OBJECTIVES

When you finish this section, you should be able to:

1. Describe surfaces defined parametrically (p. 1017)
2. Find a parametric representation of a surface (p. 1018)
3. Find equations for a tangent plane and a normal line (p. 1022)
4. Find the surface area of a parametrized surface (p. 1023)

spanDEFINITIONspan Parametric Surface

Suppose \[ x=x(u,v) \qquad y=y(u,v) \qquad z=z(u,v) \]

are functions of two variables that are defined on a region \(R\) in the \(uv\) -plane. Also suppose that the functions are continuous and one-to-one* on the interior of \(R.\) Then \(u\) and \(v\) are parameters; \(x=x(u,v) ,\) \(y=y(u,v) ,\) \(z=z(u,v) \) are parametric equations; and the set of points defined by the parametric equations is a parametric surface.

The vector-valued function \[ \mathbf{r}(u,v) =x(u,v) \mathbf{i}+y(u,v) \mathbf{j}+z(u,v) \mathbf{k} \]

is called a parametrization of the surface. The region \(R\) is the parameter domain of \(\mathbf{r}\), and the range of \(\mathbf{r}\) is the surface traced out by \(\mathbf{r}\).

spanDEFINITIONspan Coordinate Curves

Suppose \(S\) is a surface with parametrization \(\mathbf{r}(u,v) =x(u,v) \mathbf{i}+y(u,v) \mathbf{j}+z(u,v) \mathbf{k}\) and parameter domain \(R\). Then the curve traced out by \(\mathbf{r=r}(u,v) \) as \((u,v) \) ranges over \(R\), while holding \(v=v_{0}\) constant, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \mathbf{r}( u,v_{0}) =x\left( u,v_{0}\right) \mathbf{i}+y( u,v_{0}) \mathbf{j}+z( u,v_{0}) \mathbf{k} }} \]

is called a \(u\)-coordinate curve.

Similarly, the curve traced out as \((u,v) \) ranges over \(R\), while holding \(u=u_{0}\) constant, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \mathbf{r}( u_{0},v) =x( u_{0},v) \mathbf{i}+y( u_{0},v) \mathbf{j}+z( u_{0},v) \mathbf{k} }} \]

is called a \(v\)-coordinate curve.

Describing a Parametric Surface

1. Describe the parametric surface \(S\) given by \(\mathbf{r}(u,v) =( 1+2u+v) \mathbf{i}+(u-v) \mathbf{j} +(3+v) \mathbf{k}\).
2. Find a rectangular equation for \(S.\)

Solution (a) We begin by finding several coordinate curves. We choose to use \(v=0,\) \(v=1,\) \(u=0,\) and \(u=1,\) but any numbers will do.

When \(v=0,\) the \(u\)-coordinate curve is \(\mathbf{r} ( u,0) = (1+2u) \mathbf{i}+u\mathbf{j}+3\mathbf{k}\).

When \(v=1,\) the \(u\)-coordinate curve is \(\mathbf{r} ( u,1) = (2+2u) \mathbf{i}+ ( u-1) \mathbf{j}+4\mathbf{k}\).

Figure 41

When \(u=0,\) the \(v\)-coordinate curve is \(\mathbf{r} ( 0,v) = ( 1+v) \mathbf{i}-v\mathbf{j}+(3+v) \mathbf{k}\).

When \(u=1,\) the \(v\)-coordinate curve is \(\mathbf{r} ( 1,v) =(3+v) \mathbf{i}+ ( 1-v) \mathbf{j}+(3+v) \mathbf{k}\).

The graph of the parametric surface is a plane as shown in Figure 41.

(b) To find a rectangular equation for the surface, we write the components of \(\mathbf{r=r}(u,v) \) as a system of equations. Since \(\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}=x(u,v) \mathbf{i}+y(u,v) \mathbf{j}+z(u,v) \mathbf{k},\) we have\[ \left\{ \begin{array}{ll@{\qquad}l} x=2u+v+1 & &{\color{#0066A7}{(1)}}\\ y=u-v & & {\color{#0066A7}{(2)}}\\ z= v+3 & & {\color{#0066A7}{(3)}} \end{array} \right. \]

