{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

16.6withScannedExamples

# 16.6withScannedExamples - 16.6 PARAMETRIC SURFACES AND...

This preview shows pages 1–4. Sign up to view the full content.

16.6: PARAMETRIC SURFACES AND THEIR AREAS KIAM HEONG KWA 1. Parametric Surfaces Let (1.1) r ( u, v ) = x ( u, v ) i + y ( u, v ) j + z ( u, v ) k be a vector-valued function of two variables on a plane region D in the uv -plane. The collection of all points ( x, y, z ) R 3 such that (1.2) x = x ( u, v ) , y = y ( u, v ) , z = z ( u, v ) , ( u, v ) D, i.e., (1.3) S = { ( x ( u, u ) , y ( u, v ) , z ( u, v )) R 3 | ( u, v ) D } , is called a parametric surface . Equations (1.1) and (1.2) are respec- tively called the vector equation and the parametric equations of S . The variables u and v are usually referred to as parameters. The sur- face S is traced out by the tip of the position vector r ( u, v ) as ( u, v ) varies throughout the parameter domain D . Example 1 (Problem 3 in the text) . The vector equation r ( u, v ) = ( u + v ) i + (3 - v ) j + (1 + 4 u + 5 v ) k determines a plane in R 3 . To see this, write the vector equation in its equivalent parametric equations x = u + v, y = 3 - v, z = 1 + 4 u + 5 v. Solving for u and v from the first two parametric equations yields u = x + y - 3 , v = 3 - y. Substituting for u and v in terms of x and y in the last parametric equation gives z = 1 + 4( x + y - 3) + 5(3 - v ) or 4 x - y - z = - 4 , which is an equation of a plane. In general, a vector equation of the form r ( u, v ) = r 0 + u a + v b , Date : November 17, 2010. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 KIAM HEONG KWA where r 0 = h x 0 , y 0 , z 0 i , a = h a 1 , a 2 , a 3 i , and b = h b 1 , b 2 , b 3 i are vectors in R 3 such that (1) a 6 = 0 , (2) b 6 = 0 , and (3) a and b are nonparallel (i.e., a × b 6 = 0 ) represents a plane in R 3 that contains the point ( x 0 , y 0 , z 0 ) and has a × b as a normal vector. To see this, write the vector equation as r ( u, v ) - r 0 = u a + v b . Then ( r ( u, v ) - r 0 ) · a × b = u a · a × b + v b · a × b = 0 because a · a × b = b · a × b = 0 . Let us apply the general statement to the vector equation r ( u, v ) = ( u + v ) i + (3 - v ) j + (1 + 4 u + 5 v ) k = 3 j + k + u ( i + 4 k ) + v ( i - j + 5 k ) . Identifying the vectors r 0 , a , and b , one has r 0 = 3 j + k , a = i + 4 k , and b = i - j + 5 k . In particular, a × b = 4 i - j - k . Denoting r ( u, v ) = x i + y j + z k , one has [ x i + ( y - 3) j + ( z - 1) k ] · (4 i - j - k ) = 4 x - y - z + 4 = 0 . Example 2 (Problem 6 in the text) . The vector equation r ( s, t ) = h s sin 2 t, s 2 , s cos 2 t i defines a circular paraboloid whose axis is the y - axis. To see this, consider the equivalent parametric equations x = s sin 2 t, y = s 2 , z = s cos 2 t. Note that x 2 + z 2 = s 2 (sin 2 2 t + cos 2 2 t ) = s 2 = y. Traces parallel to the xz -planes are circles, while traces parallel to the xy - and yz -planes are parabolas. Example 3 (Problem 20 in the text) . We would like to find a paramet- ric representation for the lower half of the ellipsoid 2 x 2 + 4 y 2 + z 2 = 1 . Note that this can be described by 2 x 2 + 4 y 2 + z 2 = 1 and z 0 and thus more compactly by z = - p 1 - 2 x 2 - 4 y 2 . Hence a parametric representation of the surface is x = x, y = y, z = - p 1 - 2 x 2 - 4 y 2 ,
16.6: PARAMETRIC SURFACES AND THEIR AREAS 3 where we have taken x and y to be the parameters with the elliptical plane region { ( x, y ) R 2 | 2 x 2 + 4 y 2 1 } as the parameter domain.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 14

16.6withScannedExamples - 16.6 PARAMETRIC SURFACES AND...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online