, v)) defined on G as well.

so What is the relation between integral f over R and integral f (g(u, v), h(u, v)) over G. If 1. f,g,h have continuous partial derivatives and 2. J(u, v) is zero only at isolated points then f (x, y)dxdy = f (g(u, v), h(u, v)) J(u, v) dudv R G

Jacobian Jacobian determinant or Jacobian of the co-ordinate transformation x = g(u, v), y = h(u, v). is J(x, y; u, v) = ( x u x u x v y v ) = (x, y) (u, v) = J(u, v)

Example Solve the system u = x y; v = 2x + y for x and y in-terms of u and v.

Example Solve the system u = x y; v = 2x + y for x and y in-terms of u and v. Find the Jacobian.

Example Solve the system u = x y; v = 2x + y for x and y in-terms of u and v. Find the Jacobian. Let R be the triangle region with vertices (0,0), (1,1) and (1, -2) in the xy-plane. Find the image of R and sketch the transformed region in uv-plane.

Example Solve the system u = x y; v = 2x + y for x and y in-terms of u and v. Find the Jacobian. Let R be the triangle region with vertices (0,0), (1,1) and (1, -2) in the xy-plane. Find the image of R and sketch the transformed region in uv-plane. Evaluate R (2x 2 xy y 2 )dxdy enclosed by the triangular region.

Example Solve the system u = x y; v = 2x + y for x and y in-terms of u and v. Find the Jacobian. Let R be the triangle region with vertices (0,0), (1,1) and (1, -2) in the xy-plane. Find the image of R and sketch the transformed region in uv-plane. Evaluate R (2x 2 xy y 2 )dxdy enclosed by the triangular region. Evaluate the same integral in the region bounded by the lines R : y = 2x + 4, y = 2x + 7, y = x 2, y = x + 1

u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3

u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 ( x J(u, v) = (x,y) = (u,v) u y y u v x v y v ) = (0,3) ( 1 3 2 3 1 3 1 3 ) = 1 9 + 2 9 = 1 3 x G u + v = 3 (3,0) u

u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 ( x J(u, v) = (x,y) = (u,v) u y y u v x v y v ) = (0,3) ( 1 3 2 3 1 3 1 3 ) = 1 9 + 2 9 = 1 3 x G u + v = 3 (3,0) u

u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 ( x J(u, v) = (x,y) = (u,v) u y (1,1) y u v x v y v ) = ( 1 3 2 3 1 3 1 3 ) = 1 9 + 2 9 = 1 3 R x (1,-2) u

u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 ( x J(u, v) = (x,y) = (u,v) u y (1,1) y u v x v y v ) = (0,3) ( 1 3 2 3 1 3 1 3 ) = 1 9 + 2 9 = 1 3 R x G u + v = 3 (1,-2) (3,0) u

u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 ( x J(u, v) = (x,y) = (u,v) y (1,1) y u u v x v y v ) = (0,3) ( 1 3 2 3 1 3 1 3 ) = 1 9 + 2 9 = 1 3 R x G u + v = 3 (1,-2) (3,0) u ˆ 3 ˆ v 3 0 0 (2x 2 xy y 2 ) 1 3 dudv

Example(cont.) R (2x 2 xy y 2 )dxdy u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3

Example(cont.) R (2x 2 xy y 2 )dxdy u = x y; v = 2x + y x = u+v 3 ; y = 2u+v 3 R : y = 2x + 4, y = 2x + 7, y = x + 1, y = x 2 G : v = 4, v = 7, u = 1, u = 2

Example(cont.) R (2x 2 xy y 2 )dxdy u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 R : y = 2x + 4, y = 2x + 7, y = x + 1, y = x 2 G : v = 4, v = 7, u = 1, u = 2 dxdy = J(x, y; u, v)dudv = 1 3 dudv

Example(cont.) R (2x 2 xy y 2 )dxdy u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 R : y = 2x + 4, y = 2x + 7, y = x + 1, y = x 2 G : v = 4, v = 7, u = 1, u = 2 dxdy = J(x, y; u, v)dudv = 1dudv 3 ( ) 2 ( ) ( ) ( u + v u + v 2u + v 2u + v 2 2 2 3 3 G ) 2 1 3 dudv

Example(cont.) R (2x 2 xy y 2 )dxdy u = x y; v = 2x + y x = u+v ; y = 2u+v 3 3 R : y = 2x + 4, y = 2x + 7, y = x + 1, y = x 2 G : v = 4, v = 7, u = 1, u = 2 dxdy = J(x, y; u, v)dudv = 1dudv 3 ˆ 7 ( ) 2 ( ) ( ) ( u + v u + v 2u + v 2u + v 2 2 2 3 3 ˆ 2 1 4 ) 2 1 3 dud 33 4

Evaluate 1. Find the Jacobian (x,y,z) (u,v,w) 1.1 x = u cos v, y = u sin v, z = w 1.2 x = 2u 1, y = 3v 4, z = 1 2 (w 4) 2. Evaluate R (3x 2 + 14xy + 8y 2 )dxdy on the region R : y = 3 3 1 1 x + 1, y = x + 3, y = x, y = x + 1 2 2 4 4 using the substitution u = 3x + 2y, v = x + 4y 64/5