If we eliminate \(u\) and \(v\) from this system, we will have an expression involving \(x,\) \(y,\) and \(z.\) We eliminate \(v\) by adding equations \((1) \) and \((2) \) and equations \((2) \) and \((3) .\) \[ \begin{equation*} \begin{array}{ll@{\qquad\qquad}l} x+y=3u+1 & & \color{#0066A7}{{\hbox{(4) Add \(1\) and \(2\).}}}\\ y+z= \hspace{6pt}u+3 & &\color{#0066A7}{{\hbox{(5) Add \(2\) and \(3\).}}} \end{array} \end{equation*} \]

Now we eliminate \(u\): \[ \begin{equation*} \begin{array}{ll@{\qquad}l} \hspace{48pt}x+ y = 3u+1 & & \color{#0066A7}{(4)} \\[5pt] \hspace{27pt}\underline{\,\ -3y-3z=-3u-9} & \,\ \ \ \ \ \ \ \ \ \ & \color{#0066A7}{{\hbox{Multiply (5) by \(-3\).}}} \\ \hspace{21pt}x-2y-3z=-8 & & \color{#0066A7}{{\hbox{Add.}}} \end{array} \end{equation*} \]

This is the equation of a plane whose normal vector is \(\mathbf{i}-2\mathbf{j}-3\mathbf{k}.\)

Describing a Parametric Surface

Describe the parametric surface \(S\) parametrized by \[ \begin{equation*} \mathbf{r}(u,v) =( 3+\cos v) \cos u\,\mathbf{i} +( 3+\cos v) \sin u\,\mathbf{j}+\sin v\mathbf{k} \end{equation*} \]

where \(0\leq u\leq 2\pi \) and \(0\leq v\leq 2\pi .\)

Solution We begin by finding some \(u\)-coordinate curves. Since the parametrization involves trigonometric functions, we choose \(0,\dfrac{\pi }{2},\pi ,\) and \(\dfrac{3\pi }{2}.\) Then the \(u\)-coordinate curves are \[ \begin{equation*} \begin{array}{rcl@{\qquad}rrcl} \mathbf{r}(u,0) &=&4\cos u\,\mathbf{i}+4\sin u\,\mathbf{j} & \mathbf{r}\left( u,\dfrac{\pi }{2}\right) &=&3\cos u\,\mathbf{i}+3\sin u\,\mathbf{j+k}\\[5pt] \mathbf{r}( u,\pi ) &=&2\cos u\,\mathbf{i}+2\sin u\,\mathbf{j} & \mathbf{r}\left( u,\dfrac{3\pi }{2}\right) &=&3\cos u\,\mathbf{i}+3\sin u\,\mathbf{j-k} \end{array} \end{equation*} \]

We identify the graphs of each of these curves by using parametric equations. For example, for \(\mathbf{r}(u,0) \), we have \(x=4\cos u,\) \(y=4\sin u,\) and \[ x^{2}+y^{2}=( 4\cos u) ^{2}+( 4\sin u)^{2}=4^{2}( \cos ^{2}u+\sin ^{2}u) =4^{2} \]

Figure 42 A torus centered at the origin.

a circle of radius \(4\) centered at the origin. Similarly, \(\mathbf{r=r} ( u,\pi ) \) is a circle of radius \(2\) centered at the origin. The graphs of \(\mathbf{r}\left( u,\dfrac{\pi }{2}\right) \) and \(\mathbf{r}\left( u,\dfrac{3\pi }{2}\right) \) are both circles of radius \(3,\) but the center of \(\mathbf{r=r}\left( u,\dfrac{\pi }{2}\right) \) is at \(( 0,0,1)\) and the center of \(\mathbf{r=r}\left( u,\dfrac{3\pi }{2}\right) \) is at \(( 0,0,-1) .\) Do you see why? See the blue circles in Figure 42.

Using \(0,\dfrac{\pi }{2},\pi ,\) and \(\dfrac{3\pi }{2}\) for \(u,\) we obtain the \(v\)-coordinate curves: \[ \begin{array}{rcl@{\qquad}llcl} \mathbf{r}( 0,v) &=&( 3+\cos v) \mathbf{i}+\sin v\,\mathbf{k} & \mathbf{r}\left( \dfrac{\pi }{2},v\right) =( 3+\cos v) \,\mathbf{j}+\sin v\,\mathbf{k} \\[5pt] \mathbf{r}( \pi ,v) &=&-( 3+\cos v) \,\mathbf{i}+\sin v\, \mathbf{k} & \mathbf{r}\left( \dfrac{3\pi }{2},v\right) =-( 3+\cos v) \,\mathbf{j}+\sin v\,\mathbf{k} \end{array} \]

Again, we identify the graphs of these curves as circles. Each graph is a circle of radius \(1,\) but the centers of the circles \(\mathbf{r=r}(0,v) ,\) \(\mathbf{r=r}( \pi ,v) ,\) \(\mathbf{r=r}\left( \dfrac{\pi }{2},v\right)\), and \(\mathbf{r=r}\left( \dfrac{3\pi }{2},v\right) \) are \(( 3,0,0) \), \(( -3,0,0) ,\) \((0,3,0) ,\) and \(( 0,-3,0) ,\) respectively. See the red circles in Figure 42.

