Parametrization Surface Integrals Dr. Allen Back Nov. 17, 2014 - PowerPoint PPT Presentation

Parametrization V1 Surface Parametriza- tion Parametrization Surface Integrals Dr. Allen Back Nov. 17, 2014

Paraboloid z = x 2 + 4 y 2 Parametrization V1 Surface Parametriza- tion Surface The graph z = F ( x , y ) can always be parameterized by Integrals Φ( u , v ) = < u , v , F ( u , v ) > .

Paraboloid z = x 2 + 4 y 2 Parametrization V1 Surface Parametriza- tion The graph z = F ( x , y ) can always be parameterized by Surface Integrals Φ( u , v ) = < u , v , F ( u , v ) > . Parameters u and v just different names for x and y resp.

Paraboloid z = x 2 + 4 y 2 Parametrization V1 Surface Parametriza- tion The graph z = F ( x , y ) can always be parameterized by Surface Integrals Φ( u , v ) = < u , v , F ( u , v ) > . Use this idea if you can’t think of something better.

Paraboloid z = x 2 + 4 y 2 Parametrization The graph z = F ( x , y ) can always be parameterized by V1 Φ( u , v ) = < u , v , F ( u , v ) > . Surface Parametriza- tion Surface Integrals

Paraboloid z = x 2 + 4 y 2 The graph z = F ( x , y ) can always be parameterized by Parametrization V1 Φ( u , v ) = < u , v , F ( u , v ) > . Surface Parametriza- tion Surface Integrals Note the curves where u and v are constant are visible in the wireframe.

Paraboloid z = x 2 + 4 y 2 Parametrization V1 Surface Parametriza- tion Surface Integrals A trigonometric parametrization will often be better if you have to calculate a surface integral.

Paraboloid z = x 2 + 4 y 2 Parametrization V1 Surface Parametriza- tion A trigonometric parametrization will often be better if you have Surface Integrals to calculate a surface integral. Φ( u , v ) = < 2 u cos v , u sin v , 4 u 2 > .

Paraboloid z = x 2 + 4 y 2 Parametrization A trigonometric parametrization will often be better if you have V1 to calculate a surface integral. Surface Parametriza- Φ( u , v ) = < 2 u cos v , u sin v , 4 u 2 > . tion Surface Integrals

Paraboloid z = x 2 + 4 y 2 Parametrization A trigonometric parametrization will often be better if you have V1 to calculate a surface integral. Surface Parametriza- Φ( u , v ) = < 2 u cos v , u sin v , 4 u 2 > . tion Surface Integrals Algebraically, we are rescaling the algebra behind polar coordinates where x = r cos θ y = r sin θ leads to r 2 = x 2 + y 2 .

Paraboloid z = x 2 + 4 y 2 Parametrization A trigonometric parametrization will often be better if you have V1 to calculate a surface integral. Surface Parametriza- Φ( u , v ) = < 2 u cos v , u sin v , 4 u 2 > . tion Surface Integrals Here we want x 2 + 4 y 2 to be simple. So x = 2 r cos θ y = r sin θ will do better.

Paraboloid z = x 2 + 4 y 2 A trigonometric parametrization will often be better if you have Parametrization to calculate a surface integral. V1 Surface Φ( u , v ) = < 2 u cos v , u sin v , 4 u 2 > . Parametriza- tion Surface Integrals Here we want x 2 + 4 y 2 to be simple. So x = 2 r cos θ y = r sin θ will do better. Plug x and y into z = x 2 + 4 y 2 to get the z-component.

Parabolic Cylinder z = x 2 Parametrization V1 Surface Parametriza- tion Surface Integrals Graph parametrizations are often optimal for parabolic cylinders.

Parabolic Cylinder z = x 2 Parametrization V1 Surface Parametriza- tion Surface Integrals Φ( u , v ) = < u , v , u 2 >

Parabolic Cylinder z = x 2 Parametrization Φ( u , v ) = < u , v , u 2 > V1 Surface Parametriza- tion Surface Integrals

Parabolic Cylinder z = x 2 Parametrization V1 Φ( u , v ) = < u , v , u 2 > Surface Parametriza- tion Surface Integrals One of the parameters (v) is giving us the “extrusion” direction. The parameter u is just being used to describe the curve z = x 2 in the zx plane.

Elliptic Cylinder x 2 + 2 z 2 = 6 Parametrization V1 Surface Parametriza- tion Surface Integrals The trigonometric trick is often good for elliptic cylinders

Elliptic Cylinder x 2 + 2 z 2 = 6 Parametrization V1 Surface Parametriza- tion Surface Integrals √ √ √ √ √ Φ( u , v ) = < 3 · 2 cos v , u , 3 sin v > = < 6 cos v , u , 3 sin v >

Elliptic Cylinder x 2 + 2 z 2 = 6 Parametrization V1 Surface Parametriza- tion Surface Integrals √ √ Φ( u , v ) = < 6 cos v , u , 3 sin v >

Elliptic Cylinder x 2 + 2 z 2 = 6 Parametrization √ √ Φ( u , v ) = < 6 cos v , u , 3 sin v > V1 Surface Parametriza- tion Surface Integrals

Elliptic Cylinder x 2 + 2 z 2 = 6 Parametrization V1 Surface Parametriza- tion √ √ Φ( u , v ) = < 6 cos v , u , 3 sin v > Surface Integrals

Elliptic Cylinder x 2 + 2 z 2 = 6 Parametrization √ √ Φ( u , v ) = < 6 cos v , u , 3 sin v > V1 Surface Parametriza- tion Surface Integrals What happened here is we started with the polar coordinate idea x = r cos θ z = r sin θ but noted that the algebra wasn’t right for x 2 + 2 z 2 so shifted to √ x = 2 r cos θ z = r sin θ

Elliptic Cylinder x 2 + 2 z 2 = 6 Parametrization V1 √ √ Φ( u , v ) = < 6 cos v , u , 3 sin v > Surface Parametriza- tion Surface Integrals √ x = 2 r cos θ z = r sin θ makes the left hand side work out to 2 r 2 which will be 6 when √ r = 3 .

