we have now watched many kinds of functions. As soon as we talked aboutparametric curves, we defined castle as features native Rcome R2 (plane curves) or Rcome R3 (Space curves). Because eachof this has itns doKey R, castle are a dimensional(you can only go front or backward). In thins section, us investigate howcome parameterize two dimensionatogether surfaces. Below is ns definition.
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meaning the Parametric Surfaces
A parametric surface ins a duty with doKey R2 and also variety R3.
Remark: we typically use ns variablens uand also v because that the domainand x, y,and z for therange. Us regularly use vector notation come exhibit parametric surfaces.
Example
A sphere of radius 7 have the right to it is in parameterized by
r(u,v)= 7cons u sin v i + 7sin u sin v j + 7 cos v k
notification that us have simply used spherical works with with ns radiuns organized at7.
us deserve to usage a computer come graph a parametric surface. Listed below is ns graphof the surface
r(u,v) = sin u i + cos v j+ exp(2u1/3 + 2v1/3) k
Example
Recurrent ns surface
z= ex cos(x  y)
parametrically
Solution
the principle ins equivalent come parametric curves. Us simply lens x= u and y = v, to get
r(u,v)= u i + v j + eu cos(u  v) k
Example
A surconfront ins developed by revolvinns ns curve
y= cons x
about the xaxis. Discover parametric equations for this surface.
Solution
because that a resolved worth the x, we get a one ofradiuns cos x. Now use polar works with (inthe yzplane) to get
r(u,v)= u i + r cons v j + r sin v k
Due to the fact that u = x and r = cons x, we deserve to substitute cos u for r in ns over equation to get
r(u,v)= u i + cons u cons v j + cos u sin v k
Normal Vectors and also Tangenns Planes
us have actually currently learn just how come uncover a Typical vector that a surchallenge the ispresented as a duty that 2 variables, namely uncover ns gradienns vector.to discover the Common vector come a surface r(u,v)the is defined parametrically, we continue as follows.
the partiatogether derivatives
ru(u0,v0)and also rv(u0,v0)
will lin other words top top ns tangent airplane to the surconfront at ns allude (u0,v0).Thins ins true, Since resolving a variable consistent and also lettinns ns various other vary,developed a curve on the surconfront via (u0,v0).ru(u0,v0) willbe tangent come this curve. Ns tangent plane has every vectorns tangentcome curves passong with ns point.
come discover a Normal vector, we just cross the 2 tangent vectors.
Example
uncover ns equati~ above the ns tangenns plane to the surface
r(u,v) = (u2  v2)i + (u + v) j + (uv) k
in ~ ns point (1,2).
Solution
us have
ru(u,v) = (2u) i+ j + v k
rv(u,v) = (2v) i+ j + u k
so that
ru(1,2) = 2 i +j + 2 k
rv(1,2) = 4 i+ j + k
r(1,2) = 3 i + 3 j + 2 k
now cross these vectors Together to get
Wenow have actually ns Regular vector and a point (3,3,2).we usage ns Typical vectorsuggest equati~ above for a plane
1(x + 3)  10(y  3) + 6(z  2) = 0
x  10y + 6z = 15 orx + 10y  6z = 15
Surface Area
to uncover the surchallenge location the a parametricallied defined surface, wecontinue in a similar method as in the instance as a surchallenge defined through afunction. InsteADVERTISEMENT that projecting down come ns area in the xyplane, weproject earlier to a region in the uvplane. Us reduced ns region right into smallrectangle which mans roughly come small parallelogramns with adjacentspecifying vectors ru and also rv.the area the these parallelogramns will same the size of the cross productthe ru and also rv.finally include ns locations uns and also take it the limit together the rectangles get small.This will produce a dual integral.
area the a Parametric Surface Lens ns be a smooth surconfront characterized parametrically by r(u,v) = x(u,v) i + y(u,v) j + z(u,v) k wbelow u and also v to be contained in an area R. Then ns surchallenge location that ns ins provided by

Since the size the a overcome product requires a squto be root,the integral in ns surconfront location recipe ins normally impossible or nearlydifficult to evaluate withthe end power series or by approximationtechniques.
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Example
find ns surconfront area the the surconfront given by
r(u,v) = (v2)i + (u  v) j + (u2) k0 u 2 1 v 4
Solution
we calculate
ru(u,v) = j +2u k
rv(u,v) = (2v) i j
ns overcome producns is
the surface area formula gives
Thisintegral ins more than likely impossible to compute exactly. Instead, a calculatorhave the right to be provided to attain a surchallenge location that 70.9.
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