The surface \(S\) is a torus. This particular torus results from revolving a circle of radius \(1\) about a circle of radius \(3\) centered at the origin.

NEED TO REVIEW?

A torus is discussed in Section 6.8, p. 464.

NOW WORK

Problem 5.

NEED TO REVIEW?

Finding parametric representations of functions of one variable is discussed in Section 9.1, pp. 641-642.

Representing a Surface Parametrically

Find a parametrization of the surface \(S\) given by \[ x+y^{2}+z=5\qquad 0\leq x\leq 1\qquad -1\leq y\leq 1\]

Solution We begin by expressing \(S\) as an explicit function of \(x\) and \(y.\) \[ z=f(x,y) =5-x-y^{2} \]

Since \(z\) is a function of \(x\) and \(y,\) we let \(x=u\) and \(y=v.\) Then the parametric equations of \(S\) are \[ x=u\qquad y=v\qquad z=5-u-v^{2} \]

Figure 43 \(\mathbf{r}(u,v) =u\mathbf{i}+v\mathbf{j}+( 5-u-v^{2}) \mathbf{k},\\ 0\leq u\leq 1, -1\leq v\leq 1\)

Using these parametric equations, we obtain \[ \mathbf{r}(u,v) =u\mathbf{i}+v\mathbf{j}+( 5-u-v^{2}) \mathbf{k}\qquad 0\leq u\leq 1\qquad -1\leq v\leq 1 \]

NOW WORK

Problem 9.

Parametrizing a Sphere

Find a parametrization for a sphere \(S\) of radius \(2\) centered at the origin.

Solution The equation of the sphere, \(x^{2}+y^{2}+z^{2}=4,\) in rectangular coordinates cannot be expressed explicitly as a function of two of its variables. However, in spherical coordinates, the sphere has the explicit equation \(\rho =2.\) So if we use the spherical coordinates \(\theta \) and \(\phi \) as parameters, then parametric equations of the sphere are given by

and a parametrization of \(S\) is \[ \begin{equation*} \mathbf{r}\left( \theta ,\phi \right) =2\cos \theta \sin \phi \,\mathbf{i} +2\sin \theta \sin \phi \,\mathbf{j}+2\cos \phi \,\mathbf{k} \end{equation*} \]

where \(0\leq \theta \leq 2\pi \) and \(0\leq \phi \leq \pi . \)

NEED TO REVIEW?

Spherical coordinates are discussed in Section 14.8, pp. 956-959.

NOW WORK

Problems 11 and 27.

Parametrizing a Cone Using Different Coordinate Systems

Find a parametrization of the surface \(S\) defined by the cone \[ z=\sqrt{3x^{2}+3y^{2}}\qquad x^{2}+y^{2}\leq 16 \]

1. Using rectangular coordinates.
2. Using cylindrical coordinates.
3. Using spherical coordinates.

Solution (a) The equation is given in rectangular coordinates and is written explicitly as a function \(z=f(x,y) .\) So, we define \(x=u,\) \(y=v,\) \(z=\sqrt{3u^{2}+3v^{2}}.\) Then a parametrization of the cone is \[ \mathbf{r}(u,v) =u\mathbf{i}+v\mathbf{j}+\sqrt{3u^{2}+3v^{2}}\, \mathbf{k}\qquad u^{2}+v^{2}\leq 16 \]

(b) In cylindrical coordinates, the cone is given by \[ z=\sqrt{3x^{2}+3y^{2}}=\sqrt{3}\sqrt{( r\cos \theta ) ^{2}+(r\sin \theta ) ^{2}}=\sqrt{3}\sqrt{r^{2}}=\sqrt{3}\,r \]

where \(z\) is expressed explicitly in terms of \(r\) and \(\theta .\) So if we use \(r\) and \(\theta \) as parameters, then a parametrization of the cone is \[ \begin{eqnarray*} \mathbf{r}( r,\theta ) =r\cos \theta \,\mathbf{i}+ \underset{\underset{\color{#0066A7}{x=r\cos \theta ,y= r\sin \theta , z=\sqrt{3}r}}{\color{#0066A7}{\uparrow}}}{r\sin \theta \, \mathbf{j}}+\sqrt{3}r\,\mathbf{k}\qquad 0\leq r\leq 4\quad 0\leq \theta \leq 2\pi \end{eqnarray*} \]