Ellipsoid x 2 + 2 y 2 + 3 z 2 = 4 Parametrization V1 Surface Parametriza- tion Surface Integrals A similar trick occurs for using spherical coordinate ideas in parameterizing ellipsoids.

Ellipsoid x 2 + 2 y 2 + 3 z 2 = 4 Parametrization V1 Surface Parametriza- tion A similar trick occurs for using spherical coordinate ideas in Surface Integrals parameterizing ellipsoids. √ � 4 Φ( u , v ) = < 2 sin u cos v , 2 sin u sin v , 3 cos u >

Ellipsoid x 2 + 2 y 2 + 3 z 2 = 4 Parametrization √ � 4 V1 Φ( u , v ) = < 2 sin u cos v , 2 sin u sin v , 3 cos u > Surface Parametriza- tion Surface Integrals

Hyperbolic Cylinder x 2 − z 2 = − 4 Parametrization V1 Surface Parametriza- tion You may have run into the hyperbolic functions Surface Integrals e x + e − x cosh x = 2 e x − e − x sinh x = 2

Hyperbolic Cylinder x 2 − z 2 = − 4 Parametrization V1 Surface You may have run into the hyperbolic functions Parametriza- tion e x + e − x Surface cosh x = Integrals 2 e x − e − x sinh x = 2 Just as cos 2 θ + sin 2 θ = 1 helps with ellipses, the hyperbolic version cosh 2 θ − sinh 2 θ = 1 leads to the nicest hyperbola parameterizations.

Hyperbolic Cylinder x 2 − z 2 = − 4 Parametrization V1 Surface Parametriza- Just as cos 2 θ + sin 2 θ = 1 helps with ellipses, the hyperbolic tion version cosh 2 θ − sinh 2 θ = 1 leads to the nicest hyperbola Surface Integrals parameterizations. Φ( u , v ) = < 2 sinh v , u , 2 cosh v >

Hyperbolic Cylinder x 2 − z 2 = − 4 Parametrization V1 Surface Φ( u , v ) = < 2 sinh v , u , 2 cosh v > Parametriza- tion Surface Integrals

Saddle z = x 2 − y 2 Parametrization V1 Surface Parametriza- tion Surface Integrals The hyperbolic trick also works with saddles

Saddle z = x 2 − y 2 Parametrization V1 Surface Parametriza- tion Surface Integrals Φ( u , v ) = < u cosh v , u sinh v , u 2 >

Saddle z = x 2 − y 2 Parametrization Φ( u , v ) = < u cosh v , u sinh v , u 2 > V1 Surface Parametriza- tion Surface Integrals

Hyperboloid of 1 Sheet x 2 + y 2 − z 2 = 1 Parametrization V1 Surface Parametriza- tion Surface Integrals The spherical coordinate idea for ellipsoids with sin φ replaced by cosh u works well here.

Hyperboloid of 1 Sheet x 2 + y 2 − z 2 = 1 Parametrization V1 Surface Parametriza- tion Surface Integrals Φ( u , v ) = < cosh u cos v , cosh u sin v , sinh u >

Hyperboloid of 1 Sheet x 2 + y 2 − z 2 = 1 Parametrization V1 Surface Parametriza- tion Surface Integrals

Hyperboloid of 2-Sheets x 2 + y 2 − z 2 = − 1 Parametrization V1 Surface Parametriza- tion Surface Integrals Φ( u , v ) = < sinh u cos v , sinh u sin v , cosh u >

Hyperboloid of 2-Sheets x 2 + y 2 − z 2 = − 1 Parametrization V1 Surface Parametriza- tion Surface Integrals

Top Part of Cone z 2 = x 2 + y 2 Parametrization V1 Surface Parametriza- tion Surface Integrals x 2 + y 2 . � So z =

Top Part of Cone z 2 = x 2 + y 2 Parametrization V1 Surface Parametriza- tion x 2 + y 2 . � So z = Surface Integrals The polar coordinate idea leads to Φ( u , v ) = < u cos v , u sin v , u >

Top Part of Cone z 2 = x 2 + y 2 x 2 + y 2 . � So z = Parametrization The polar coordinate idea leads to V1 Surface Φ( u , v ) = < u cos v , u sin v , u > Parametriza- tion Surface Integrals

Mercator Parametrization of the Sphere Parametrization For 0 ≤ v ≤ ∞ , 0 ≤ u ≤ 2 π V1 Φ( u , v ) = ( sech ( v ) cos u , sech ( v ) sin u , tanh ( v )) . Surface Parametriza- tion Surface (Note tanh 2 ( v ) + sech 2 ( v ) = 1) Integrals

Surface Integrals Parametrization Picture of � T u , � T v for a Lat/Long Param. of the Sphere. V1 Surface Parametriza- tion Surface Integrals

Surface Integrals Parametrization Basic Parametrization Picture V1 Surface Parametriza- tion Surface Integrals