PARAMETRIZED SURFACES; SURFACE AREA

We have seen that a space curve $C$ can be parametrized by a vector function $r=r(u)$ where $u$ ranges over some interval $I$ of the $u$-axis. In an analogous manner we can parametrize a surface $S$ in space by a vector function $r=r(u,v)$ where $(u,v)$ ranges over some region $\Omega $ of the $uv$-plane.

Example 1. (The graph of a function) The graph of a function
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can be parametrized by setting
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In the same vein the graph of a function
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can be parametrized by setting
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As $(u,v)$ ranges over $\Omega $, the tip of $r(u,v)$ traces out the graph of $f$.

Example 2. (A plane) If two vectors $a$ and $b$ are not parallel, then the set of all linear combinations $ua+vb$ generate a plane $p_0$ that passes through the origin. We can parametrize this plane by setting
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The plane $p$ that is parallel to $p_0$ and passes through the tip of $c$ can be parametrized by setting
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Note that the plane contains the lines
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Example 3. (A sphere) The sphere of radius $a$ centered at the origin can be parametrized by
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with $(u,v)$ ranging over the rectangle MATH

Derive this parametrization. The points of latitude $v$ form a circle of radius $a\cos \,v$ on the horizontal plane $z=a\sin \,v$. This circle can be parametrized by
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This expands to give
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Letting $v$ range from $-\frac \pi 2$ to $\frac \pi 2$, we obtain the entire sphere. The $xyz$-equation for this same sphere is $x^2+y^2+z^2=a^2.$ It is easy to verify that the parametrization satisfies this equation:
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Example 4. (A cone) Considers a cone with apex semiangle $\alpha $ and slant height $s$. The points of slant height $v$ form a circle of radius $v\sin \,\alpha $ on the horizontal plane $z=v\cos \,a.$ This circle can be parametrized by
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Since we can obtain the entire cone by letting $v$ range from $0$ to $s$, the cone is parametrized by
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with MATH

Example 5. (A spiral ramp) A rod of length $l$ initially resting on the $x$-axis and attached at one end to the $z$-axis sweeps out a surface by rotating about the $z$-axis at constant rate $\omega $ while climbing at a constant rate $b$.

To parametrize this surface we mark the point of the rod at a distance $u$ from the $z$-axis ($0\le u\le l$ ) and ask for the position of this point at time $v$. At time $v$ the rod will have climbed a distance $bv$ and rotated through an angle $\omega v$. Thus the point will be found at the tip of the vector
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The entire surface can be parametrized by
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The Fundamental Vector Product

Let $S$ be a surface parametrized by a differentiable vector function $r=r(u,v).$ For simplicity let us suppose that $(u,v)$ varies over an open rectangle
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Since $r$ is a function of two variables, we can take two partial derivatives: $r_u^{\prime }$, the partial of $r$ with respect to $u$, and $r_v^{\prime }$, the partial with respect to $v$. Now let $(u_0,v_0)$ be a point of $R$ for which
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The vector function
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(here we are keeping $v$ fixed at $v_0$) traces a differentiable curve $C_{\QTR{group}{1}}$ that lies on $S$. The vector function
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(this time we are keeping $u$ fixed at $u_0$) traces a differentiable curve $C_2$ that also lies on $S$. Both curves pass through the tip of $r(u_0,v_0):$

$C_{\QTR{group}{1}}$ with tangent vector MATH

$C_{\QTR{group}{2}}$ with tangent vector MATH

The cross product MATH which we have assumed to be different from zero, is thus perpendicular to both curves at the tip of $r(u_0,v_0)$ and can be taken as a normal to the surface at that point.

We record the result as follows:

if $S$ is a surface given by a differentiable function $r=r(u,v),$ then the vector MATH is perpendicular to the surface at the tip of $r(u,v)$ and, if different from zero, can be taken as a normal to the surface at that point.

The cross product MATH is called the fundamental vector product of the surface.

Example 6. For the plane $r(u,v)=ua+vb+c$ we have

MATH, MATH and therefore $N(u,v)=a\times b$. The vector $a\times b$ is normal to the plane.

Example 7. We parametrized the sphere $x^2+y^2+z^2=a^2$ by setting
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with MATH In this case
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and
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Thus
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As was to be expected, the fundamental vector product of a sphere is parallel to the radius vector $r(u,v).$

The Area of a Parametrized Surface

A linear function

$r(u,v)=ua+vb+c$ ($a$ and $b$ not parallel)

parametrizes a plane $p$. Horizontal lines from the $uv$-plane, lines with equations of the form $v=v_{0}$, are carried onto lines parallel to $a$, and vertical lines, $u=u_{0}$, are carried onto lines parallel to $b$:
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Thus a rectangle $R$ in the $uv$-plane with sides parallel to the $u$ and $v$ axes,
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is carried onto a parallelogram on $p$ with sides parallel to $a$ and $b$. What is important to us here is that

the area of the parallelogram = MATH (the area of $R$).