Figure 45

(c) In spherical coordinates, the equation of a half cone, where \(z\geq 0,\) is \(\phi =a,\) \(0< a <\dfrac{\pi }{2}.\) We use Figure 45 to find \(a.\) \[ \begin{eqnarray*} \tan \phi &=&\dfrac{r}{z}=\dfrac{r}{\sqrt{3}r}=\dfrac{1}{\sqrt{3}}\qquad \color{#0066A7}{{\hbox{\(z=\sqrt{3}r\)}}}\\[6pt] \phi &=&\tan ^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi }{6}=a \end{eqnarray*} \]

Using \(\rho \) and \(\theta \) as parameters, parametric equations for the cone in spherical coordinates are \[ \begin{eqnarray*} x\underset{\underset{\color{#0066A7}{\sin \tfrac{\pi }{6}=\tfrac{1}{2}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{ \rho }{2}\cos \theta \qquad y\underset{\underset{\color{#0066A7}{\sin \tfrac{\pi }{6}=\tfrac{1}{2}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{\rho }{2}\sin \theta \qquad z \underset{\underset{\color{#0066A7}{\cos \tfrac{\pi }{6}=\tfrac{\sqrt{3}}{2}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{\sqrt{3}}{2}\rho \end{eqnarray*} \]

Now we find the parameter domain. Since \[ z^{2}=3x^{2}+3y^{2} \ \hbox{and}\ \rho ^{2}=x^{2}+y^{2}+z^{2}=x^{2}+y^{2}+ (\underset{\color{#0066A7}{z^{2}}}{\underbrace{ 3x^{2}+3y^{2}}})=4x^{2}+4y^{2} \]

we have \[ 0\leq \rho ^{2}=4(x^{2}+y^{2}) \leq 4\,{\cdot}\, 16=64,\ \hbox{or equivalently,}\ 0\leq \rho \leq 8.\]

The parameter domain of the cone is \(0\leq \rho \,\leq 8\) and \(0\leq \theta \leq 2\pi .\) Then a parametrization of the cone is \[ \mathbf{r}( \rho ,\theta ) =\dfrac{\rho }{2}\cos \theta \,\mathbf{ i}+\dfrac{\rho }{2}\sin \theta \,\mathbf{j}+\dfrac{\sqrt{3}}{2}\rho \, \mathbf{k}\qquad 0\leq \rho \leq 8\quad 0\leq \theta \leq 2\pi \]

NOW WORK

Problem 33.

NEED TO REVIEW?

Tangent vectors to a curve are discussed in Section 11.2, pp. 767-769.

spanDEFINITIONspan Smooth Parametrization

A parametrization \(\mathbf{r}=\mathbf{r}(u,v) \) is called a smooth parametrization if

• \(\mathbf{r}\) is continuous and has continuous partial derivatives in the parameter domain \(R,\) and
• \(\mathbf{r}_{u}\times \mathbf{r}_{v}\neq \mathbf{0}\) in \(R.\)

A surface \(S\) that can be given a smooth parametrization is called a smooth surface.

Find Equations for a Tangent Plane and a Normal Line

1. Find an equation of the tangent plane to the surface \(S\) parametrized by \(\mathbf{r=r}(u,v) =3u\mathbf{i}+( 6-u^{2}-v^{2}) \mathbf{j}+2v \mathbf{k\ }\) at the point \(( -3,1,4) .\)
2. Find an equation of the normal line to the tangent plane at \((-3,1,4) .\)

Solution (1) We begin by finding the values of the parameters at the point \(( -3,1,4) .\) That is, we solve the system of equations \[ \left\{ \begin{array}{rcl@{\qquad}r} 3u &=&-3 & \color{#0066A7}{{\hbox{\((1)\)}}}\\ 6-u^{2}-v^{2}&=& 1 & \color{#0066A7}{{\hbox{\((2)\)}}}\\ 2v&=& 4 & \color{#0066A7}{{\hbox{\((3)\)}}} \end{array} \right. \]

From \(( 1) \) we find \(u=-1,\) and from (3) we find \(v=2.\) [Checking, we find these values also satisfy (2).]

