Parametric Surfaces

Definition of a Parametric Surface

We have now seen many kinds of functions.  When we talked about parametric curves, we defined them as functions from R to R² (plane curves) or R to R³ (space curves).  Because each of these has its domain R, they are one dimensional (you can only go forward or backward).  In this section, we investigate how to parameterize two dimensional surfaces.  Below is the definition.

Remark:  We typically use the variables u and v for the domain and x, y, and z for the range.  We often use vector notation to exhibit parametric surfaces.

Example 

A sphere of radius 7 can be parameterized by 

        r(u,v)  =  7cos u sin v i + 7sin u sin v j + 7 cos v k

Notice that we have just used spherical coordinates with the radius held at 7.  

We can use a computer to graph a parametric surface.  Below is the graph of the surface

         r(u,v)  =  sin u i + cos v j + exp(2u^1/3 + 2v^1/3) k

Example 

Represent the surface 

        z  =  e^x cos(x - y)

parametrically

Solution

The idea is similar to parametric curves.  We just let x  =  u and y  =  v, to get

         r(u,v)  =  u i + v j + e^u cos(u - v) k

Example

A surface is created by revolving the curve 

        y  =  cos x

about the x-axis.  Find parametric equations for this surface.

Solution

For a fixed value of x, we get a circle of radius cos x.  Now use polar coordinates (in the yz-plane) to get

         r(u,v)  =  u i + r cos v j + r sin v k

Since u = x and r = cos x, we can substitute cos u for r in the above equation to get

         r(u,v)  =  u i + cos u cos v j + cos u sin v k

Normal Vectors and Tangent Planes

We have already learned how to find a normal vector of a surface that is presented as a function of two variables, namely find the gradient vector.  To find the normal vector to a surface r(u,v) that is defined parametrically, we proceed as follows. 

The partial derivatives 

        r_u(u₀,v₀)         and        r_v(u₀,v₀) 

will lie on the tangent plane to the surface at the point (u₀,v₀).  This is true, because fixing one variable constant and letting the other vary, produced a curve on the surface through (u₀,v₀). r_u(u₀,v₀) will be tangent to this curve.  The tangent plane contains all vectors tangent to curves passing through the point.  

To find a normal vector, we just cross the two tangent vectors.

Example

Find the equation of the tangent plane to the surface 

        r(u,v)  =  (u² - v²) i + (u + v) j + (uv) k

at the point (1,2).

Solution

We have 

        r_u(u,v)  =  (2u) i +  j + v k

        r_v(u,v)  =  (-2v) i +  j + u k

so that 

        r_u(1,2)  =  2 i +  j + 2 k

        r_v(1,2)  =  -4 i +  j + k

        r(1,2)  =  -3 i + 3 j + 2 k

Now cross these vectors together to get

We now have the normal vector and a point (-3,3,2).  We use the normal vector-point equation for a plane

        -1(x + 3) - 10(y - 3) + 6(z - 2)  =  0

        -x - 10y + 6z  =  -15        or        x + 10y - 6z  =  15

Surface Area

To find the surface area of a parametrically defined surface, we proceed in a similar way as in the case as a surface defined by a function.  Instead of projecting down to the region in the xy-plane, we project back to a region in the uv-plane.  We cut the region into small rectangles which map approximately to small parallelograms with adjacent defining vectors r_u and r_v.  The area of these parallelograms will equal the magnitude of the cross product of r_u and r_v.  Finally add the areas up and take the limit as the rectangles get small.  This will produce a double integral.

Since the magnitude of a cross product involves a square root, the integral in the surface area formula is usually impossible or nearly impossible to evaluate without power series or by approximation techniques.  

Example

Find the surface area of the surface given by 

        r(u,v)  =  (v²) i + (u - v) j + (u²) k        0  <  u  <  2        1  <  v  <  4

Solution

We calculate

        r_u(u,v)  =   j + 2u k

        r_v(u,v)  =  (2v) i -  j

The cross product is 

The surface area formula gives

This integral is probably impossible to compute exactly.  Instead, a calculator can be used to obtain a surface area of 70.9.

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