The parallelogram is generated by the vectors
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The area of the parallelogram is thus
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We can summarize as follows:

Let $R$ be a rectangle in the $uv$-plane with sides parallel to the coordinate axes. If $a$ and $b$ are not parallel, the linear function
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parametrizes a parallelogram with sides parallel to $a$ and $b$ and

the area of the parallelogram MATH(area of $R$).

More generally, let's suppose that we have a surface $S$ parametrized by a continuously differentiable function
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We will assume that MATH is a basic region in the $uv$-plane and that $r$ is one-to-one on the interior of $\Omega $. Also we will assume that the fundamental vector product MATH is never zero on the interior of $\Omega $. Under these conditions we call $S$ a continuously differentiable surface and define
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We will try to show the reasoning behind this definition in the case that $\Omega $ is a rectangle $R$ with sides parallel to the coordinate axes. We begin by breaking up $R$ into $N$ little rectangles $R_1,...,R_N$. This induces a decomposition of $S$ into little pieces $S_1,...,S_N.$ Taking $(u_i^{*},v_i^{*})$ as the center of $R_i$, we have the tip of $r(u_i^{*},v_i^{*})$ in $S_i$. Since the vector MATH is normal to the surface at the tip of $r(u_i^{*},v_i^{*})$, we can parametrize the tangent plane at this point by the linear function
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$S_i$ is the portion of $S$ that corresponds to $R_i$. The portion of the tangent plane that corresponds to this same $R_i$ is a parallelogram with area
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Taking this as our estimate for the area of $S_i$, we have

area of $S=\sum_{i=1}^N$ area of MATH(area of $R_i$).

This is a Riemann sum for
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and tends to this integral as the maximum diameter of the $R_i$ tends to zero.

Example 8. (The surface area of a sphere) The function
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with $(u,v)$ ranging over the set MATH parametrizes a sphere of radius $a$. For this parametrization

MATH and MATH

According to the new formula

area of the sphere MATH
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which is known to be correct.

Example 9. (The area of a region) If $S$ is a plane region $\Omega $, then $S$ can be parametrized by setting
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Here MATH and MATH In this case we have the familiar formula
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Example 10. (The area of a surface of revolution) Let $S$ be the surface generated by revolving the graph of a function
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about the $x$-axis. We will assume that $f$ is positive and continuously differentiable.

We can parametrize $S$ by setting
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with $(u,v)$ ranging over the set MATH In this case
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Therefore MATH and

area of MATH
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Example 11. (Spiral ramp) One turn of the spiral ramp of Example 5 is the surface
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with $(u,v)$ ranging over the set $\Omega :$MATH In this case
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Therefore
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and
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Thus

area of MATH
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The integral can be evaluated by setting MATH

The Area of a Surface $z=f(x,y)$

Let $S$ be the surface of a function $f(x,y):$
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We are to show that if $f$ is continuously differentiable, then
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Derivation. We can parametrize $S$ by setting
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We may just as well use $x$ and $y$ and write
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Clearly
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Thus
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Therefore MATH and the formula is verified.

Problem 12. Find the surface area of that part of the parabolic cylinder $z=y^2$ that lies over the triangle with vertices $(0,0),(0,1),(1,1) $ in the $xy$-plane.

Solution. Here $f(x,y)=y^2$ so that
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The base triangle can be expressed by writing
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The surface has area
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Problem 13. Find the surface area of that part of the hyperbolic paraboloid $z=xy$ that lies inside the cylinder $x^2+y^2=a^2.$

Solution. Here $f(x,y)=xy$ so that
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The formula gives
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In polar coordinates the base region takes the form
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Thus we have
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There is an elegant version of this last area formula that is geometrically vivid. We know that the vector
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is normal to the surface at the point $(x,y,f(x,y)).$ The unit vector in that direction, the vector
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is called the upper unit normal (It is the unit normal with a nonnegative $k$-component.)

Now let $\gamma (x,y)$ be the angle between $n(x,y)$ and $k.$ Since $n(x,y)$ and $k$ are both unit vectors,
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Taking reciprocals we have
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The area formula can therefore be written
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SURFACE INTEGRALS

The Mass of a Material Surface

Imagine a thin distribution of matter spread out over a surface S. We call this a material surface.