We now find the tangent vectors \(\mathbf{r}_{u}\) and \(\mathbf{r}_{v}.\) \[ \begin{array}{@{\hspace*{-2.2pc}}rcl@{\qquad}lrll} \mathbf{r}_{u} &=&\dfrac{\partial }{\partial u}( 3u) \mathbf{i}+ \dfrac{\partial }{\partial u}( 6-u^{2}-v^{2}) \mathbf{j}+\dfrac{ \partial }{\partial u}\left( 2v\right) \mathbf{k}=3\mathbf{i}-2u\,\mathbf{j} & \mathbf{r}_{u}( -1,2) &=&3\mathbf{i}+2\,\mathbf{j}\\[4pt] \mathbf{r}_{v} &=&\dfrac{\partial }{\partial v}( 3u) \mathbf{i}+ \dfrac{\partial }{\partial v}( 6-u^{2}-v^{2}) \mathbf{j}+\dfrac{ \partial }{\partial v}( 2v) \mathbf{k}=-2v\mathbf{j}+2\,\mathbf{k} & \mathbf{r}_{v}( -1,2) &=&-4\mathbf{j}+2\, \mathbf{k} \end{array} \]

Figure 48 The tangent plane to \(S\) at \((-3,1,4)\).

The normal vector \(\mathbf{n}\) to the tangent plane at \(( -3,1,4) \) is \[ \mathbf{n} = \mathbf{r}_u ( -1,2) \times \mathbf{r}_v ( -1,2) =\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 3 & 2 & 0 \\[3pt] 0 & -4 & 2 \end{array} \right\vert =4\mathbf{i}-6\mathbf{j}-12\mathbf{k} \]

An equation of the tangent plane is \[ 4( x+3) -6( y-1) -12( z-4) =0\qquad\hbox{or equivalently}\qquad 4x-6y-12z=-66 \]

(b) The normal line to the tangent plane contains the point \(( -3,1,4) \) and is parallel to the vector \(\mathbf{n}=4\mathbf{i}-6\mathbf{j} -12\mathbf{k}\). A parametrization of the normal line is \[ \mathbf{r}(t) = ( -3+4t ) \mathbf{i}+ ( 1-6t ) \mathbf{j} + ( 4-12t ) \mathbf{k} \]

See Figure 48.

NOW WORK

Problem 13.

Finding a Tangent Plane and a Normal Line

A surface \(S\) is defined by \(x^{2}+z^{2}=25\) and \(0\leq y\leq 10.\)

1. Find a smooth parametrization for \(S.\)
2. Find an equation of the tangent plane to \(S\) at the point \(( -3,5,4) .\)
3. Find an equation of the normal line to the tangent plane at \((-3,5,4) .\)

Solution (a) Since \(S\) is a circular cylinder, we choose a parametrization that uses a variation of cylindrical coordinates. If \(x=r\cos \theta ,\) \(y=y,\) and \(z=r\sin \theta ,\) then \(x^{2}+z^{2}=r^{2}=25,\) or equivalently, \(r=5\). Then parametric equations for the cylinder are \[ x=5\cos \theta \qquad y=y\qquad z=5\sin \theta \]

and a parametrization of \(S\) is \[ \begin{equation*} \mathbf{r}( \theta ,y) =5\cos \theta\ \mathbf{i}+y\mathbf{j}+5\sin \theta\ \mathbf{k} \end{equation*} \]

where the parameter domain is \(0\leq \theta \leq 2\pi \) and \(0\leq y\leq 10.\)

Figure 49 The tangent plane \(\mathbf{r}( \theta ,y) =5\cos \theta\ \mathbf{i}+y\mathbf{j} +5\sin \theta\ \mathbf{k}\) at the point \(( -3,5,4)\).

To determine whether this is a smooth parametrization, we find the partial derivatives of \(\mathbf{r=\mathbf{r}}( \theta ,y)\). \[ \mathbf{r}_{\theta }( \theta ,y) =-5\sin \theta\ \mathbf{i}+5\cos \theta\ \mathbf{k}\qquad \mathbf{r}_{y}( \theta ,y) =\mathbf{j} \]

Each of these is continuous. Now \[ \mathbf{r}_{\theta }\times \mathbf{r}_{y}= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -5\sin \theta & 0 & 5\cos \theta \\ 0 & 1 & 0 \end{array}\right| =-5\cos \theta\ \mathbf{i}-5\sin \theta\ \mathbf{k} \]

Since \(\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{y}\right\Vert =\left\Vert -5\cos \theta\ \mathbf{i}-5\sin \theta\ \mathbf{k}\right\Vert =\sqrt{25\cos ^{2}\theta +25\sin ^{2}\theta }=5,\) then \(\mathbf{r}_{\theta }\times \mathbf{r}_{y}\neq \mathbf{0}\). So, \(\mathbf{r}=\mathbf{r}( \theta ,y) \mathbf{\ }\) is a smooth parametrization, and \(S\) is a smooth surface.

