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Engineering Mathematics (MCS-21007) Dr. Ir. Harinaldi, M.Eng

MCS21007-22: Parametric Surfaces- 1

UNIVERSITY OF INDONESIA FACULTY OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING

ENGINEERING MATHEMATICS (MCS-21007) 1. Course Name/Units : Engineering Mathematics/4 2. Department/Semester : Mechanical Engineering /3 3. Program/Period : S1/ First semester 2004-2005 4. Date/Time/Room : I. Tuesday : 08.00 – 09.50 (K105) : II. Thursday : 08.00 -09.50 (K206) 5. Instructor : Dr. Ir. Harinaldi, M.Eng 6. Topics : MCS-21007-22: Parametric Surfaces

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. Definition of Parametric Surfaces A parametric surface is a function with domain R2 and range R3. 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(2u1/3 + 2v1/3) k



Example Represent the surface z = ex 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 + eu 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 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(t) that is defined parametrically, we proceed as follows. The partial derivatives ru(u0,v0) and rv(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).



ru(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 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

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 ru and rv. The area of these parallelograms will equal the magnitude of the cross product of ru and rv. Finally add the areas up and take the limit as the rectangles get small. This will produce a double integral. Area of a Parametric Surface Let S be a smooth surface defined parametrically by r(u,v) = x(u,v) i + y(u,v) j + z(u,v) k where u and v are contained in a region R. Then the surface area of S is given by



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 ru(u,v) = j + 2u k rv(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. Surface Integrals for Parametric Surfaces In the last section, we learned how to find the surface area for parametric surfaces. We cut the region in the uv-plane into tiny rectangles and added up the area of the corresponding tiny parallelograms in the xy-plane. The area of these parallelograms was If we think of the surface as having varying density f(x,y,z), then the mass of this parallelogram will be and adding up all these masses and taking the limit as the rectangle sizes approach zero, gives the definition of the surface integral. Definition of the Surface Integral Let S be a smooth surface given by the vector valued function r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k and f(x,y,z) be a continuous function. Then the surface integral of f over S is

As with finding the surface area the integral typically results in an impossible integral.



Example Find

where S is the surface r(u,v) = ui + u2j + (u+ v)k 0 < u < 2 1 < v < 4 and f(x,y,z) = x + 2z Solution We find ru = i + (2u)j + k rv = k and take the cross product

We have f(x(u,v),y(u,v),z(u,v)) = x(u,v) +2z(u,v) = u +2(u + v) = 3u + v We find

Although this integral is possible, its solution is quite involved. You can verify that the surface integral evaluates to approximately 525.27. Surface Integrals for Surfaces that are Functions of Two Variables We have seen before that if z = g(x,y) is a surface such that g has continuous first order partial derivatives, then the parameterization r(u,v) = ui + vj + g(u,v)k has the property that This leads to the formula for surface integrals. Theorem: The Surface Integral for Surfaces of the Form z = g(x,y) Let S be a surface given by z = g(x,y) over a region R such that both first order partial derivatives of g are continuous and let f(x,y,z) be a continuous function. Then the surface integral of f over S is



Example Find

where S is the part of the paraboloid z = x2 + y2

that lies inside the cylinder x2 + y2 = 1

and

f(x,y,z) = z

Solution

We have

and f(x,y,z) = z =

x2 + y2

At this point, you should be thinking, "This looks like a job for polar coordinates." And we get

Let u = 1 + 4r2 du = 8r dr r2 = 1/4 u - 1/4

and the substitution gives us

Oriented Surfaces and Flux We have seen how a region R with boundary curve C can be oriented. Traveling along C, we look to see if the region is on the right or left. Unfortunately, this definition does not work will for surfaces in three dimensions. The idea of right and left is not well defined. In fact not all surfaces can be oriented. We say that a surface is orientable if a unit normal vector can be defined on the surface such that it varies continuously over the surface. Below is an example of a non-orientable surface (called the Mobius Strip)



You can see that there is no front or back of this surface. Recall that a unit normal vector to a surface can be given by

There is another choice for the normal vector to the surface, namely the vector in the opposite direction, -N. By this point, you may have noticed the similarity between the formulas for the unit normal vector and the surface integral. This idea leads us to the definition of the Flux Integral. Consider a fluid flowing through a surface S. The Flux of the fluid across S measures the amount of fluid passing through the surface per unit time. If the fluid flow is represented by the vector field F, then for a small piece with area DS of the surface the flux will equal to DFlux = F . N DS Adding up all these together and taking a limit, we get Definition of the Flux Integral Let F be a differentiable vector field on a surface S oriented by a unit normal vector N. The flux integral of F across N is given by

Notice that the denominator of N and the formula for dS both involve ||ru x rv||. Canceling, we get NdS = ru x rv dvdu for a surface that is defined by the function z = g(x,y), we get the nice formula NdS = -gx(x,y)i - gy(x,y)j + k (oriented upward) or NdS = gx(x,y)i + gy(x,y)j - k (oriented downward) Example



Find the flux of F(x,y,z) = xi + 2yj + zk across the part of the surface z = x + y2 with upward pointing normal that lies within the box 0 < x < 3 2 < y < 5 Solution We compute NdS = -i - 2yj + k dydx and F . N dS = -x - 4y2 + x + y2 = -3y2 The flux integral is

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