If the mass density (the mass per unit area) is constant throughout, then the total mass of the material surface is the density $\lambda $ times the area of $S:$

$M=\lambda $(area of $S$).

If, however, the mass density varies continuously from point to point, MATH then the total mass must be calculated by integration.

To develop the appropriate integral we suppose that
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is a continuously differentiable surface, a surface that meets the conditions for area formula. Our first step is to break up MATH into $N$ little basic regions MATH This decomposes the surface into $N$ little pieces $S_1,...,S_N$. The area of $S_i$ is given by the integral
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By the mean-value theorem for double integrals, there exists a point $(u_i^{*},v_i^{*})$ in $\Omega _i$ for which
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It follows that

area of $S_i$ = MATH(area of $\Omega _i$).

Since the point $(u_i^{*},v_i^{*})$ is in $\Omega _i$, the tip of $r(u_i^{*},v_i^{*})$ is on $S_i.$ The mass density at this point is MATH. If $S_i$ is small (which we can guarantee by choosing $\Omega _i$, small), then the mass density on $S_i$ is approximately the same throughout. Thus we can estimate $M_i$, the mass contribution of $S_i$, by writing
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Adding up these estimates we have an estimate for the total mass of the surface:
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This last expression is a Riemann sum for
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Surface Integrals

The above double integral can be calculated not only for a mass density function $\lambda $ but for any scalar field $H$ continuous over $S$. We call this integral the surface integral of $H$ over $S$ and write
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Note that, if $H(x,y,z)$ is identically $1$, then the right-hand side gives the area of $S$. Thus
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Problem 1. Calculate
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Solution. Call the parameter set $\Omega $. Then
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By a previous calculation MATH. Thus
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To find $x(u,v)$ and $y(u,v)$ we need the $i$ and $j$ components of $r(u,v).$ We can get these as follows:
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Therefore MATH and $y(u,v)=a_2u+b_2v.$ We can now write
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Problem 2. Calculate
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where $S$ is the ramp of Example 11 of the previous section:


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Solution. Call the parameter set $\Omega $. By a previous calculation
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Therefore
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Like the other integrals we have studied, the surface integral satisfies a mean-value condition; namely, if $H$ is continuous, then there is a point $(x_0,y_0,z_0)$ on $S$ for which
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We call $H(x_0,y_0,z_0)$ the average value of $H$ on $S$. We can thus write
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We can also take weighted averages: if $H$ and $G$ are continuous on $S$, then there is a point $(x_0,y_0,z_0)$ on $S$ for which
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$H(x_0,y_0,z_0)$ is called the $G$-weighted average of $H$ on $S$.

The coordinates of the centroid MATH of a surface are simple averages over the surface: for a surface $S$ of area $A$
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In the case of a material surface of mass density MATH the coordinates of the center of mass $(x_M,y_M,z_M)$ are density-weighted averages: for a surface $S$ of total mass $M$
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Problem 3. Locate the center of mass of a material surface in the form of a hemisphere $x^2+y^2+z^2=a^2$ with $z\ge 0$ given that the mass density varies directly as the distance from the $xy$-plane.

Solution. The surface $S$ can be parametrized by
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Call the parameter set $\Omega $ and recall that MATH. The density function can be written MATH We can calculate the mass as follows:
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By symmetry $x_{M}=0$ and $y_{M}=0$. To find $z_{M}$ we write
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Since $M=\pi ka^{3}$ we see that MATH The center of mass is the point MATH

Suppose that a material surface $S$ rotates about an axis. The moment of inertia of the surface about that axis is given by the formula
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where MATH is the mass density function and $R(x,y,z)$ is the distance from the axis to the point $(x,y,z)$. (As with other configurations the moments of inertia about the $x,y,z$ axes are denoted by $I_x,I_y,I_z.$)

Problem 4. Calculate the moment of inertia about the $z$-axis of a material sphere
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Assume that the surface is homogeneous and has constant mass density $1$.

Solution. We parametrize $S$ by setting
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Call the parameter set $\Omega $ and recall that MATH. We can calculate the moment of inertia as follows:
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Since the surface has mass $M=V=4\pi a^2,$ we can write $I_z=\frac 23Ma^2.$

A surface
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can be parametrized by
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Since MATH where $\gamma (x,y)$ is the angle between $k$ and the upper unit normal. Therefore
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In evaluating this last integral we use the fact that
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Problem 5. Calculate
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Solution. The base region $\Omega $ is the unit disc. The function $z=f(x,y)=xy$ has partial derivatives MATH. Therefore
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and
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We evaluate this last integral by changing to polar coordinates. $\Omega $ is the set of all $(x,y)$ with polar coordinates $(r,\theta )$ in the set
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Therefore
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