(b) At the point \(( -3,5,4) ,\) we have \(5\cos \theta =-3,\) \(y=5,\) and \(5\sin \theta =4.\) Then the normal vector \(\mathbf{n}\) to the plane at \(( -3,5,4) \) is \(\mathbf{r}_{\theta }\times \mathbf{r} _{y}=-5\left( -\dfrac{3}{5}\right) \mathbf{i}-5\left( \dfrac{4}{5}\right) \mathbf{k}=3\mathbf{i}-4\mathbf{k}.\) An equation of the tangent plane is \[ 3( x+3) -4( z-4) =0\qquad \hbox{or equivalently}\qquad 3x-4z=-25 \]

(c) The normal line to the tangent plane contains the point \(( -3,5,4) \) and is parallel to \(3\mathbf{i}-4\mathbf{k}\). A parametrization of the normal line is \(\ \mathbf{r}(t) =(-3+3t) \mathbf{i}+5\mathbf{j}+( 4-4t) \mathbf{k}\). See Figure 49.

NOTE

Whenever a surface \(S\) has a smooth parametrization, we can find its surface area using a double integral.

Finding the Surface Area of a Torus

Find the surface area of the torus parametrized by \[ \begin{equation*} \mathbf{r}(u,v) =( 3+\cos v) \cos u\,\mathbf{i} +( 3+\cos v) \sin u\,\mathbf{j}+\sin v\mathbf{k} \end{equation*} \]

where \(0\leq u\leq 2\pi \) and \(0\leq v\leq 2\pi .\)

Solution This is the torus graphed in Figure 42. To use formula (1) to find the area of a parametrized surface, we need to find the partial derivatives of \(\mathbf{r=r}(u,v) .\) \[ \begin{eqnarray*} \mathbf{r}_{u}(u,v) &=&( 3+\cos v) \left( -\sin u\right) \,\mathbf{i}+( 3+\cos v) \cos u\,\mathbf{j} \\[4pt] \mathbf{r}_{v}(u,v) &=& (-\sin v) \cos u \mathbf{i} -\sin v\sin u\,\mathbf{j}+\cos v\mathbf{k} \end{eqnarray*} \]

Next we find the cross product: \[ \begin{eqnarray*} \mathbf{r}_{u}\times \mathbf{r}_{v} &=& \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -( 3+\cos v) \sin u & ( 3+\cos v) \cos u & 0 \\[3pt] -\sin v\cos u & -\sin v\sin u & \cos v \end{array}\right|\\[5pt] &=&( 3+\cos v) \left( \cos u\cos v\mathbf{i}+\sin u\cos v\mathbf{j }+\sin v\mathbf{k}\right) \end{eqnarray*} \]

Then \[ \begin{eqnarray*} \hspace{-2pc}\left\Vert \mathbf{r}_{u}\times \mathbf{r}_{v}\right\Vert &=&\sqrt{( 3+\cos v) ^{2}( \cos ^{2}u\cos ^{2}v+\sin ^{2}u\cos ^{2}v+\sin ^{2}v) } \\ &=&( 3+\cos v) \sqrt{\cos ^{2}v( \cos ^{2}u+\sin ^{2}u) +\sin ^{2}v}=( 3+\cos v) \sqrt{\cos ^{2}v+\sin ^{2}v}\\ &=&3+\cos v \end{eqnarray*} \]

Now using formula (1), the surface area of \(S\) is \[ \begin{eqnarray*} \iint\limits_{\kern-3ptR}( 3+\cos v)\, du\,dv&=&\int_{0}^{2\pi }\int_{0}^{2\pi }( 3+\cos v)\, dv\,du=\int_{0}^{2\pi }\big[ 3v+\sin v \big] _{0}^{2\pi }du\\[4pt] &=&\int_{0}^{2\pi }6\pi du=12\pi ^{2} \end{eqnarray*} \]

NOW WORK

Problem 17.